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[{"Name":"Solution Concentration","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Solution Concentration","Duration":"5m 17s","ChapterTopicVideoID":25623,"CourseChapterTopicPlaylistID":245373,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25623.jpeg","UploadDate":"2021-06-10T06:58:15.4730000","DurationForVideoObject":"PT5M17S","Description":null,"MetaTitle":"Solution Concentration: Video + Workbook | Proprep","MetaDescription":"Physical Properties of Solutions - Solution Concentration. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/general-chemistry/physical-properties-of-solutions/solution-concentration/vid30925","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.025","Text":"In this video, we\u0027ll discuss various ways of expressing concentrations of solutions."},{"Start":"00:06.025 ","End":"00:10.450","Text":"We\u0027ll start by going over some terminology that we\u0027ve met before."},{"Start":"00:10.450 ","End":"00:16.260","Text":"A solution is a homogeneous mixture of substances,"},{"Start":"00:16.260 ","End":"00:18.075","Text":"and a solution contains a solvent,"},{"Start":"00:18.075 ","End":"00:20.430","Text":"and 1 or more solutes."},{"Start":"00:20.430 ","End":"00:23.919","Text":"The solvent is the major component of the solution."},{"Start":"00:23.919 ","End":"00:25.825","Text":"For example, in a salt solution,"},{"Start":"00:25.825 ","End":"00:28.225","Text":"the solvent would be water."},{"Start":"00:28.225 ","End":"00:31.060","Text":"The solutes are dissolved in the solvent."},{"Start":"00:31.060 ","End":"00:34.615","Text":"For example, if we again had a salt solution,"},{"Start":"00:34.615 ","End":"00:38.550","Text":"then salt would be the solute."},{"Start":"00:38.550 ","End":"00:43.140","Text":"Solutions can either be dilute or concentrated."},{"Start":"00:43.140 ","End":"00:45.565","Text":"Here\u0027s some examples."},{"Start":"00:45.565 ","End":"00:48.550","Text":"We can have gas solutions such as air,"},{"Start":"00:48.550 ","End":"00:53.290","Text":"which is oxygen and nitrogen with a majority of nitrogen,"},{"Start":"00:53.290 ","End":"00:56.575","Text":"and a few other gases, or natural gas,"},{"Start":"00:56.575 ","End":"01:00.515","Text":"which is mainly methane and other gases."},{"Start":"01:00.515 ","End":"01:04.270","Text":"Or we could have liquid solutions such as seawater,"},{"Start":"01:04.270 ","End":"01:08.740","Text":"which is water with a number of salts dissolved in it,"},{"Start":"01:08.740 ","End":"01:12.875","Text":"or vinegar, which is acetic acid in water."},{"Start":"01:12.875 ","End":"01:17.140","Text":"Or we can have solid solutions like yellow brass,"},{"Start":"01:17.140 ","End":"01:19.135","Text":"which is copper and zinc,"},{"Start":"01:19.135 ","End":"01:24.130","Text":"or palladium-hydrogen solution, which is hydrogen dissolved in,"},{"Start":"01:24.130 ","End":"01:27.420","Text":"or absorbed into palladium."},{"Start":"01:27.420 ","End":"01:31.150","Text":"One way of expressing concentration is mass percent,"},{"Start":"01:31.150 ","End":"01:34.385","Text":"or volume percent, or mass volume percent."},{"Start":"01:34.385 ","End":"01:36.520","Text":"Here are some examples."},{"Start":"01:36.520 ","End":"01:42.155","Text":"Mass percent, an example of this is 5 percent sodium chloride solution by mass,"},{"Start":"01:42.155 ","End":"01:47.420","Text":"that would be 5 grams of sodium chloride and 100 grams of aqueous solution."},{"Start":"01:47.420 ","End":"01:49.805","Text":"Or we could have volume percent."},{"Start":"01:49.805 ","End":"01:53.550","Text":"For example, 5 percent vinegar by volume is"},{"Start":"01:53.550 ","End":"01:58.580","Text":"5 milliliters of acetic acid in 100 milliliters of aqueous solution."},{"Start":"01:58.580 ","End":"02:00.860","Text":"Or we could have mass volume percent."},{"Start":"02:00.860 ","End":"02:03.290","Text":"This is often used in medicine."},{"Start":"02:03.290 ","End":"02:07.820","Text":"For example, 5 percent sodium chloride solution by mass volume is"},{"Start":"02:07.820 ","End":"02:13.190","Text":"5 grams of sodium chloride in 100 milliliters of aqueous solution."},{"Start":"02:13.190 ","End":"02:17.720","Text":"Another way of expressing concentration is parts per million,"},{"Start":"02:17.720 ","End":"02:18.860","Text":"or parts per billion,"},{"Start":"02:18.860 ","End":"02:21.079","Text":"or parts per trillion."},{"Start":"02:21.079 ","End":"02:23.560","Text":"This is used for low mass,"},{"Start":"02:23.560 ","End":"02:26.245","Text":"or volume percent concentration."},{"Start":"02:26.245 ","End":"02:28.760","Text":"Now very low concentrations,"},{"Start":"02:28.760 ","End":"02:33.680","Text":"the density of the solution is the same as that of water so we can replace kilograms"},{"Start":"02:33.680 ","End":"02:39.620","Text":"by liter because the density of water is 1 kilogram per liter."},{"Start":"02:41.570 ","End":"02:51.120","Text":"1 ppm means 1 milligram per liter and 1 ppb is 1 microgram per liter."},{"Start":"02:51.120 ","End":"02:58.425","Text":"1 ppt is 1 nanogram per liter of solution of course."},{"Start":"02:58.425 ","End":"03:01.660","Text":"We met mole fraction and mole percent when we"},{"Start":"03:01.660 ","End":"03:04.615","Text":"talked about gases so we\u0027re just going over it here."},{"Start":"03:04.615 ","End":"03:06.040","Text":"The mole fraction,"},{"Start":"03:06.040 ","End":"03:09.100","Text":"chi of a substance i is the number of"},{"Start":"03:09.100 ","End":"03:12.790","Text":"moles of that substance divided by the total number of moles."},{"Start":"03:12.790 ","End":"03:16.510","Text":"If we add up all the mole fractions, we get 1."},{"Start":"03:16.510 ","End":"03:19.270","Text":"The sum of all the mole fractions is"},{"Start":"03:19.270 ","End":"03:24.125","Text":"1 and the mole percent is the mole fraction times 100 percent."},{"Start":"03:24.125 ","End":"03:27.425","Text":"Now we\u0027ve also met molarity before."},{"Start":"03:27.425 ","End":"03:33.260","Text":"That\u0027s the number of moles of solute divided by the volume of solution in liters."},{"Start":"03:33.260 ","End":"03:36.080","Text":"Important to note that the volume of solution,"},{"Start":"03:36.080 ","End":"03:38.795","Text":"not the volume of the solvent."},{"Start":"03:38.795 ","End":"03:45.050","Text":"If we want to make up 250 milliliters of a particular solution,"},{"Start":"03:45.050 ","End":"03:50.270","Text":"then we put the calculated mass of the solute in the bottom"},{"Start":"03:50.270 ","End":"03:55.445","Text":"of the flask and fill up until we get to 250 milliliters."},{"Start":"03:55.445 ","End":"03:58.640","Text":"There are flasks in all different volumes."},{"Start":"03:58.640 ","End":"04:03.655","Text":"Now another concept is not molarity but molality."},{"Start":"04:03.655 ","End":"04:05.355","Text":"Why do we need this?"},{"Start":"04:05.355 ","End":"04:10.670","Text":"Now the volume changes with temperature so that molarity is temperature-dependent."},{"Start":"04:10.670 ","End":"04:12.710","Text":"If we heat a substance,"},{"Start":"04:12.710 ","End":"04:17.945","Text":"it generally takes up a larger volume."},{"Start":"04:17.945 ","End":"04:25.405","Text":"Molality is more accurate because it deals with mass rather than volume."},{"Start":"04:25.405 ","End":"04:27.335","Text":"Here\u0027s the definition."},{"Start":"04:27.335 ","End":"04:30.485","Text":"The molality is the number of moles of solute"},{"Start":"04:30.485 ","End":"04:34.715","Text":"divided by the mass of the solvent in kilograms."},{"Start":"04:34.715 ","End":"04:36.200","Text":"Here\u0027s an example."},{"Start":"04:36.200 ","End":"04:41.600","Text":"If we dissolve 0.1 moles of ethanol in 200 grams of water,"},{"Start":"04:41.600 ","End":"04:44.044","Text":"what is the molality of the solution?"},{"Start":"04:44.044 ","End":"04:48.150","Text":"Now the mass of water is 0.2 kilograms."},{"Start":"04:48.150 ","End":"04:51.960","Text":"200 grams of water is 0.2 kilograms."},{"Start":"04:51.960 ","End":"05:01.010","Text":"The molality is equal to 0.1 moles of ethanol divided by 0.2 kilograms of water."},{"Start":"05:01.010 ","End":"05:05.825","Text":"We divide that, we get 0.5 moles per kilogram."},{"Start":"05:05.825 ","End":"05:10.175","Text":"The units of molality are moles per kilogram."},{"Start":"05:10.175 ","End":"05:17.130","Text":"In this video, we learned about various ways to express the concentration of solutions."}],"ID":30925},{"Watched":false,"Name":"Exercise 1","Duration":"5m 31s","ChapterTopicVideoID":28903,"CourseChapterTopicPlaylistID":245373,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.390","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.390 ","End":"00:05.085","Text":"In order to make juice,"},{"Start":"00:05.085 ","End":"00:11.025","Text":"1 part by volume of concentrated juice solution needs to be mixed with 7 parts water."},{"Start":"00:11.025 ","End":"00:14.580","Text":"How many milliliters of concentrated juice solution and"},{"Start":"00:14.580 ","End":"00:19.785","Text":"water need to be mixed to make 56 milliliters of juice?"},{"Start":"00:19.785 ","End":"00:22.365","Text":"Let\u0027s take a look at what we know."},{"Start":"00:22.365 ","End":"00:24.570","Text":"We know that we want to make juice."},{"Start":"00:24.570 ","End":"00:27.360","Text":"Juice is going to be our general solution."},{"Start":"00:27.360 ","End":"00:32.895","Text":"We know that we make it out of concentrated juice solution and water."},{"Start":"00:32.895 ","End":"00:39.225","Text":"We\u0027re going to write concentrated juice solution and water."},{"Start":"00:39.225 ","End":"00:42.300","Text":"Again, our concentrated solution is our first liquid."},{"Start":"00:42.300 ","End":"00:43.680","Text":"Water is the second liquid."},{"Start":"00:43.680 ","End":"00:48.975","Text":"We mix these 2 to receive our general solution, which is juice."},{"Start":"00:48.975 ","End":"00:55.875","Text":"Now, we also know that our concentrated juice is 1 part by volume of the total solution."},{"Start":"00:55.875 ","End":"00:57.660","Text":"If we look at parts,"},{"Start":"00:57.660 ","End":"01:00.375","Text":"it\u0027s 1 part volume,"},{"Start":"01:00.375 ","End":"01:04.210","Text":"and the water is 7 parts volume."},{"Start":"01:04.640 ","End":"01:08.160","Text":"Now, again, since we\u0027re mixing the concentrated juice and"},{"Start":"01:08.160 ","End":"01:12.025","Text":"the water and receiving our general solution which is juice,"},{"Start":"01:12.025 ","End":"01:16.370","Text":"we can also calculate the number of parts of volume of our juice."},{"Start":"01:16.370 ","End":"01:21.020","Text":"All we need to do is sum up 1 part plus 7 parts,"},{"Start":"01:21.020 ","End":"01:23.240","Text":"we get 8 parts."},{"Start":"01:23.240 ","End":"01:29.980","Text":"We know that we have 8 parts of volume in our juice."},{"Start":"01:29.980 ","End":"01:37.995","Text":"Now, we also know from our question that we have 56 milliliters of our total solution."},{"Start":"01:37.995 ","End":"01:41.550","Text":"That\u0027s 56 milliliters."},{"Start":"01:41.550 ","End":"01:47.340","Text":"We\u0027re asked to find how many milliliters of concentrated juice and how many milliliters"},{"Start":"01:47.340 ","End":"01:53.370","Text":"of water needs to be mixed to receive 56 milliliters of our total solution."},{"Start":"01:53.370 ","End":"01:56.250","Text":"Before we go on, let\u0027s take a look again at what we know."},{"Start":"01:56.250 ","End":"01:57.450","Text":"We have Liquid 1,"},{"Start":"01:57.450 ","End":"01:58.740","Text":"which is our concentrated juice,"},{"Start":"01:58.740 ","End":"02:00.240","Text":"Liquid 2, which is our water."},{"Start":"02:00.240 ","End":"02:03.465","Text":"We know that our concentrated juices 1 part by volume,"},{"Start":"02:03.465 ","End":"02:07.230","Text":"and we know that our water is 7 parts by volume."},{"Start":"02:07.230 ","End":"02:10.050","Text":"We have our total solution, which is our juice."},{"Start":"02:10.050 ","End":"02:15.540","Text":"First of all, we summed up 1 plus 7 parts and we got that juice is 8 parts by volume."},{"Start":"02:15.540 ","End":"02:18.585","Text":"We also know that it\u0027s 56 milliliters."},{"Start":"02:18.585 ","End":"02:22.140","Text":"To find the number of milliliters of concentrated juice and water that we need to"},{"Start":"02:22.140 ","End":"02:26.685","Text":"mix to get 56 milliliters of our total solution juice,"},{"Start":"02:26.685 ","End":"02:28.700","Text":"first, what we need to do is find out"},{"Start":"02:28.700 ","End":"02:33.110","Text":"how many milliliters we have in each part of volume."},{"Start":"02:33.110 ","End":"02:37.115","Text":"We take our total volume for our total solution,"},{"Start":"02:37.115 ","End":"02:41.360","Text":"which is 56 milliliters and divide it by the total number of parts."},{"Start":"02:41.360 ","End":"02:43.940","Text":"That\u0027s total number of parts."},{"Start":"02:43.940 ","End":"02:46.730","Text":"The total number of parts we have is 8 parts."},{"Start":"02:46.730 ","End":"02:49.225","Text":"This equals 7."},{"Start":"02:49.225 ","End":"02:52.250","Text":"It\u0027s 7 milliliters per parts,"},{"Start":"02:52.250 ","End":"02:55.760","Text":"meaning that in each 1 part of volume,"},{"Start":"02:55.760 ","End":"02:58.475","Text":"we have 7 milliliters."},{"Start":"02:58.475 ","End":"03:02.255","Text":"Now that we know how many milliliters we have in each part of volume,"},{"Start":"03:02.255 ","End":"03:04.700","Text":"all we need to do is take the number of milliliters for"},{"Start":"03:04.700 ","End":"03:07.355","Text":"each part and multiply it by the number of parts we"},{"Start":"03:07.355 ","End":"03:13.265","Text":"have in order to find the number of milliliters that we have for each liquid."},{"Start":"03:13.265 ","End":"03:14.630","Text":"Let\u0027s take the first liquid."},{"Start":"03:14.630 ","End":"03:16.040","Text":"We have concentrated juice."},{"Start":"03:16.040 ","End":"03:19.600","Text":"Now, we know we have 1 part by volume of concentrated juice."},{"Start":"03:19.600 ","End":"03:22.325","Text":"To find the number of milliliters of concentrated juice,"},{"Start":"03:22.325 ","End":"03:24.785","Text":"all we need to do is take the number of milliliters per part,"},{"Start":"03:24.785 ","End":"03:31.840","Text":"which is 7 milliliters per part of volume,"},{"Start":"03:31.840 ","End":"03:34.940","Text":"and multiply it by the number of parts of volumes."},{"Start":"03:34.940 ","End":"03:37.210","Text":"We\u0027re going to multiply it by 1 part."},{"Start":"03:37.210 ","End":"03:39.270","Text":"Now the parts cancel out,"},{"Start":"03:39.270 ","End":"03:42.765","Text":"and 7 times 1 is going to give us 7 milliliters."},{"Start":"03:42.765 ","End":"03:47.330","Text":"We know that the number of milliliters of concentrated juice equals 7 milliliters."},{"Start":"03:47.330 ","End":"03:50.465","Text":"Now let\u0027s calculate the number of milliliters of water."},{"Start":"03:50.465 ","End":"03:52.610","Text":"Again, we take the number of milliliters per part,"},{"Start":"03:52.610 ","End":"03:56.250","Text":"which is 7 millimeters for each part of volume,"},{"Start":"03:56.250 ","End":"03:58.490","Text":"and in the water, we know we have 7 parts,"},{"Start":"03:58.490 ","End":"04:01.070","Text":"so we\u0027re going to multiply this by 7 parts."},{"Start":"04:01.070 ","End":"04:03.535","Text":"Again, our parts is going to cancel out,"},{"Start":"04:03.535 ","End":"04:07.250","Text":"and 7 times 7 is going to give us 49 milliliters of"},{"Start":"04:07.250 ","End":"04:11.105","Text":"water and 7 milliliters of concentrated juice."},{"Start":"04:11.105 ","End":"04:17.685","Text":"Again, we need to mix 7 milliliters of concentrated juice in 49 milliliters of water,"},{"Start":"04:17.685 ","End":"04:24.755","Text":"in order to make 56 milliliters of our total solution, which is juice."},{"Start":"04:24.755 ","End":"04:27.950","Text":"Now, before we conclude, I just want to mention that,"},{"Start":"04:27.950 ","End":"04:33.230","Text":"our total solution could be made up of a number of liquids."},{"Start":"04:33.230 ","End":"04:35.330","Text":"In our case, it\u0027s made up of 2 liquids,"},{"Start":"04:35.330 ","End":"04:36.875","Text":"concentrated juice and water."},{"Start":"04:36.875 ","End":"04:41.965","Text":"However, it could be made out of 3 liquids or 4 liquids and so on."},{"Start":"04:41.965 ","End":"04:44.870","Text":"In that case, we do exactly the same thing."},{"Start":"04:44.870 ","End":"04:47.450","Text":"We take our total number of parts of"},{"Start":"04:47.450 ","End":"04:51.305","Text":"solution by summing up all of the parts of our liquids,"},{"Start":"04:51.305 ","End":"04:54.970","Text":"we take our total number of milliliters of solution,"},{"Start":"04:54.970 ","End":"04:59.180","Text":"we calculate the number of milliliters per part, just as we did here,"},{"Start":"04:59.180 ","End":"05:01.670","Text":"we take the total number of milliliters"},{"Start":"05:01.670 ","End":"05:04.415","Text":"of the total solution divided by the total number of parts,"},{"Start":"05:04.415 ","End":"05:07.415","Text":"and we get the number of milliliters per parts."},{"Start":"05:07.415 ","End":"05:10.070","Text":"After we know the number of milliliters per part,"},{"Start":"05:10.070 ","End":"05:14.195","Text":"we just multiply this by the number of parts in each liquid."},{"Start":"05:14.195 ","End":"05:17.635","Text":"That way we find the number of milliliters of each liquid."},{"Start":"05:17.635 ","End":"05:21.560","Text":"Again, 7 milliliters of concentrated juice solution needs to be mixed with"},{"Start":"05:21.560 ","End":"05:26.120","Text":"49 milliliters of water to achieve 56 milliliters of juice,"},{"Start":"05:26.120 ","End":"05:28.040","Text":"which is our total solution."},{"Start":"05:28.040 ","End":"05:29.330","Text":"That is our final answer."},{"Start":"05:29.330 ","End":"05:32.010","Text":"Thank you very much for watching."}],"ID":30926},{"Watched":false,"Name":"Exercise 2","Duration":"1m 39s","ChapterTopicVideoID":28904,"CourseChapterTopicPlaylistID":245373,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.480 ","End":"00:10.005","Text":"A solution contains 2.67 grams of sodium chloride and 100 grams of water."},{"Start":"00:10.005 ","End":"00:14.220","Text":"Calculate the percent by mass of sodium chloride in the solution."},{"Start":"00:14.220 ","End":"00:20.085","Text":"We need to calculate the percent by mass of sodium chloride in our solution."},{"Start":"00:20.085 ","End":"00:25.050","Text":"The percent by mass equals the mass of the solute,"},{"Start":"00:25.050 ","End":"00:27.750","Text":"and the solute is what we\u0027re dissolving,"},{"Start":"00:27.750 ","End":"00:30.975","Text":"divide by the mass of the whole solution."},{"Start":"00:30.975 ","End":"00:36.625","Text":"We multiply this by 100 percent in order to reach the percent by mass."},{"Start":"00:36.625 ","End":"00:38.022","Text":"In our case, the solute,"},{"Start":"00:38.022 ","End":"00:40.700","Text":"it\u0027s what we\u0027re dissolving is the sodium chloride."},{"Start":"00:40.700 ","End":"00:43.760","Text":"That\u0027s the mass of sodium chloride,"},{"Start":"00:43.760 ","End":"00:46.940","Text":"divided by again the mass of the solution."},{"Start":"00:46.940 ","End":"00:50.645","Text":"Now our solution is water plus sodium chloride."},{"Start":"00:50.645 ","End":"00:53.590","Text":"That\u0027s the mass of our sodium chloride,"},{"Start":"00:53.590 ","End":"00:56.820","Text":"plus the mass of water."},{"Start":"00:56.820 ","End":"01:00.325","Text":"This is all multiplied by 100 percent."},{"Start":"01:00.325 ","End":"01:04.940","Text":"The mass of the sodium chloride equals 2.67 grams,"},{"Start":"01:04.940 ","End":"01:08.890","Text":"divided by the mass of the sodium chloride plus the mass of water."},{"Start":"01:08.890 ","End":"01:12.590","Text":"The mass of sodium chloride again is 2.67 grams,"},{"Start":"01:12.590 ","End":"01:17.000","Text":"plus the mass of water which equals 100 grams."},{"Start":"01:17.000 ","End":"01:21.450","Text":"This is all multiplied by 100 percent."},{"Start":"01:21.530 ","End":"01:26.935","Text":"After dividing and multiplying we get 2.6 percent."},{"Start":"01:26.935 ","End":"01:34.815","Text":"The percent by mass of sodium chloride in the solution equals 2.6 percent."},{"Start":"01:34.815 ","End":"01:36.150","Text":"That is our final answer,"},{"Start":"01:36.150 ","End":"01:38.710","Text":"thank you very much for watching."}],"ID":30927},{"Watched":false,"Name":"Exercise 3","Duration":"3m 4s","ChapterTopicVideoID":28905,"CourseChapterTopicPlaylistID":245373,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.510","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.510 ","End":"00:06.540","Text":"An aqueous glucose solution is 27 percent glucose by mass."},{"Start":"00:06.540 ","End":"00:11.220","Text":"Calculate the amount of glucose and water in 100 grams of solution."},{"Start":"00:11.220 ","End":"00:16.050","Text":"We have to calculate the amount of glucose and water in a 100 grams of solution and"},{"Start":"00:16.050 ","End":"00:20.955","Text":"we know that the solution is 27 percent glucose by mass."},{"Start":"00:20.955 ","End":"00:24.930","Text":"Now, in general, if we\u0027re given a certain percentage by mass,"},{"Start":"00:24.930 ","End":"00:27.960","Text":"which is in this case 27 percent,"},{"Start":"00:27.960 ","End":"00:31.905","Text":"we can assume that we have 100 grams of solution,"},{"Start":"00:31.905 ","End":"00:34.280","Text":"27 grams it\u0027s going to be the solute,"},{"Start":"00:34.280 ","End":"00:35.570","Text":"which is glucose,"},{"Start":"00:35.570 ","End":"00:37.670","Text":"and let\u0027s explain this for a minute."},{"Start":"00:37.670 ","End":"00:43.820","Text":"We have percent by mass and the percent by mass equals the mass of the solute,"},{"Start":"00:43.820 ","End":"00:45.815","Text":"which is what we\u0027re dissolving,"},{"Start":"00:45.815 ","End":"00:48.695","Text":"divided by the mass of the solution,"},{"Start":"00:48.695 ","End":"00:52.430","Text":"and this is multiplied by 100 percent to get our percent."},{"Start":"00:52.430 ","End":"00:56.885","Text":"In our case, we know that this equals 27 percent."},{"Start":"00:56.885 ","End":"01:00.185","Text":"Now let\u0027s divide both sides by 100 percent."},{"Start":"01:00.185 ","End":"01:09.905","Text":"We get that the mass of the solute divided by the mass of the solution equals 0.27."},{"Start":"01:09.905 ","End":"01:16.870","Text":"Now, 0.27 also equals 27 divided by 100 and in our case,"},{"Start":"01:16.870 ","End":"01:19.300","Text":"we\u0027re talking about the percent by mass."},{"Start":"01:19.300 ","End":"01:22.870","Text":"Therefore, it\u0027s going to be 27 grams in 100 grams."},{"Start":"01:22.870 ","End":"01:28.690","Text":"Now, the 100 grams is the solution and 27 grams is the solute."},{"Start":"01:28.690 ","End":"01:32.349","Text":"This is also going to equal the mass of the glucose"},{"Start":"01:32.349 ","End":"01:37.274","Text":"divided by the mass of the whole solution,"},{"Start":"01:37.274 ","End":"01:40.240","Text":"assuming we have 100 grams of solution."},{"Start":"01:40.240 ","End":"01:42.520","Text":"In 100 grams of solution, first of all,"},{"Start":"01:42.520 ","End":"01:45.520","Text":"we know that we have 27 grams of glucose."},{"Start":"01:45.520 ","End":"01:47.450","Text":"In 100 grams of solution,"},{"Start":"01:47.450 ","End":"01:50.400","Text":"we know that we have 27 grams of glucose."},{"Start":"01:50.400 ","End":"01:55.235","Text":"The mass of glucose equals 27 grams."},{"Start":"01:55.235 ","End":"01:58.295","Text":"Now we have to calculate the mass of the water."},{"Start":"01:58.295 ","End":"02:02.240","Text":"Now, as we said, the mass of a whole solution is 100 grams."},{"Start":"02:02.240 ","End":"02:04.850","Text":"But remember the mass of the solution"},{"Start":"02:04.850 ","End":"02:11.270","Text":"equals the mass of the glucose plus the mass of the water."},{"Start":"02:11.270 ","End":"02:16.180","Text":"The mass of the glucose plus the mass of the water equals 100 grams."},{"Start":"02:16.180 ","End":"02:18.160","Text":"To find the mass of the water,"},{"Start":"02:18.160 ","End":"02:21.235","Text":"would just going to subtract the mass of the glucose from both sides."},{"Start":"02:21.235 ","End":"02:25.690","Text":"So 100 grams minus the mass of glucose,"},{"Start":"02:25.690 ","End":"02:33.523","Text":"which equals 100 grams minus 27 grams,"},{"Start":"02:33.523 ","End":"02:37.580","Text":"and this equals 73 grams."},{"Start":"02:37.580 ","End":"02:39.470","Text":"In 100 grams of solution,"},{"Start":"02:39.470 ","End":"02:41.300","Text":"we found that the mass of the glucose equals"},{"Start":"02:41.300 ","End":"02:46.683","Text":"27 grams and the mass of the water equals 73 grams,"},{"Start":"02:46.683 ","End":"02:48.500","Text":"and again, remember that the mass of"},{"Start":"02:48.500 ","End":"02:54.290","Text":"the solution is the mass of the solvent plus the solute."},{"Start":"02:54.290 ","End":"03:00.185","Text":"Remember that the mass of the solution equals the mass of the solvent plus the solute."},{"Start":"03:00.185 ","End":"03:01.700","Text":"That is our final answer."},{"Start":"03:01.700 ","End":"03:04.110","Text":"Thank you very much for watching."}],"ID":30928},{"Watched":false,"Name":"Exercise 4","Duration":"4m 17s","ChapterTopicVideoID":28906,"CourseChapterTopicPlaylistID":245373,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.610","Text":"How are we going to solve the following exercise?"},{"Start":"00:02.610 ","End":"00:06.540","Text":"You are asked to prepare 270 grams of an aqueous solution of"},{"Start":"00:06.540 ","End":"00:11.295","Text":"glucose with a 16 percent glucose by mass concentration."},{"Start":"00:11.295 ","End":"00:14.415","Text":"How much solute and solvent do you need?"},{"Start":"00:14.415 ","End":"00:18.180","Text":"We need to calculate how much solute and how much solvent we need to prepare"},{"Start":"00:18.180 ","End":"00:22.500","Text":"270 grams of solution."},{"Start":"00:22.500 ","End":"00:27.660","Text":"First of all, we know that our solution is going to be 16 percent glucose by mass."},{"Start":"00:27.660 ","End":"00:35.099","Text":"16 percent glucose by mass means that if we have 100 grams of a solution,"},{"Start":"00:35.099 ","End":"00:38.355","Text":"we\u0027re going to have 16 grams of glucose."},{"Start":"00:38.355 ","End":"00:41.840","Text":"Because remember that percent by mass equals the mass"},{"Start":"00:41.840 ","End":"00:45.200","Text":"of the solute divided by the mass of the solution,"},{"Start":"00:45.200 ","End":"00:49.255","Text":"and this is multiplied by 100 percent."},{"Start":"00:49.255 ","End":"00:52.675","Text":"Now in our case, this equals 16 percent."},{"Start":"00:52.675 ","End":"00:55.040","Text":"Now if we divide both sides by 100 percent,"},{"Start":"00:55.040 ","End":"00:59.090","Text":"we get that the mass of the solute divided by the mass of"},{"Start":"00:59.090 ","End":"01:08.670","Text":"the solution equals 0.16 and 0.16 equals 16 divided by a 100."},{"Start":"01:08.670 ","End":"01:13.385","Text":"If we have 100 grams of solution,"},{"Start":"01:13.385 ","End":"01:15.845","Text":"we\u0027re going to have 16 grams of the solute,"},{"Start":"01:15.845 ","End":"01:19.085","Text":"which in our case is glucose."},{"Start":"01:19.085 ","End":"01:25.888","Text":"Now, we also know that we have 270 grams of our solution,"},{"Start":"01:25.888 ","End":"01:31.375","Text":"and we have to calculate how much of this 270 grams is glucose and how much is water."},{"Start":"01:31.375 ","End":"01:34.090","Text":"Because remember, a solute is what we\u0027re dissolving,"},{"Start":"01:34.090 ","End":"01:38.180","Text":"meaning our glucose and the solvent is the water."},{"Start":"01:38.180 ","End":"01:41.920","Text":"If we look at our solute or our glucose,"},{"Start":"01:41.920 ","End":"01:44.860","Text":"we know that if we have 100 grams of solution,"},{"Start":"01:44.860 ","End":"01:48.200","Text":"we have 16 grams of glucose."},{"Start":"01:48.200 ","End":"01:51.295","Text":"If we have 100 grams of solution,"},{"Start":"01:51.295 ","End":"01:53.890","Text":"we know that we have 16 grams of glucose,"},{"Start":"01:53.890 ","End":"02:02.825","Text":"and we\u0027re asked to find the amount of glucose when we have 270 grams of solution."},{"Start":"02:02.825 ","End":"02:06.390","Text":"In order to find the mass of glucose,"},{"Start":"02:06.390 ","End":"02:09.130","Text":"we\u0027re going to take the mass of the solution which we"},{"Start":"02:09.130 ","End":"02:13.495","Text":"know and multiply it by a conversion factor."},{"Start":"02:13.495 ","End":"02:19.540","Text":"We know that we have 16 grams of glucose for every 100 grams of"},{"Start":"02:19.540 ","End":"02:26.170","Text":"solution and the amount of solution that we want is 270 grams."},{"Start":"02:26.170 ","End":"02:33.445","Text":"This equals 270 grams of solution times"},{"Start":"02:33.445 ","End":"02:41.820","Text":"16 grams of glucose for every 100 grams of solution."},{"Start":"02:41.820 ","End":"02:47.024","Text":"The grams of solution cancel out and this equals"},{"Start":"02:47.024 ","End":"02:53.429","Text":"43.2 grams of glucose."},{"Start":"02:53.429 ","End":"02:57.680","Text":"We know the amount of glucose that we need for"},{"Start":"02:57.680 ","End":"03:00.275","Text":"the 270 grams solution and"},{"Start":"03:00.275 ","End":"03:04.885","Text":"now we want to find the amount of water that we have in the solution."},{"Start":"03:04.885 ","End":"03:07.939","Text":"The mass of the solution, remember,"},{"Start":"03:07.939 ","End":"03:12.810","Text":"equals the mass of the solute plus the mass of the solvent,"},{"Start":"03:12.810 ","End":"03:16.850","Text":"so that\u0027s the mass of the glucose plus the mass of the water."},{"Start":"03:16.850 ","End":"03:19.910","Text":"If we want to find the mass of the water,"},{"Start":"03:19.910 ","End":"03:22.880","Text":"we\u0027re going to subtract the mass of glucose from both sides."},{"Start":"03:22.880 ","End":"03:24.050","Text":"The mass of water,"},{"Start":"03:24.050 ","End":"03:28.835","Text":"and we\u0027re just moving water to the left and mass of the solution to the right,"},{"Start":"03:28.835 ","End":"03:31.675","Text":"equals the mass of glucose,"},{"Start":"03:31.675 ","End":"03:33.895","Text":"so you want to find the mass of the water."},{"Start":"03:33.895 ","End":"03:36.995","Text":"We\u0027re just going to subtract the mass of glucose from both sides."},{"Start":"03:36.995 ","End":"03:41.435","Text":"The mass of water equals the mass of the solution,"},{"Start":"03:41.435 ","End":"03:45.440","Text":"I just switch sides, minus the mass of glucose."},{"Start":"03:45.440 ","End":"03:52.820","Text":"The mass of the solution equals 270 grams minus the mass of glucose that we found,"},{"Start":"03:52.820 ","End":"03:54.860","Text":"which equals 43.2 grams,"},{"Start":"03:54.860 ","End":"04:03.390","Text":"and this equals 226.8 grams."},{"Start":"04:03.390 ","End":"04:07.775","Text":"Again, the mass of glucose that we found equals 43.2 grams and the mass"},{"Start":"04:07.775 ","End":"04:13.355","Text":"of water that we found equals 226.8 grams."},{"Start":"04:13.355 ","End":"04:15.140","Text":"That is our final answer."},{"Start":"04:15.140 ","End":"04:17.550","Text":"Thank you very much for watching."}],"ID":30929},{"Watched":false,"Name":"Exercise 5","Duration":"5m 7s","ChapterTopicVideoID":28907,"CourseChapterTopicPlaylistID":245373,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:03.404","Text":"How are we going to solve the following exercise?"},{"Start":"00:03.404 ","End":"00:07.620","Text":"How many grams of sodium chloride need to be added to 250 grams of"},{"Start":"00:07.620 ","End":"00:12.945","Text":"water to make a 3.2 percent sodium chloride by mass solution?"},{"Start":"00:12.945 ","End":"00:17.280","Text":"We have to calculate the grams of sodium chloride and we know that"},{"Start":"00:17.280 ","End":"00:21.555","Text":"the solution is 3.2 percent sodium chloride by mass."},{"Start":"00:21.555 ","End":"00:26.115","Text":"We also know that we have 250 grams of water."},{"Start":"00:26.115 ","End":"00:31.965","Text":"Percent by mass equals the mass of the solute,"},{"Start":"00:31.965 ","End":"00:38.215","Text":"divided by the mass of the solution times 100 percent."},{"Start":"00:38.215 ","End":"00:42.880","Text":"In our case, the solute is sodium chloride."},{"Start":"00:42.880 ","End":"00:45.780","Text":"That\u0027s the mass of sodium chloride divided by,"},{"Start":"00:45.780 ","End":"00:50.765","Text":"and the solution is the sodium chloride plus the water."},{"Start":"00:50.765 ","End":"00:53.240","Text":"That\u0027s the mass of the solution equals the mass of"},{"Start":"00:53.240 ","End":"00:58.495","Text":"sodium chloride plus the mass of water."},{"Start":"00:58.495 ","End":"01:02.670","Text":"This is multiplied by 100 percent."},{"Start":"01:02.670 ","End":"01:07.710","Text":"We also know that we have 3.2 percent sodium chloride by mass,"},{"Start":"01:07.710 ","End":"01:11.590","Text":"so this equals 3.2 percent."},{"Start":"01:11.590 ","End":"01:14.202","Text":"Now we also know the mass of the water which equals"},{"Start":"01:14.202 ","End":"01:19.850","Text":"250 grams and we have to calculate the mass of sodium chloride."},{"Start":"01:20.360 ","End":"01:25.130","Text":"At this point, what we\u0027re going to do is we\u0027re going to divide both sides by 100 percent."},{"Start":"01:25.130 ","End":"01:29.165","Text":"We\u0027re left with the mass of the sodium chloride"},{"Start":"01:29.165 ","End":"01:35.330","Text":"divided by the mass of sodium chloride plus the mass"},{"Start":"01:35.330 ","End":"01:44.340","Text":"of water equals 0.032 because"},{"Start":"01:44.340 ","End":"01:46.290","Text":"that\u0027s 3.2 divided by 100 percent,"},{"Start":"01:46.290 ","End":"01:48.480","Text":"so we get 0.032."},{"Start":"01:48.480 ","End":"01:51.380","Text":"Now, at this point, we\u0027re just going to multiply both sides by"},{"Start":"01:51.380 ","End":"01:54.065","Text":"the mass of NaCl plus the mass of water."},{"Start":"01:54.065 ","End":"01:55.880","Text":"We\u0027re left with, on the left side,"},{"Start":"01:55.880 ","End":"01:57.071","Text":"the mass of NaCl equals"},{"Start":"01:57.071 ","End":"02:06.730","Text":"0.032 times the mass of NaCl,"},{"Start":"02:06.730 ","End":"02:12.825","Text":"sodium chloride plus the mass of water."},{"Start":"02:12.825 ","End":"02:22.265","Text":"Again, the mass of NaCl equals 0.032 times the mass of NaCl,"},{"Start":"02:22.265 ","End":"02:29.250","Text":"sodium chloride plus 0.032 times the mass of water."},{"Start":"02:29.250 ","End":"02:35.470","Text":"Now we\u0027re going to subtract 0.032 times the mass of sodium chloride from both sides."},{"Start":"02:35.470 ","End":"02:43.150","Text":"We have the mass of sodium chloride times 1 minus 0.032 times"},{"Start":"02:43.150 ","End":"02:52.530","Text":"the mass of sodium chloride equals 0.032 times the mass of water."},{"Start":"02:52.530 ","End":"03:02.200","Text":"On the left side, we have the mass of sodium chloride times 0.968,"},{"Start":"03:02.870 ","End":"03:07.290","Text":"since we have here 1 minus 0.032."},{"Start":"03:07.290 ","End":"03:10.140","Text":"That\u0027s 0.968 times the mass of"},{"Start":"03:10.140 ","End":"03:19.450","Text":"sodium chloride equals 0.032 times the mass of water."},{"Start":"03:19.450 ","End":"03:25.130","Text":"Remember, the mass of the water that was given equals 250 grams."},{"Start":"03:25.130 ","End":"03:28.490","Text":"That\u0027s times 250 grams."},{"Start":"03:28.490 ","End":"03:32.350","Text":"This equals 8 grams."},{"Start":"03:32.350 ","End":"03:35.840","Text":"Now, we\u0027re going to divide both sides by 0.968."},{"Start":"03:35.840 ","End":"03:40.880","Text":"The mass of sodium chloride equals"},{"Start":"03:40.880 ","End":"03:47.550","Text":"8 grams divided by 0.968."},{"Start":"03:49.100 ","End":"03:53.970","Text":"This equals 8.26 grams."},{"Start":"03:53.970 ","End":"03:59.120","Text":"The amount of sodium chloride which needs to be added to 250 grams"},{"Start":"03:59.120 ","End":"04:04.394","Text":"of the water equals 8.26 grams."},{"Start":"04:04.394 ","End":"04:06.470","Text":"Now, at this point, we\u0027re done and we have our answer."},{"Start":"04:06.470 ","End":"04:08.855","Text":"However, if you want to check yourself,"},{"Start":"04:08.855 ","End":"04:12.590","Text":"you can calculate the percent by mass and see if you\u0027re right,"},{"Start":"04:12.590 ","End":"04:14.495","Text":"that you didn\u0027t make any mistakes."},{"Start":"04:14.495 ","End":"04:18.590","Text":"Again, percent by mass equals the mass of"},{"Start":"04:18.590 ","End":"04:24.785","Text":"the solute divided by the mass of the solution times 100 percent."},{"Start":"04:24.785 ","End":"04:30.955","Text":"In our case, the mass of the solute sodium chloride equals 8.26 grams,"},{"Start":"04:30.955 ","End":"04:33.875","Text":"and we divide this by the mass of the solution,"},{"Start":"04:33.875 ","End":"04:39.920","Text":"which equals the mass of the sodium chloride 8.26 grams plus the mass of the water,"},{"Start":"04:39.920 ","End":"04:42.385","Text":"which remember was 250 grams."},{"Start":"04:42.385 ","End":"04:45.900","Text":"We have to multiply this by 100 percent."},{"Start":"04:45.900 ","End":"04:49.000","Text":"If you divide then multiply,"},{"Start":"04:49.340 ","End":"04:52.055","Text":"this equals 3.2 percent."},{"Start":"04:52.055 ","End":"04:54.560","Text":"This was the percent by mass that was given in"},{"Start":"04:54.560 ","End":"04:58.420","Text":"our questions so we know that we calculated everything correct."},{"Start":"04:58.420 ","End":"05:03.395","Text":"Again, the mass of the sodium chloride that we found equals 8.26 grams."},{"Start":"05:03.395 ","End":"05:04.820","Text":"That is our final answer."},{"Start":"05:04.820 ","End":"05:07.290","Text":"Thank you very much for watching."}],"ID":30930},{"Watched":false,"Name":"Exercise 6","Duration":"5m 41s","ChapterTopicVideoID":28908,"CourseChapterTopicPlaylistID":245373,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.300","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.300 ","End":"00:08.295","Text":"A solution is prepared by dissolving 9.94 milliliters"},{"Start":"00:08.295 ","End":"00:13.590","Text":"of ethanol in enough water to produce 80 milliliters of solution."},{"Start":"00:13.590 ","End":"00:17.745","Text":"We\u0027re given the density of ethanol and the density of the solution."},{"Start":"00:17.745 ","End":"00:22.615","Text":"In a, we need to calculate the volume percent of ethanol in the solution."},{"Start":"00:22.615 ","End":"00:26.055","Text":"Let\u0027s begin with a and then we\u0027ll go on to b and c. We"},{"Start":"00:26.055 ","End":"00:30.080","Text":"need to calculate the volume percent of the ethanol in the solution."},{"Start":"00:30.080 ","End":"00:38.840","Text":"The volume percent equals the volume of the solute divided by the volume of the solution."},{"Start":"00:38.840 ","End":"00:42.155","Text":"Now, since the volume of solute and solution are given,"},{"Start":"00:42.155 ","End":"00:45.165","Text":"a is very easy to calculate."},{"Start":"00:45.165 ","End":"00:54.245","Text":"This is going to be 9.94 milliliters of ethanol divided by the volume of the solution,"},{"Start":"00:54.245 ","End":"00:56.815","Text":"which equals 80 milliliters."},{"Start":"00:56.815 ","End":"00:59.790","Text":"This is divided by 80 milliliters."},{"Start":"00:59.790 ","End":"01:02.270","Text":"Since we want to find the percent,"},{"Start":"01:02.270 ","End":"01:06.370","Text":"we need to multiply this by 100 percent."},{"Start":"01:06.370 ","End":"01:10.050","Text":"This is multiplied by 100 percent."},{"Start":"01:10.050 ","End":"01:13.405","Text":"After dividing and multiplying by 100 percent,"},{"Start":"01:13.405 ","End":"01:19.965","Text":"this equals 12.42 percent."},{"Start":"01:19.965 ","End":"01:22.740","Text":"That was a, in a,"},{"Start":"01:22.740 ","End":"01:27.450","Text":"we find that the volume percent equals 12.42 percent."},{"Start":"01:27.450 ","End":"01:30.450","Text":"Now we\u0027re going to go on to b."},{"Start":"01:30.450 ","End":"01:36.145","Text":"In b, we need to calculate the mass percent of ethanol in the solution."},{"Start":"01:36.145 ","End":"01:40.620","Text":"So b, the mass percent"},{"Start":"01:40.620 ","End":"01:47.395","Text":"equals the mass of the solute divided by the mass of the solution,"},{"Start":"01:47.395 ","End":"01:50.635","Text":"and again, times 100 percent."},{"Start":"01:50.635 ","End":"01:52.510","Text":"Now, in the question,"},{"Start":"01:52.510 ","End":"01:55.420","Text":"we were given the volume of the ethanol,"},{"Start":"01:55.420 ","End":"01:57.025","Text":"which is 9.94 milliliters,"},{"Start":"01:57.025 ","End":"01:58.300","Text":"and the volume of the solution,"},{"Start":"01:58.300 ","End":"02:00.624","Text":"which equals 80 milliliters."},{"Start":"02:00.624 ","End":"02:03.985","Text":"You were not given the masses of each."},{"Start":"02:03.985 ","End":"02:09.040","Text":"However, we do have the density of ethanol and the density of the solution."},{"Start":"02:09.040 ","End":"02:11.800","Text":"Remember the d, the density equals m,"},{"Start":"02:11.800 ","End":"02:13.875","Text":"which is the mass divided by v,"},{"Start":"02:13.875 ","End":"02:15.815","Text":"which equals the volume."},{"Start":"02:15.815 ","End":"02:18.140","Text":"We\u0027re looking for the mass, so m,"},{"Start":"02:18.140 ","End":"02:22.210","Text":"the mass equals d times v,"},{"Start":"02:22.210 ","End":"02:25.105","Text":"we just multiplied both sides by the volume."},{"Start":"02:25.105 ","End":"02:27.710","Text":"If we\u0027re looking for the mass of the ethanol,"},{"Start":"02:27.710 ","End":"02:29.630","Text":"we\u0027re going to use the density of the ethanol,"},{"Start":"02:29.630 ","End":"02:37.730","Text":"which equals 0.789 grams per milliliters and that\u0027s times the volume of the ethanol,"},{"Start":"02:37.730 ","End":"02:40.490","Text":"which equals 9.94 milliliters."},{"Start":"02:40.490 ","End":"02:44.875","Text":"We can see that the milliliters cancel out and we\u0027re going to be left with grams,"},{"Start":"02:44.875 ","End":"02:48.255","Text":"which is the units of our mass."},{"Start":"02:48.255 ","End":"02:55.820","Text":"This equals 7.84 grams."},{"Start":"02:55.820 ","End":"02:57.800","Text":"That\u0027s the mass of the ethanol,"},{"Start":"02:57.800 ","End":"02:59.645","Text":"which is our solute."},{"Start":"02:59.645 ","End":"03:02.225","Text":"Now we need to look for the mass of the solution."},{"Start":"03:02.225 ","End":"03:04.580","Text":"We need to calculate the mass of the solution."},{"Start":"03:04.580 ","End":"03:06.820","Text":"We\u0027re going to do that in the same way,"},{"Start":"03:06.820 ","End":"03:12.020","Text":"so that density of the solution equals 0.981 grams per milliliters."},{"Start":"03:12.020 ","End":"03:13.940","Text":"Again, the mass, however,"},{"Start":"03:13.940 ","End":"03:16.340","Text":"this time of the solution equals"},{"Start":"03:16.340 ","End":"03:19.960","Text":"the density of the solution times the volume of the solution."},{"Start":"03:19.960 ","End":"03:26.780","Text":"The density of the solution equals 0.981 grams per milliliters,"},{"Start":"03:26.780 ","End":"03:29.605","Text":"this was given in the question,"},{"Start":"03:29.605 ","End":"03:36.260","Text":"times 80 milliliters, which is the volume of the solution also given in the question."},{"Start":"03:36.260 ","End":"03:46.340","Text":"The milliliters cancel out and this equals 78.48 grams."},{"Start":"03:46.340 ","End":"03:49.175","Text":"Now, we can continue."},{"Start":"03:49.175 ","End":"03:51.890","Text":"Remember, the mass percent equals the mass of a solute"},{"Start":"03:51.890 ","End":"03:55.205","Text":"divided by mass of the solution times 100 percent."},{"Start":"03:55.205 ","End":"03:58.550","Text":"In our case, we\u0027re just going to continue here."},{"Start":"03:58.550 ","End":"04:01.850","Text":"This equals the mass of the ethanol,"},{"Start":"04:01.850 ","End":"04:04.670","Text":"which equals 7.84 grams,"},{"Start":"04:04.670 ","End":"04:07.250","Text":"divided by the mass of the solution,"},{"Start":"04:07.250 ","End":"04:11.220","Text":"which equals 78.48 grams."},{"Start":"04:11.220 ","End":"04:14.625","Text":"This is multiplied by 100 percent."},{"Start":"04:14.625 ","End":"04:16.640","Text":"After dividing and multiplying,"},{"Start":"04:16.640 ","End":"04:22.610","Text":"this equals 9.99 percent."},{"Start":"04:22.610 ","End":"04:29.510","Text":"In b, we found the mass percent equals 9.99 percent."},{"Start":"04:29.510 ","End":"04:32.870","Text":"Now, if we look at c, we can see that we need to calculate the mass per"},{"Start":"04:32.870 ","End":"04:36.715","Text":"volume percent of ethanol in the solution."},{"Start":"04:36.715 ","End":"04:43.375","Text":"In c, we need to calculate the mass per volume percent."},{"Start":"04:43.375 ","End":"04:47.430","Text":"This equals the mass of the solute"},{"Start":"04:47.430 ","End":"04:53.860","Text":"divided by the volume of the solution times 100 percent."},{"Start":"04:57.620 ","End":"05:01.655","Text":"We already calculated the mass of the solute,"},{"Start":"05:01.655 ","End":"05:04.325","Text":"which equals 7.84 grams."},{"Start":"05:04.325 ","End":"05:12.319","Text":"This needs to be divided by the volume of the solution,"},{"Start":"05:12.319 ","End":"05:13.730","Text":"which was given in the question,"},{"Start":"05:13.730 ","End":"05:15.680","Text":"and equals 80 milliliters,"},{"Start":"05:15.680 ","End":"05:23.050","Text":"that\u0027s divided by 80 milliliters and this is times 100 percent."},{"Start":"05:23.050 ","End":"05:24.935","Text":"Now after dividing and multiplying,"},{"Start":"05:24.935 ","End":"05:29.905","Text":"this equals 9.8 percent."},{"Start":"05:29.905 ","End":"05:37.010","Text":"The mass per volume percent of the ethanol in the solution equals 9.8 percent."},{"Start":"05:37.010 ","End":"05:38.930","Text":"These are our final answers."},{"Start":"05:38.930 ","End":"05:41.580","Text":"Thank you very much for watching."}],"ID":30931},{"Watched":false,"Name":"Exercise 7","Duration":"7m 39s","ChapterTopicVideoID":28909,"CourseChapterTopicPlaylistID":245373,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.795","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.795 ","End":"00:07.950","Text":"Calculate the amount of acetic acid CH_3COOH in"},{"Start":"00:07.950 ","End":"00:12.300","Text":"grams contained in 520 milliliters of vinegar."},{"Start":"00:12.300 ","End":"00:17.370","Text":"The mass percent of the acetic acid in vinegar equals 5.01 percent,"},{"Start":"00:17.370 ","End":"00:22.100","Text":"and the density of vinegar equals 1.01 grams per milliliter."},{"Start":"00:22.100 ","End":"00:24.710","Text":"Let\u0027s start with our 5.01 percent,"},{"Start":"00:24.710 ","End":"00:25.895","Text":"which is our mass percent."},{"Start":"00:25.895 ","End":"00:34.890","Text":"The mass percent of acetic acid in vinegar equals 5.01 percent."},{"Start":"00:34.890 ","End":"00:36.960","Text":"When we have 5.01 percent,"},{"Start":"00:36.960 ","End":"00:42.425","Text":"we can say that we have 5.01 grams of the solute,"},{"Start":"00:42.425 ","End":"00:44.900","Text":"which in our case is acetic acid,"},{"Start":"00:44.900 ","End":"00:50.955","Text":"divided by a 100 grams of our solution, which is vinegar."},{"Start":"00:50.955 ","End":"00:53.110","Text":"Here we have our vinegar,"},{"Start":"00:53.110 ","End":"00:56.795","Text":"and up here we have acetic acid."},{"Start":"00:56.795 ","End":"01:03.200","Text":"Again, we have 5.01 grams of acetic acid for every 100 grams of vinegar."},{"Start":"01:03.200 ","End":"01:06.830","Text":"Now we want to calculate the amount of acetic acid in grams"},{"Start":"01:06.830 ","End":"01:09.800","Text":"that we have in 520 milliliters of vinegar,"},{"Start":"01:09.800 ","End":"01:15.310","Text":"meaning we have a volume of the solution and we need to find the number of grams."},{"Start":"01:15.310 ","End":"01:17.975","Text":"Let\u0027s just take a look before we start calculating."},{"Start":"01:17.975 ","End":"01:20.300","Text":"Again, we have 5.01 grams."},{"Start":"01:20.300 ","End":"01:22.745","Text":"This is acetic acid,"},{"Start":"01:22.745 ","End":"01:25.520","Text":"and this is our solution which is vinegar."},{"Start":"01:25.520 ","End":"01:31.380","Text":"We have 5.01 gram for every 100 grams of vinegar,"},{"Start":"01:31.380 ","End":"01:38.955","Text":"and now we\u0027re given that the amount of vinegar is 520 milliliters,"},{"Start":"01:38.955 ","End":"01:42.410","Text":"and we want to find the grams of acetic acid."},{"Start":"01:42.410 ","End":"01:44.540","Text":"There are number of ways to start calculating,"},{"Start":"01:44.540 ","End":"01:50.165","Text":"but the first thing we need to do is make the amount of vinegar comparable in units,"},{"Start":"01:50.165 ","End":"01:55.260","Text":"because right now, we have 1 in grams and 1 in milliliters."},{"Start":"01:55.360 ","End":"01:58.915","Text":"We want to change 1 into the other,"},{"Start":"01:58.915 ","End":"02:01.505","Text":"and I think that the easiest would be to take"},{"Start":"02:01.505 ","End":"02:05.330","Text":"our milliliters of vinegar and just change them into grams."},{"Start":"02:05.330 ","End":"02:08.600","Text":"Now we\u0027re also given, remember the density of our vinegar,"},{"Start":"02:08.600 ","End":"02:13.040","Text":"and that\u0027s how we\u0027re going to change our milliliters into grams, so d,"},{"Start":"02:13.040 ","End":"02:15.635","Text":"the density equals m divided by v,"},{"Start":"02:15.635 ","End":"02:17.210","Text":"m is the mass divided by v,"},{"Start":"02:17.210 ","End":"02:18.970","Text":"which is the volume."},{"Start":"02:18.970 ","End":"02:22.400","Text":"We\u0027re given the volume and we want to find the mass."},{"Start":"02:22.400 ","End":"02:24.470","Text":"We\u0027re going to multiply both sides by the volume."},{"Start":"02:24.470 ","End":"02:25.910","Text":"Mass, m=d,"},{"Start":"02:25.910 ","End":"02:28.505","Text":"the density times v the volume."},{"Start":"02:28.505 ","End":"02:34.025","Text":"The density equals 1.01 grams per milliliter,"},{"Start":"02:34.025 ","End":"02:37.160","Text":"and we\u0027re multiplying this by the volume that we were given,"},{"Start":"02:37.160 ","End":"02:40.655","Text":"which is 520 milliliters."},{"Start":"02:40.655 ","End":"02:43.250","Text":"The milliliters cancel out immediately and we\u0027re"},{"Start":"02:43.250 ","End":"02:46.585","Text":"going to be left with grams for our mass."},{"Start":"02:46.585 ","End":"02:53.210","Text":"This comes to 525.2 grams."},{"Start":"02:53.210 ","End":"03:00.210","Text":"We actually converted our volume into mass."},{"Start":"03:00.210 ","End":"03:04.265","Text":"I\u0027m going to change this volume, going to erase it,"},{"Start":"03:04.265 ","End":"03:13.130","Text":"and change it and just write 525.2 grams right over here."},{"Start":"03:13.130 ","End":"03:17.749","Text":"Now that we have the same units for the vinegar,"},{"Start":"03:17.749 ","End":"03:20.585","Text":"we can continue calculating."},{"Start":"03:20.585 ","End":"03:23.780","Text":"Now, as we said, we have the amount of acetic acid in"},{"Start":"03:23.780 ","End":"03:27.460","Text":"100 grams and we want to find the amount of acetic acid,"},{"Start":"03:27.460 ","End":"03:33.160","Text":"which is the mass, which is not known in 525.2 grams."},{"Start":"03:41.630 ","End":"03:45.220","Text":"To find the mass of the acetic acid,"},{"Start":"03:45.220 ","End":"03:48.250","Text":"we\u0027re going to start with the mass of the vinegar,"},{"Start":"03:48.250 ","End":"03:55.710","Text":"525.2 grams, then we\u0027re going to multiply this."},{"Start":"03:55.710 ","End":"03:57.720","Text":"That\u0027s the mass of the vinegar,"},{"Start":"03:57.720 ","End":"04:03.280","Text":"so we\u0027re just going to write vinegar right here and multiply this by a conversion factor."},{"Start":"04:03.280 ","End":"04:08.305","Text":"We\u0027re going to put the mass of the acetic acid in the numerator,"},{"Start":"04:08.305 ","End":"04:09.850","Text":"which is grams, of course,"},{"Start":"04:09.850 ","End":"04:14.340","Text":"divided by 100 grams of vinegar."},{"Start":"04:14.340 ","End":"04:18.620","Text":"Then our grams of vinegar is going to cancel out."},{"Start":"04:18.620 ","End":"04:21.455","Text":"That\u0027s why we needed in our numerator."},{"Start":"04:21.455 ","End":"04:31.520","Text":"When we multiply, this equals 26.31 grams."},{"Start":"04:31.520 ","End":"04:36.980","Text":"The mass of the acetic acid equals 26.31 grams."},{"Start":"04:36.980 ","End":"04:45.590","Text":"Remember, 26.31 grams is the amount of acetic acid in 520 milliliters of solution."},{"Start":"04:45.590 ","End":"04:47.690","Text":"Now, I use the conversion factor,"},{"Start":"04:47.690 ","End":"04:49.745","Text":"but not everyone likes to do that though."},{"Start":"04:49.745 ","End":"04:55.485","Text":"People will take the mass times the 100 grams,"},{"Start":"04:55.485 ","End":"05:02.370","Text":"and this is going to equal our 5.01 grams times 525.2 grams."},{"Start":"05:06.190 ","End":"05:08.540","Text":"For a second, let\u0027s just look at this."},{"Start":"05:08.540 ","End":"05:12.320","Text":"It\u0027s going to be our mass? This is something else."},{"Start":"05:12.320 ","End":"05:19.560","Text":"There\u0027s going to be our mass times a 100 grams equals 5.01 grams."},{"Start":"05:19.560 ","End":"05:23.050","Text":"100 grams, obviously is of the vinegar."},{"Start":"05:23.630 ","End":"05:26.460","Text":"Equals our 5.01 grams,"},{"Start":"05:26.460 ","End":"05:29.095","Text":"which is of the acetic acid."},{"Start":"05:29.095 ","End":"05:31.930","Text":"I\u0027ll just write AA,"},{"Start":"05:31.930 ","End":"05:38.509","Text":"times 525.2 grams of vinegar."},{"Start":"05:38.509 ","End":"05:42.400","Text":"Let\u0027s take this down here because this is getting too complicated."},{"Start":"05:42.400 ","End":"05:44.320","Text":"Okay. I just wanted to write this again."},{"Start":"05:44.320 ","End":"05:48.295","Text":"We have our mass times a 100 grams,"},{"Start":"05:48.295 ","End":"05:50.110","Text":"which is grams of the solution,"},{"Start":"05:50.110 ","End":"05:52.165","Text":"which is the vinegar,"},{"Start":"05:52.165 ","End":"05:57.240","Text":"equals 5.01 grams, which is acetic acids,"},{"Start":"05:57.240 ","End":"06:05.380","Text":"so grams of acetic acid times 525.2 grams of vinegar."},{"Start":"06:05.380 ","End":"06:10.290","Text":"The grams of vinegar cancel out and we divide now by 100,"},{"Start":"06:10.290 ","End":"06:15.080","Text":"so that\u0027s going to be the mass equals 5.01"},{"Start":"06:15.080 ","End":"06:25.825","Text":"grams of acetic acid divided by 100 times 525.2."},{"Start":"06:25.825 ","End":"06:29.290","Text":"Now, if we take a look at what we\u0027re calculating right"},{"Start":"06:29.290 ","End":"06:33.310","Text":"now and the calculation with the conversion factor,"},{"Start":"06:33.310 ","End":"06:35.110","Text":"essentially, we\u0027re doing the same thing."},{"Start":"06:35.110 ","End":"06:38.725","Text":"We\u0027re taking 525.2 times 5.01 divided by 100,"},{"Start":"06:38.725 ","End":"06:40.390","Text":"and the units are also the same,"},{"Start":"06:40.390 ","End":"06:46.540","Text":"and this is also going to give us 26.31 grams."},{"Start":"06:46.540 ","End":"06:49.085","Text":"It\u0027s just another way to look at it."},{"Start":"06:49.085 ","End":"06:52.060","Text":"Again, I calculate it with a conversion factor"},{"Start":"06:52.060 ","End":"06:54.820","Text":"but there are people who prefer this method,"},{"Start":"06:54.820 ","End":"06:57.565","Text":"which essentially comes out the exact same."},{"Start":"06:57.565 ","End":"06:59.875","Text":"Now, I want to mention one last thing."},{"Start":"06:59.875 ","End":"07:04.790","Text":"In this question, I chose to convert the volume of the vinegar,"},{"Start":"07:04.790 ","End":"07:12.005","Text":"which was 520 milliliters into grams so that our units of the vinegar will be equal."},{"Start":"07:12.005 ","End":"07:16.250","Text":"But I could have also taken our 100 grams of"},{"Start":"07:16.250 ","End":"07:21.395","Text":"solution of vinegar and converted it into milliliters, into volume."},{"Start":"07:21.395 ","End":"07:28.215","Text":"Then continued the same way as we did just with a different conversion factor."},{"Start":"07:28.215 ","End":"07:35.690","Text":"Again, the mass of the acetic acid that we found that we calculated equals 26.31 grams."},{"Start":"07:35.690 ","End":"07:37.220","Text":"That is our final answer."},{"Start":"07:37.220 ","End":"07:39.720","Text":"Thank you very much for watching."}],"ID":30932},{"Watched":false,"Name":"Exercise 8","Duration":"1m 55s","ChapterTopicVideoID":28910,"CourseChapterTopicPlaylistID":245373,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.270","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.270 ","End":"00:10.290","Text":"22 grams of sugar are dissolved in enough water to produce 250 milliliters of solution."},{"Start":"00:10.290 ","End":"00:16.890","Text":"Calculate the mass to volume concentration of sugar in the solution in grams per liter."},{"Start":"00:16.890 ","End":"00:21.455","Text":"We\u0027re asked to calculate the mass to volume concentration in grams per liter,"},{"Start":"00:21.455 ","End":"00:24.250","Text":"and we know that we have 22 grams of sugar,"},{"Start":"00:24.250 ","End":"00:28.900","Text":"and we know that we have 250 milliliters of solution."},{"Start":"00:29.270 ","End":"00:34.740","Text":"If we take the mass to volume that we\u0027re given,"},{"Start":"00:34.740 ","End":"00:38.970","Text":"this equals 22 grams of sugar,"},{"Start":"00:38.970 ","End":"00:40.620","Text":"which is a solute,"},{"Start":"00:40.620 ","End":"00:47.400","Text":"divided by 250 milliliters of the solution."},{"Start":"00:47.400 ","End":"00:52.115","Text":"We have our mass to volume in grams per milliliters units,"},{"Start":"00:52.115 ","End":"00:54.415","Text":"and we want it in grams per liters."},{"Start":"00:54.415 ","End":"00:59.165","Text":"All we need to do is to convert our 250 milliliters into liters,"},{"Start":"00:59.165 ","End":"01:01.505","Text":"and then we will have grams per liter."},{"Start":"01:01.505 ","End":"01:05.945","Text":"In order to convert our 250 milliliters into liters,"},{"Start":"01:05.945 ","End":"01:08.875","Text":"we\u0027re going to multiply by a conversion factor."},{"Start":"01:08.875 ","End":"01:13.350","Text":"Remember that in 1 liter we have 1,000 milliliters,"},{"Start":"01:13.350 ","End":"01:17.825","Text":"so the milliliters will cancel out and we\u0027ll be left with liters."},{"Start":"01:17.825 ","End":"01:23.990","Text":"This equals 22 grams of sugar, after dividing,"},{"Start":"01:23.990 ","End":"01:30.390","Text":"we get 0.250 liters of solution."},{"Start":"01:30.390 ","End":"01:33.305","Text":"Now that we have our units and grams per liter,"},{"Start":"01:33.305 ","End":"01:37.070","Text":"we can just divide 22 by 0.25,"},{"Start":"01:37.070 ","End":"01:40.105","Text":"and this comes to 88."},{"Start":"01:40.105 ","End":"01:45.170","Text":"The units that we\u0027re left with is grams per liter."},{"Start":"01:45.170 ","End":"01:51.935","Text":"The mass to volume equals 88 grams per liter."},{"Start":"01:51.935 ","End":"01:53.330","Text":"That is our final answer,"},{"Start":"01:53.330 ","End":"01:55.920","Text":"thank you very much for watching."}],"ID":30933},{"Watched":false,"Name":"Exercise 9","Duration":"5m 4s","ChapterTopicVideoID":28911,"CourseChapterTopicPlaylistID":245373,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.760","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.760 ","End":"00:10.410","Text":"We need to prepare 250 milliliters of a 20 grams per liter solution of sodium chloride."},{"Start":"00:10.410 ","End":"00:14.380","Text":"How much sodium chloride in grams do we need?"},{"Start":"00:14.420 ","End":"00:19.170","Text":"First of all, since we know that we need a 20 grams per liter solution"},{"Start":"00:19.170 ","End":"00:23.550","Text":"it means that we have 20 grams per liter"},{"Start":"00:23.550 ","End":"00:33.540","Text":"equals 20 grams of sodium chloride for every 1 liter of solution."},{"Start":"00:33.540 ","End":"00:38.880","Text":"We have our sodium chloride and we have our solution."},{"Start":"00:38.880 ","End":"00:42.945","Text":"Again, we know that we have 20 grams of sodium chloride."},{"Start":"00:42.945 ","End":"00:46.230","Text":"It means we need 1 liter of solution."},{"Start":"00:46.230 ","End":"00:50.790","Text":"In our case we\u0027re given 250 milliliters,"},{"Start":"00:50.790 ","End":"00:54.185","Text":"we need to prepare a 250-milliliter solution."},{"Start":"00:54.185 ","End":"00:55.940","Text":"This will go here,"},{"Start":"00:55.940 ","End":"00:58.578","Text":"250 milliliters of solution."},{"Start":"00:58.578 ","End":"01:02.960","Text":"We\u0027re asked, how many grams of sodium chloride do we need?"},{"Start":"01:02.960 ","End":"01:07.310","Text":"Meaning what mass of sodium chloride do we need to make"},{"Start":"01:07.310 ","End":"01:13.670","Text":"this 250 milliliters of solution which have a concentration of 20 grams per liter."},{"Start":"01:13.670 ","End":"01:16.430","Text":"As I explained before in other videos,"},{"Start":"01:16.430 ","End":"01:17.600","Text":"we have 2 ways of doing this."},{"Start":"01:17.600 ","End":"01:20.095","Text":"We can actually use a conversion factor"},{"Start":"01:20.095 ","End":"01:23.120","Text":"or some people find it more comfortable just to do"},{"Start":"01:23.120 ","End":"01:26.375","Text":"20 grams times the 250 milliliters of solution"},{"Start":"01:26.375 ","End":"01:30.830","Text":"equals 1 liter of solution times the mass of the sodium chloride,"},{"Start":"01:30.830 ","End":"01:33.125","Text":"which again essentially is the same thing."},{"Start":"01:33.125 ","End":"01:35.185","Text":"Let\u0027s just, we\u0027ll do them both right"},{"Start":"01:35.185 ","End":"01:39.540","Text":"now and in our next video is where probably going to just choose 1."},{"Start":"01:39.540 ","End":"01:41.387","Text":"The first method,"},{"Start":"01:41.387 ","End":"01:46.110","Text":"we\u0027re going to say 1 liter times the mass of"},{"Start":"01:46.110 ","End":"01:56.030","Text":"sodium chloride equals 20 grams of sodium chloride times 250 milliliters."},{"Start":"01:56.030 ","End":"02:00.440","Text":"Remember, the volume is of the solution so here we have"},{"Start":"02:00.440 ","End":"02:05.180","Text":"the solution and here we also have our solution volume."},{"Start":"02:05.180 ","End":"02:08.720","Text":"Now let\u0027s divide both sides by 1 liter of solution."},{"Start":"02:08.720 ","End":"02:17.180","Text":"The mass of our sodium chloride equals 20 grams of"},{"Start":"02:17.180 ","End":"02:22.875","Text":"sodium chloride times 250 milliliters"},{"Start":"02:22.875 ","End":"02:30.444","Text":"of our solution and this is all divided by 1 liter of solution."},{"Start":"02:30.444 ","End":"02:37.190","Text":"Since we have our milliliters in our numerator and our liters in our denominator,"},{"Start":"02:37.190 ","End":"02:39.478","Text":"we have to convert 1 into the other."},{"Start":"02:39.478 ","End":"02:43.340","Text":"So let\u0027s just convert our liters into milliliters."},{"Start":"02:43.340 ","End":"02:45.490","Text":"We have to multiply by a conversion factor,"},{"Start":"02:45.490 ","End":"02:49.570","Text":"so we\u0027re going to multiply by 1,000 milliliters for every 1 liter,"},{"Start":"02:49.570 ","End":"02:51.485","Text":"that way our liters cancel out."},{"Start":"02:51.485 ","End":"02:56.370","Text":"Of course, this is milliliters of solution divided by liters of solution."},{"Start":"02:56.380 ","End":"03:01.700","Text":"Milliliters of solution also cancel out and all we have to do is"},{"Start":"03:01.700 ","End":"03:07.445","Text":"20 grams times 250 divided by 1,000 so this equals 20 grams"},{"Start":"03:07.445 ","End":"03:15.755","Text":"of sodium chloride times 250 divided by"},{"Start":"03:15.755 ","End":"03:19.997","Text":"1,000 and this equals"},{"Start":"03:19.997 ","End":"03:26.350","Text":"20 grams of sodium chloride divided by 4."},{"Start":"03:26.350 ","End":"03:30.490","Text":"Obviously, you could do this whole thing in your calculator and this equals"},{"Start":"03:30.490 ","End":"03:35.499","Text":"5 grams of sodium chloride."},{"Start":"03:35.499 ","End":"03:40.930","Text":"The mass of sodium chloride that we need for"},{"Start":"03:40.930 ","End":"03:46.550","Text":"our 250 milliliters solution equals 5 grams of sodium chloride."},{"Start":"03:46.550 ","End":"03:48.090","Text":"That was 1 method."},{"Start":"03:48.090 ","End":"03:50.860","Text":"Now the second method which again it\u0027s essentially the same,"},{"Start":"03:50.860 ","End":"03:54.190","Text":"we\u0027re doing the same thing it\u0027s just another way to look at it is to"},{"Start":"03:54.190 ","End":"03:58.010","Text":"take our 250 milliliters which is the volume of the solution that we"},{"Start":"03:58.010 ","End":"04:04.360","Text":"need to find the mass of sodium chloride for and multiply this by a conversion factor"},{"Start":"04:04.360 ","End":"04:07.070","Text":"and our conversion factor is going to be our mass to"},{"Start":"04:07.070 ","End":"04:11.490","Text":"volume that we\u0027re given which is 20 grams for 1 liter."},{"Start":"04:11.490 ","End":"04:17.405","Text":"Our second method is just to take our volume which is 250 milliliters of solution."},{"Start":"04:17.405 ","End":"04:24.475","Text":"Again, the mass of sodium chloride equals 250 milliliters of solution times our 20 grams"},{"Start":"04:24.475 ","End":"04:32.015","Text":"of sodium chloride for every 1 liter of solution."},{"Start":"04:32.015 ","End":"04:35.765","Text":"If you compare this calculation is exactly the same."},{"Start":"04:35.765 ","End":"04:39.950","Text":"We have 20 grams of sodium chloride times"},{"Start":"04:39.950 ","End":"04:46.280","Text":"250 milliliters of solution and in the end we divide by 1 liter of solution."},{"Start":"04:46.280 ","End":"04:49.370","Text":"This is also going to come out 5 grams,"},{"Start":"04:49.370 ","End":"04:52.370","Text":"obviously of sodium chloride."},{"Start":"04:52.370 ","End":"04:55.730","Text":"Again, the mass of sodium chloride that we found that we need for"},{"Start":"04:55.730 ","End":"05:01.070","Text":"our 250-milliliter solution equals 5 grams of sodium chloride."},{"Start":"05:01.070 ","End":"05:02.540","Text":"That is our final answer."},{"Start":"05:02.540 ","End":"05:05.100","Text":"Thank you very much for watching."}],"ID":30934}],"Thumbnail":null,"ID":245373},{"Name":"Solution Process","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Dissolving Ionic Crystals","Duration":"4m 26s","ChapterTopicVideoID":25624,"CourseChapterTopicPlaylistID":245374,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.560","Text":"In the previous video,"},{"Start":"00:01.560 ","End":"00:06.170","Text":"we talked about the enthalpy of solutions and the like dissolves like rule."},{"Start":"00:06.170 ","End":"00:10.530","Text":"In this video, we\u0027ll talk about dissolving ionic crystals in water."},{"Start":"00:10.530 ","End":"00:16.800","Text":"The question we want to ask is how is an ionic solution water formed?"},{"Start":"00:16.800 ","End":"00:21.510","Text":"What happens is it the ionic crystal decomposes to form ions,"},{"Start":"00:21.510 ","End":"00:26.205","Text":"which are then hydrated surrounded by water molecules."},{"Start":"00:26.205 ","End":"00:29.670","Text":"Here\u0027s a picture. Here\u0027s an NaCl crystal."},{"Start":"00:29.670 ","End":"00:31.275","Text":"We\u0027ve seen that many times,"},{"Start":"00:31.275 ","End":"00:35.425","Text":"Na plus ions next to Cl minus ions."},{"Start":"00:35.425 ","End":"00:38.910","Text":"That is broken up, that costs energy."},{"Start":"00:38.910 ","End":"00:41.595","Text":"That\u0027s the endothermic process."},{"Start":"00:41.595 ","End":"00:46.900","Text":"Then the ions are hydrated by water molecules,"},{"Start":"00:46.900 ","End":"00:52.175","Text":"and that\u0027s an exothermic process because it stabilizes the ions."},{"Start":"00:52.175 ","End":"00:54.110","Text":"We look at Cl minus."},{"Start":"00:54.110 ","End":"00:56.465","Text":"It has H next to it,"},{"Start":"00:56.465 ","End":"00:59.110","Text":"because H is Delta plus."},{"Start":"00:59.110 ","End":"01:01.305","Text":"If we look at the Na plus,"},{"Start":"01:01.305 ","End":"01:06.189","Text":"we have O next to it because O is Delta minus."},{"Start":"01:06.189 ","End":"01:11.260","Text":"Let\u0027s look at the energy balance between these 2 processes."},{"Start":"01:11.260 ","End":"01:15.770","Text":"First, NaCl solid is being broken up"},{"Start":"01:15.770 ","End":"01:20.165","Text":"to Na plus the gas phase plus Cl minus in the gas phase."},{"Start":"01:20.165 ","End":"01:23.120","Text":"This is just to help us understand the energy balance,"},{"Start":"01:23.120 ","End":"01:25.295","Text":"it doesn\u0027t exactly happen like that."},{"Start":"01:25.295 ","End":"01:27.320","Text":"If we look at this expression,"},{"Start":"01:27.320 ","End":"01:31.070","Text":"we can recognize it as minus Delta H lattice"},{"Start":"01:31.070 ","End":"01:36.830","Text":"minus the enthalpy of minus the lattice energy."},{"Start":"01:36.830 ","End":"01:42.860","Text":"Since lattice energy is negative minus Delta H lattice will be positive."},{"Start":"01:42.860 ","End":"01:46.535","Text":"So this is an endothermic process."},{"Start":"01:46.535 ","End":"01:48.680","Text":"Then Na plus,"},{"Start":"01:48.680 ","End":"01:53.870","Text":"the gas phase becomes Na plus in the aqueous solution,"},{"Start":"01:53.870 ","End":"01:57.455","Text":"so it\u0027s being hydrated. That\u0027s here."},{"Start":"01:57.455 ","End":"02:07.415","Text":"Delta H of hydration of Na plus is an exothermic process because it stabilizes the ion."},{"Start":"02:07.415 ","End":"02:09.890","Text":"Similarly, for Cl minus,"},{"Start":"02:09.890 ","End":"02:14.885","Text":"we have Cl minus in the gas phase going to Cl minus in aqueous solution."},{"Start":"02:14.885 ","End":"02:20.130","Text":"Again, that is an exothermic process."},{"Start":"02:21.010 ","End":"02:25.430","Text":"Now, we can sum up these processes."},{"Start":"02:25.430 ","End":"02:30.860","Text":"We have NaCl in the solid phase, NaCl solid,"},{"Start":"02:30.860 ","End":"02:34.525","Text":"and Na plus the gas phase cancels,"},{"Start":"02:34.525 ","End":"02:37.415","Text":"and so the Cl minus in the gas phase."},{"Start":"02:37.415 ","End":"02:44.899","Text":"So we\u0027re left with NaCl solid going to Na plus in aqueous plus Cl minus aqueous."},{"Start":"02:44.899 ","End":"02:46.625","Text":"Then here\u0027s our equation."},{"Start":"02:46.625 ","End":"02:51.364","Text":"This is the equation for the crystal going into solution."},{"Start":"02:51.364 ","End":"02:55.160","Text":"Delta H, dissolution is equal to the sum of"},{"Start":"02:55.160 ","End":"03:02.450","Text":"these enthalpies minus Delta H lattice plus Delta H of hydration of Na plus,"},{"Start":"03:02.450 ","End":"03:05.300","Text":"plus Delta H of hydration of Cl minus."},{"Start":"03:05.300 ","End":"03:10.310","Text":"What we have is a balance between these processes,"},{"Start":"03:10.310 ","End":"03:14.540","Text":"between the endothermic process of breaking up the crystal and"},{"Start":"03:14.540 ","End":"03:20.930","Text":"the exothermic process of stabilizing the ions by hydration."},{"Start":"03:20.930 ","End":"03:25.250","Text":"Depending on whether this is positive or negative,"},{"Start":"03:25.250 ","End":"03:28.535","Text":"Delta H will be positive or negative."},{"Start":"03:28.535 ","End":"03:32.420","Text":"The dissolution will be exothermic for example,"},{"Start":"03:32.420 ","End":"03:37.885","Text":"if more energy is produced in salvation than is used to breaking up the crystal."},{"Start":"03:37.885 ","End":"03:40.240","Text":"Now, if we look NaCl,"},{"Start":"03:40.240 ","End":"03:43.205","Text":"the process is slightly endothermic."},{"Start":"03:43.205 ","End":"03:48.320","Text":"Delta H of dissolution is plus 5 kilo joules per mole."},{"Start":"03:48.320 ","End":"03:55.010","Text":"But we know from previous work that endothermic processes can be spontaneous."},{"Start":"03:55.010 ","End":"04:03.980","Text":"That means Delta G can be negative provided the entropy change is sufficiently positive."},{"Start":"04:03.980 ","End":"04:06.200","Text":"Now, in this case,"},{"Start":"04:06.200 ","End":"04:11.915","Text":"we have an ordered crystal going into disordered solution."},{"Start":"04:11.915 ","End":"04:15.170","Text":"Obviously, Delta S is positive,"},{"Start":"04:15.170 ","End":"04:21.350","Text":"and it\u0027s sufficiently positive that Delta G becomes negative."},{"Start":"04:21.350 ","End":"04:26.490","Text":"In this video, we talked about dissolving ionic crystals."}],"ID":30937},{"Watched":false,"Name":"Solution Enthalpy and Like-Dissolves-Like Rule","Duration":"6m 21s","ChapterTopicVideoID":25625,"CourseChapterTopicPlaylistID":245374,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"In the previous video,"},{"Start":"00:02.040 ","End":"00:05.805","Text":"we talked about different ways to express solution concentrations."},{"Start":"00:05.805 ","End":"00:08.340","Text":"In this video, we\u0027ll learn about the enthalpy of"},{"Start":"00:08.340 ","End":"00:11.385","Text":"solution.and the like dissolves like rule."},{"Start":"00:11.385 ","End":"00:15.840","Text":"The question we\u0027re asking is how is the solution formed?"},{"Start":"00:15.840 ","End":"00:18.300","Text":"The solvent molecules have to separate."},{"Start":"00:18.300 ","End":"00:21.135","Text":"This is an endothermic process."},{"Start":"00:21.135 ","End":"00:24.255","Text":"Delta H solvent is positive."},{"Start":"00:24.255 ","End":"00:27.235","Text":"The solute molecules also have to separate."},{"Start":"00:27.235 ","End":"00:29.730","Text":"Again, this is an endothermic process."},{"Start":"00:29.730 ","End":"00:32.340","Text":"Delta H solute is positive."},{"Start":"00:32.340 ","End":"00:37.635","Text":"Then the solute molecules slip in-between the solvent molecules to form a solution."},{"Start":"00:37.635 ","End":"00:40.650","Text":"This is an exothermic process."},{"Start":"00:40.650 ","End":"00:43.610","Text":"Delta H interaction between the solvent,"},{"Start":"00:43.610 ","End":"00:48.095","Text":"and solute is negative, so that\u0027s exothermic."},{"Start":"00:48.095 ","End":"00:51.215","Text":"If we have our solvent that\u0027s red,"},{"Start":"00:51.215 ","End":"00:54.710","Text":"and the solute has to go in-between, thereof,"},{"Start":"00:54.710 ","End":"00:57.545","Text":"the solvent and solute are liquids,"},{"Start":"00:57.545 ","End":"00:59.465","Text":"then Delta H for the solvent,"},{"Start":"00:59.465 ","End":"01:04.040","Text":"or Delta H for the solute will be the same as Delta H vaporization."},{"Start":"01:04.040 ","End":"01:08.225","Text":"If the solvent and or solute are atomic or molecular solids,"},{"Start":"01:08.225 ","End":"01:14.705","Text":"the Delta H for the solvent or the solute will be like Delta H of sublimation."},{"Start":"01:14.705 ","End":"01:19.160","Text":"Now let\u0027s talk about the total enthalpy of solution."},{"Start":"01:19.160 ","End":"01:24.635","Text":"The enthalpy of solution or dissolution according to what you want to write,"},{"Start":"01:24.635 ","End":"01:27.080","Text":"is delta H of the solvent."},{"Start":"01:27.080 ","End":"01:30.243","Text":"First step, delta H of the solute,"},{"Start":"01:30.243 ","End":"01:33.620","Text":"second step, plus delta H of the introduction,"},{"Start":"01:33.620 ","End":"01:35.840","Text":"which is the solvent and solute."},{"Start":"01:35.840 ","End":"01:41.090","Text":"We have these 3 terms that add together to give delta H of solution."},{"Start":"01:41.090 ","End":"01:44.345","Text":"Now, if Delta H of solution is 0,"},{"Start":"01:44.345 ","End":"01:50.442","Text":"because the exothermic processes exactly balance the endothermic processes,"},{"Start":"01:50.442 ","End":"01:53.180","Text":"we say the solution is ideal."},{"Start":"01:53.180 ","End":"01:56.165","Text":"An example of this is benzene and toluene."},{"Start":"01:56.165 ","End":"02:00.470","Text":"Benzene, as you know, is C6H6."},{"Start":"02:00.470 ","End":"02:07.700","Text":"Then 1 of these hydrogens is replaced by CH3 to give toluene."},{"Start":"02:07.700 ","End":"02:09.118","Text":"This is benzene,"},{"Start":"02:09.118 ","End":"02:10.189","Text":"and this is toluene,"},{"Start":"02:10.189 ","End":"02:13.010","Text":"so they\u0027re very, very similar."},{"Start":"02:13.010 ","End":"02:16.070","Text":"They form an ideal solution."},{"Start":"02:16.070 ","End":"02:19.940","Text":"Now, when delta H is negative,"},{"Start":"02:19.940 ","End":"02:24.215","Text":"that\u0027s exothermic, the solution is non-ideal."},{"Start":"02:24.215 ","End":"02:28.865","Text":"An example, this is a solution of acetone and chloroform."},{"Start":"02:28.865 ","End":"02:33.420","Text":"Chloroform on its own has no hydrogen bonding."},{"Start":"02:33.420 ","End":"02:36.815","Text":"Acetone on its own has no hydrogen bonding."},{"Start":"02:36.815 ","End":"02:39.710","Text":"But, when the 2 are together in a solution,"},{"Start":"02:39.710 ","End":"02:42.040","Text":"there is hydrogen bonding between them."},{"Start":"02:42.040 ","End":"02:49.010","Text":"That means that delta H of the solvent and solute is larger,"},{"Start":"02:49.010 ","End":"02:55.025","Text":"is more negative than that of the solvent and solute."},{"Start":"02:55.025 ","End":"02:58.490","Text":"Here we have hydrogen bonding."},{"Start":"02:58.490 ","End":"03:03.410","Text":"Now, sometimes delta H solution is slightly positive, not very positive."},{"Start":"03:03.410 ","End":"03:05.990","Text":"Then again, the solution is non-ideal."},{"Start":"03:05.990 ","End":"03:09.875","Text":"An example of this is CS2 in acetone."},{"Start":"03:09.875 ","End":"03:16.000","Text":"Here, acetone has a dipole moment pointing in this direction."},{"Start":"03:16.000 ","End":"03:21.500","Text":"That induces a dipole moment in the non-polar CS2."},{"Start":"03:21.500 ","End":"03:23.675","Text":"This is delta minus,"},{"Start":"03:23.675 ","End":"03:27.125","Text":"delta plus, and this will be delta minus."},{"Start":"03:27.125 ","End":"03:31.775","Text":"Delta H solution is very positive, it\u0027s quite large."},{"Start":"03:31.775 ","End":"03:33.920","Text":"No solution at all is formed,"},{"Start":"03:33.920 ","End":"03:39.590","Text":"and we get just a heterogeneous mixture rather than a homogeneous mixture."},{"Start":"03:39.590 ","End":"03:42.604","Text":"Up to now, we\u0027ve already talked about enthalpy."},{"Start":"03:42.604 ","End":"03:45.865","Text":"But really we need Gibbs free energy of solution."},{"Start":"03:45.865 ","End":"03:50.675","Text":"Sometimes it happens that predictions using just entropy are wrong,"},{"Start":"03:50.675 ","End":"03:53.525","Text":"and then it\u0027s better to use the free energy of solution,"},{"Start":"03:53.525 ","End":"03:58.895","Text":"delta G solution equal to Delta H solution minus T Delta S solution."},{"Start":"03:58.895 ","End":"04:02.570","Text":"Now we\u0027re going to talk about the like dissolves like rule,"},{"Start":"04:02.570 ","End":"04:06.215","Text":"which my supervisor, when I was an undergraduate,"},{"Start":"04:06.215 ","End":"04:08.554","Text":"told me that was the art of chemistry,"},{"Start":"04:08.554 ","End":"04:14.195","Text":"finding the correct solvent to dissolve by organics crystals."},{"Start":"04:14.195 ","End":"04:18.110","Text":"Now, this is a widely used generalization."},{"Start":"04:18.110 ","End":"04:21.530","Text":"What it says is that when a solution is formed,"},{"Start":"04:21.530 ","End":"04:24.740","Text":"the solute-solvent interactions replace interactionally"},{"Start":"04:24.740 ","End":"04:28.940","Text":"the solid molecules and the solvent molecules, that we know already."},{"Start":"04:28.940 ","End":"04:32.360","Text":"We\u0027ve already said that. Like dissolves,"},{"Start":"04:32.360 ","End":"04:35.870","Text":"like means that a polar liquids such as water is"},{"Start":"04:35.870 ","End":"04:40.055","Text":"the best solvent for polar or ionic substance such as salt."},{"Start":"04:40.055 ","End":"04:43.864","Text":"Both are polar, both the solvent and the solute."},{"Start":"04:43.864 ","End":"04:47.315","Text":"Now if we have non-polar liquids like CS2,"},{"Start":"04:47.315 ","End":"04:51.725","Text":"these are the best solvents for non-polar solutes, such as sulfur."},{"Start":"04:51.725 ","End":"04:55.415","Text":"Here both are held together by London forces."},{"Start":"04:55.415 ","End":"04:57.365","Text":"Now, at each bonded substance,"},{"Start":"04:57.365 ","End":"05:01.850","Text":"a hydrogen bonded substance is most likely to dissolve at a hydrogen bonded solvent,"},{"Start":"05:01.850 ","End":"05:04.700","Text":"for example, glucose in water."},{"Start":"05:04.700 ","End":"05:08.570","Text":"Now, it\u0027s not the most accurate of rules,"},{"Start":"05:08.570 ","End":"05:10.145","Text":"but it\u0027s very, very useful."},{"Start":"05:10.145 ","End":"05:15.065","Text":"Now let\u0027s look for a moment at soaps or surfer actins,"},{"Start":"05:15.065 ","End":"05:17.705","Text":"that surface active agents."},{"Start":"05:17.705 ","End":"05:21.290","Text":"These consists of a hydrophilic polar head and"},{"Start":"05:21.290 ","End":"05:26.540","Text":"a hydrophobic nonpolar tail. Here\u0027s a picture."},{"Start":"05:26.540 ","End":"05:31.040","Text":"Here\u0027s the hydrophilic polar head."},{"Start":"05:31.040 ","End":"05:34.745","Text":"Here is the hydrophobic nonpolar tail."},{"Start":"05:34.745 ","End":"05:39.470","Text":"Supposing we have some dirt or some oil,"},{"Start":"05:39.470 ","End":"05:44.480","Text":"the nonpolar tails will point towards the dirt."},{"Start":"05:44.480 ","End":"05:50.525","Text":"The polar tails, the polar heads will point towards the water outside."},{"Start":"05:50.525 ","End":"05:52.858","Text":"A micelle is formed,"},{"Start":"05:52.858 ","End":"05:59.375","Text":"a small unit consisting of many of these soap or detergent molecules."},{"Start":"05:59.375 ","End":"06:03.770","Text":"These capture the piece of dirt or oil,"},{"Start":"06:03.770 ","End":"06:06.200","Text":"and then it\u0027s washed away."},{"Start":"06:06.200 ","End":"06:12.770","Text":"Here we have hydrophobic and hydrophilic at the same time,"},{"Start":"06:12.770 ","End":"06:14.360","Text":"in the same molecule."},{"Start":"06:14.360 ","End":"06:21.420","Text":"In this video, we talked about solution enthalpy and the like dissolves like rule."}],"ID":30938}],"Thumbnail":null,"ID":245374},{"Name":"Solution Formation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Solubility and Fractional Crystallization","Duration":"5m 8s","ChapterTopicVideoID":25626,"CourseChapterTopicPlaylistID":245375,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.550","Text":"In previous videos, we discussed"},{"Start":"00:02.550 ","End":"00:06.605","Text":"solution formation from molecular and thermodynamic perspectives."},{"Start":"00:06.605 ","End":"00:11.760","Text":"In this video, we\u0027ll discuss solution formation as observed in the lab."},{"Start":"00:11.760 ","End":"00:17.415","Text":"So what happens when we dissolve a solid solute in the liquid solvent?"},{"Start":"00:17.415 ","End":"00:20.810","Text":"At first, the solid dissolves in the liquid."},{"Start":"00:20.810 ","End":"00:23.915","Text":"Here is a solid dissolving in the liquid."},{"Start":"00:23.915 ","End":"00:26.914","Text":"These are these red arrows pointing upwards."},{"Start":"00:26.914 ","End":"00:29.510","Text":"Then the solution begins to crystallize,"},{"Start":"00:29.510 ","End":"00:32.435","Text":"that\u0027s these green arrows pointing downwards."},{"Start":"00:32.435 ","End":"00:35.000","Text":"Now, when the rates of dissolution"},{"Start":"00:35.000 ","End":"00:39.260","Text":"dissolving and the rates of crystallization are equal,"},{"Start":"00:39.260 ","End":"00:43.055","Text":"the solution reaches a state of dynamic equilibrium."},{"Start":"00:43.055 ","End":"00:46.760","Text":"The concentration of the solution is constant."},{"Start":"00:46.760 ","End":"00:50.720","Text":"We say the solution is saturated."},{"Start":"00:50.720 ","End":"00:53.995","Text":"Now, let\u0027s talk about solubility."},{"Start":"00:53.995 ","End":"00:59.495","Text":"Now, the concentration of the saturated solution is called the solubility,"},{"Start":"00:59.495 ","End":"01:04.220","Text":"and the solubility is different for each solute-solvent combination."},{"Start":"01:04.220 ","End":"01:06.845","Text":"Now, the units can be molarity,"},{"Start":"01:06.845 ","End":"01:09.470","Text":"grams of solid per 100 grams of solvent,"},{"Start":"01:09.470 ","End":"01:12.545","Text":"grams of solid per a 100 milliliters of solution,"},{"Start":"01:12.545 ","End":"01:16.055","Text":"mass percent, et cetera, all different units."},{"Start":"01:16.055 ","End":"01:21.410","Text":"So we have to be very careful to look at what units we\u0027re talking about."},{"Start":"01:21.410 ","End":"01:26.400","Text":"Now, let\u0027s consider solubility as a function of temperature."},{"Start":"01:26.450 ","End":"01:29.840","Text":"So we\u0027re talking about a graph of the solubility,"},{"Start":"01:29.840 ","End":"01:34.115","Text":"that\u0027s the concentration of a saturated solution versus temperature,"},{"Start":"01:34.115 ","End":"01:36.710","Text":"and that\u0027s called a solubility curve."},{"Start":"01:36.710 ","End":"01:42.750","Text":"So we have solubility on the y-axis and temperature on the x-axis."},{"Start":"01:42.760 ","End":"01:48.445","Text":"Now, usually, solubility increases with temperature."},{"Start":"01:48.445 ","End":"01:51.220","Text":"As we go to higher temperatures,"},{"Start":"01:51.220 ","End":"01:53.800","Text":"the solubility is increasing."},{"Start":"01:53.800 ","End":"01:57.805","Text":"However, some very hydrated ionic solids,"},{"Start":"01:57.805 ","End":"01:59.440","Text":"such as lithium carbonate,"},{"Start":"01:59.440 ","End":"02:02.545","Text":"are less soluble at high temperature than at low ones,"},{"Start":"02:02.545 ","End":"02:06.395","Text":"their solubility is decreasing,"},{"Start":"02:06.395 ","End":"02:09.760","Text":"and a few even show mixed behavior."},{"Start":"02:09.760 ","End":"02:13.240","Text":"For example, sodium sulfate hydrated with"},{"Start":"02:13.240 ","End":"02:17.770","Text":"10 water molecules has a maximum at 32 degrees Celsius."},{"Start":"02:17.770 ","End":"02:20.915","Text":"So we have a maximum."},{"Start":"02:20.915 ","End":"02:23.310","Text":"Know this isn\u0027t to scale,"},{"Start":"02:23.310 ","End":"02:26.250","Text":"it\u0027s just to indicate the direction."},{"Start":"02:26.250 ","End":"02:31.750","Text":"Here\u0027s decrease, increase, and a maximum."},{"Start":"02:31.750 ","End":"02:36.710","Text":"Now, let\u0027s consider a particular point on this red curve here."},{"Start":"02:36.710 ","End":"02:40.930","Text":"So that will be the solubility at this temperature."},{"Start":"02:40.930 ","End":"02:44.290","Text":"Now, if the concentration is lower than the solubility,"},{"Start":"02:44.290 ","End":"02:46.825","Text":"the solution is unsaturated."},{"Start":"02:46.825 ","End":"02:51.635","Text":"So we have a point down here that will be unsaturated."},{"Start":"02:51.635 ","End":"02:54.690","Text":"Now, if a saturated solution is cooled,"},{"Start":"02:54.690 ","End":"02:57.475","Text":"that means we\u0027re going to a lower point here,"},{"Start":"02:57.475 ","End":"03:02.830","Text":"the solubility will decrease so that less of the solute will"},{"Start":"03:02.830 ","End":"03:08.705","Text":"be dissolved and the excess solid will crystallize out of the solution."},{"Start":"03:08.705 ","End":"03:11.420","Text":"So the saturated solution is cooled,"},{"Start":"03:11.420 ","End":"03:17.885","Text":"the solubility decreases, so the excess solute crystallizes out of the solution."},{"Start":"03:17.885 ","End":"03:22.130","Text":"Now supposing the excess solid remains in the solution,"},{"Start":"03:22.130 ","End":"03:27.365","Text":"this can happen, we get a supersaturated solution."},{"Start":"03:27.365 ","End":"03:30.500","Text":"That means we\u0027ve gone to a lower temperature,"},{"Start":"03:30.500 ","End":"03:33.505","Text":"but the solubility has stayed the same,"},{"Start":"03:33.505 ","End":"03:36.035","Text":"so it\u0027s above here."},{"Start":"03:36.035 ","End":"03:37.850","Text":"It\u0027s above the red curve."},{"Start":"03:37.850 ","End":"03:41.210","Text":"That\u0027s called supersaturation."},{"Start":"03:46.520 ","End":"03:50.270","Text":"If the excess solute remains in solution,"},{"Start":"03:50.270 ","End":"03:52.895","Text":"we get a supersaturated solution."},{"Start":"03:52.895 ","End":"03:55.430","Text":"It won\u0027t stay like that generally,"},{"Start":"03:55.430 ","End":"04:01.430","Text":"it will crystallize if seeded with a few small crystals or even speck of dust."},{"Start":"04:01.430 ","End":"04:07.870","Text":"Now, let\u0027s talk about fractional crystallization or recrystallization."},{"Start":"04:07.870 ","End":"04:12.620","Text":"It\u0027s a method of purifying a substance synthesized in the lab,"},{"Start":"04:12.620 ","End":"04:17.765","Text":"and you\u0027ll meet many times in organic labs in particular."},{"Start":"04:17.765 ","End":"04:20.600","Text":"We prepare a concentrated solution of"},{"Start":"04:20.600 ","End":"04:25.159","Text":"an impure substance in the solvent at high temperature."},{"Start":"04:25.159 ","End":"04:31.025","Text":"Of course, finding the correct solvent is in itself sometimes rather difficult."},{"Start":"04:31.025 ","End":"04:35.705","Text":"Then you cool it generally by putting it in the fridge overnight."},{"Start":"04:35.705 ","End":"04:39.245","Text":"Often, the desired compound will crystallize,"},{"Start":"04:39.245 ","End":"04:42.624","Text":"leaving impurities in solution."},{"Start":"04:42.624 ","End":"04:45.445","Text":"Now, I just want to make a note."},{"Start":"04:45.445 ","End":"04:51.625","Text":"Now, in this video, we talked about the solubility of highly soluble substances,"},{"Start":"04:51.625 ","End":"04:53.480","Text":"and in the later chapter,"},{"Start":"04:53.480 ","End":"05:00.365","Text":"we\u0027ll talk about the solubility of ionic solids that are only slightly soluble in water."},{"Start":"05:00.365 ","End":"05:07.800","Text":"We\u0027ll talk about something called the solubility product, Ksp."}],"ID":30936}],"Thumbnail":null,"ID":245375},{"Name":"Solubility of Gases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Effect of Temperature and Pressure on Gas Solubility","Duration":"7m ","ChapterTopicVideoID":25627,"CourseChapterTopicPlaylistID":245376,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:02.640","Text":"In the previous video,"},{"Start":"00:02.640 ","End":"00:05.910","Text":"we talked about the solubility of solids in liquids."},{"Start":"00:05.910 ","End":"00:10.379","Text":"In this video, we\u0027ll talk about the solubility of gases in liquids."},{"Start":"00:10.379 ","End":"00:16.875","Text":"Let\u0027s begin by discussing the effect of temperature on the solubility of gases."},{"Start":"00:16.875 ","End":"00:21.300","Text":"The solubilities of most gases such as oxygen, nitrogen,"},{"Start":"00:21.300 ","End":"00:26.385","Text":"and carbon dioxide, and water decrease with increasing temperature."},{"Start":"00:26.385 ","End":"00:28.590","Text":"Here\u0027s the solubility curve,"},{"Start":"00:28.590 ","End":"00:35.175","Text":"and we can see that they decrease with increasing temperature,"},{"Start":"00:35.175 ","End":"00:39.480","Text":"the solubility decreases with increasing temperature."},{"Start":"00:39.480 ","End":"00:44.220","Text":"Some examples are that more air dissolves in cold water than hot water,"},{"Start":"00:44.220 ","End":"00:46.560","Text":"this is crucial for the life of fish,"},{"Start":"00:46.560 ","End":"00:52.235","Text":"and bubbles of air escape as water is heated well below its boiling point."},{"Start":"00:52.235 ","End":"00:56.090","Text":"We\u0027re all familiar with these small bubbles that escape."},{"Start":"00:56.090 ","End":"01:01.980","Text":"In organic solvents, solubility usually increases with temperature."},{"Start":"01:03.730 ","End":"01:06.735","Text":"When we talk about rare gases,"},{"Start":"01:06.735 ","End":"01:09.440","Text":"the solubility decreases with temperature,"},{"Start":"01:09.440 ","End":"01:12.755","Text":"reaches a minimum, and then increases."},{"Start":"01:12.755 ","End":"01:14.490","Text":"For example, for helium,"},{"Start":"01:14.490 ","End":"01:16.125","Text":"at 1 atm in water,"},{"Start":"01:16.125 ","End":"01:20.040","Text":"the minimum is at 35 degrees celsius,"},{"Start":"01:20.040 ","End":"01:25.135","Text":"so we have a decrease and then an increase."},{"Start":"01:25.135 ","End":"01:29.143","Text":"Now, let\u0027s talk about the effect of pressure."},{"Start":"01:29.143 ","End":"01:34.985","Text":"The solubility of a gas increases with increasing gas pressure above the solution,"},{"Start":"01:34.985 ","End":"01:38.900","Text":"and in the 18th, 19th century,"},{"Start":"01:38.900 ","End":"01:42.730","Text":"William Henry proposed Henry\u0027s law,"},{"Start":"01:42.730 ","End":"01:48.830","Text":"and that says that the solubility C is equal to k, a constant,"},{"Start":"01:48.830 ","End":"01:52.915","Text":"times the pressure of the gas above the solution,"},{"Start":"01:52.915 ","End":"01:57.470","Text":"so C is equal to k times pressure of the gas,"},{"Start":"01:57.470 ","End":"01:59.560","Text":"that\u0027s Henry\u0027s law,"},{"Start":"01:59.560 ","End":"02:02.520","Text":"and that\u0027s true at constant temperature."},{"Start":"02:02.520 ","End":"02:04.140","Text":"Here in this equation,"},{"Start":"02:04.140 ","End":"02:05.580","Text":"C is solubility,"},{"Start":"02:05.580 ","End":"02:10.070","Text":"k is a constant, and P gas is the pressure of the gas above the solution."},{"Start":"02:10.070 ","End":"02:11.510","Text":"Here\u0027s a picture,"},{"Start":"02:11.510 ","End":"02:13.555","Text":"here\u0027s our solution,"},{"Start":"02:13.555 ","End":"02:15.510","Text":"here\u0027s the gas,"},{"Start":"02:15.510 ","End":"02:20.215","Text":"and this is a piston being pressed down,"},{"Start":"02:20.215 ","End":"02:22.150","Text":"with a weight or something like that,"},{"Start":"02:22.150 ","End":"02:28.735","Text":"or force so we can increase or decrease the pressure of the gas above the liquid."},{"Start":"02:28.735 ","End":"02:30.260","Text":"Here\u0027s an example."},{"Start":"02:30.260 ","End":"02:32.060","Text":"When a bottle of soda is opened,"},{"Start":"02:32.060 ","End":"02:37.285","Text":"some carbon dioxide escapes so that P gas decreases,"},{"Start":"02:37.285 ","End":"02:40.580","Text":"and that leads to a decrease in the solubility,"},{"Start":"02:40.580 ","End":"02:43.985","Text":"so that the excess gas bubbles out of the solution,"},{"Start":"02:43.985 ","End":"02:46.480","Text":"that\u0027s the familiar fizz."},{"Start":"02:46.480 ","End":"02:48.600","Text":"If we look at the equation,"},{"Start":"02:48.600 ","End":"02:50.790","Text":"and want to find the constant, for example,"},{"Start":"02:50.790 ","End":"02:56.420","Text":"we could write the k is equal to the solubility divided by the pressure of the gas."},{"Start":"02:56.420 ","End":"03:00.370","Text":"If we want to compare the solubility at 2 different pressures,"},{"Start":"03:00.370 ","End":"03:03.040","Text":"P_1 and P_2,"},{"Start":"03:04.040 ","End":"03:11.595","Text":"we can write the k is equal to C_1 over P_1 which is also equal to C_2 over P_2."},{"Start":"03:11.595 ","End":"03:15.330","Text":"We can equate C_1 over P_1 and C_2 over P_2,"},{"Start":"03:15.330 ","End":"03:20.175","Text":"and write that C_2 is equal to C_1 times P_2 over P_1."},{"Start":"03:20.175 ","End":"03:23.100","Text":"Or if we want to reinterest the pressure,"},{"Start":"03:23.100 ","End":"03:28.200","Text":"we can write that P_2 is equal to P_1 times C_2 over C_1."},{"Start":"03:28.200 ","End":"03:31.040","Text":"In these equations, it doesn\u0027t matter what units are"},{"Start":"03:31.040 ","End":"03:34.595","Text":"used as long as the same units are used for P_1 and P_2,"},{"Start":"03:34.595 ","End":"03:37.760","Text":"and for C_1 and C_2 because we\u0027re talking about ratios of"},{"Start":"03:37.760 ","End":"03:41.120","Text":"P_1 and P_2 and ratios of C_1 and C_2."},{"Start":"03:41.120 ","End":"03:46.550","Text":"Strictly speaking, Henry\u0027s law is only valid when the system is in equilibrium,"},{"Start":"03:46.550 ","End":"03:49.550","Text":"the gas pressure is not extremely high,"},{"Start":"03:49.550 ","End":"03:53.330","Text":"and the gas and solvent do not react with each other."},{"Start":"03:53.330 ","End":"03:56.150","Text":"Let\u0027s solve an example."},{"Start":"03:56.150 ","End":"04:00.260","Text":"For a solution of oxygen in water at 25 degrees celsius,"},{"Start":"04:00.260 ","End":"04:06.590","Text":"k is equal to 1.3 times 10^ minus 3 moles per liter per atmosphere."},{"Start":"04:06.590 ","End":"04:09.935","Text":"If the pressure above the solution is 3 atmospheres,"},{"Start":"04:09.935 ","End":"04:15.505","Text":"how many milliliters of oxygen will dissolve in 1 liter of water?"},{"Start":"04:15.505 ","End":"04:18.315","Text":"We can work out the solubility,"},{"Start":"04:18.315 ","End":"04:20.220","Text":"when the pressure is 3 atmospheres."},{"Start":"04:20.220 ","End":"04:23.605","Text":"C is equal to k times the pressure of the gas,"},{"Start":"04:23.605 ","End":"04:27.490","Text":"that\u0027s 1.3 times 10 to the power minus 3 moles,"},{"Start":"04:27.490 ","End":"04:29.250","Text":"all of this is per liter of water,"},{"Start":"04:29.250 ","End":"04:33.770","Text":"because we have liters of oxygen as well in the problem,"},{"Start":"04:33.770 ","End":"04:41.420","Text":"so it\u0027s 1.3 times 10^minus 3 moles per liter of water per atmosphere times 3 atmosphere,"},{"Start":"04:41.420 ","End":"04:45.070","Text":"so atmosphere goes with atmosphere to power minus 1,"},{"Start":"04:45.070 ","End":"04:47.955","Text":"and we multiply it out,"},{"Start":"04:47.955 ","End":"04:53.145","Text":"3.9 times 10^minus 3 moles per liter of water."},{"Start":"04:53.145 ","End":"04:58.200","Text":"We can recall that the volume occupied by 1 mole of an ideal gas,"},{"Start":"04:58.200 ","End":"05:02.925","Text":"the pressure of 1 atmosphere is 22.4 liters per mole,"},{"Start":"05:02.925 ","End":"05:06.910","Text":"and if we recall the ideal gas law,"},{"Start":"05:06.910 ","End":"05:08.620","Text":"PV equals nRT,"},{"Start":"05:08.620 ","End":"05:10.895","Text":"and that gives us Boyle\u0027s law,"},{"Start":"05:10.895 ","End":"05:13.860","Text":"the P_2V_2 is equal to P_1V_1,"},{"Start":"05:13.860 ","End":"05:18.195","Text":"so V_2 is V_1 times P_1 over P_2,"},{"Start":"05:18.195 ","End":"05:23.415","Text":"and that\u0027s 22.4 liters times 1 atmosphere over 3 atmospheres,"},{"Start":"05:23.415 ","End":"05:27.030","Text":"because we know 22.4 liters relates"},{"Start":"05:27.030 ","End":"05:30.675","Text":"to 1 atmosphere and though we\u0027re interested in 3 atmospheres."},{"Start":"05:30.675 ","End":"05:32.325","Text":"If we work this out,"},{"Start":"05:32.325 ","End":"05:35.985","Text":"we get 7.47 liters."},{"Start":"05:35.985 ","End":"05:41.090","Text":"Again, we can, we go back to the ideal gas law, PV equals nRT,"},{"Start":"05:41.090 ","End":"05:45.845","Text":"we can see that V_2 over n_2 is equal to V_1 over n_1,"},{"Start":"05:45.845 ","End":"05:50.140","Text":"so that V_2 is equal to V_1 times n_2 over n_1,"},{"Start":"05:50.140 ","End":"05:53.540","Text":"and all of this is related to Avogadro\u0027s law."},{"Start":"05:53.540 ","End":"05:56.090","Text":"We can write V_2 is equal to V_1,"},{"Start":"05:56.090 ","End":"05:58.280","Text":"times n_2 over n_1."},{"Start":"05:58.280 ","End":"06:02.875","Text":"V_1 is 7.47 liters,"},{"Start":"06:02.875 ","End":"06:05.220","Text":"actually it\u0027s into 1 mole,"},{"Start":"06:05.220 ","End":"06:11.360","Text":"and we\u0027re interested in n_2 equal to 3.9 times 10^minus 3,"},{"Start":"06:11.360 ","End":"06:16.325","Text":"so it\u0027s 3.9 times 10^minus 3 moles per liter of water,"},{"Start":"06:16.325 ","End":"06:19.840","Text":"divided by n_1, 1 mole,"},{"Start":"06:19.840 ","End":"06:25.335","Text":"and that gives us 29.12 times 10^minus 3 liters,"},{"Start":"06:25.335 ","End":"06:30.480","Text":"that\u0027s liters of oxygen per liter of water."},{"Start":"06:30.480 ","End":"06:34.250","Text":"Since the problem asked for the answer in milliliters,"},{"Start":"06:34.250 ","End":"06:42.435","Text":"we can work out that that\u0027s 29.12 milliliters of oxygen per liter of water,"},{"Start":"06:42.435 ","End":"06:46.504","Text":"so that\u0027s the amount in milliliters of oxygen"},{"Start":"06:46.504 ","End":"06:51.720","Text":"that will dissolve in 1 liter of water at 3 atmospheres."},{"Start":"06:53.110 ","End":"06:56.900","Text":"In this video, we learned about the effect of temperature"},{"Start":"06:56.900 ","End":"07:00.720","Text":"and pressure and a gas dissolved in the liquid."}],"ID":30924}],"Thumbnail":null,"ID":245376},{"Name":"Colligative Properties","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Fractional Distillation of Ideal and Nonideal Solutions","Duration":"5m 27s","ChapterTopicVideoID":25628,"CourseChapterTopicPlaylistID":245377,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.680","Text":"In the previous video,"},{"Start":"00:01.680 ","End":"00:05.280","Text":"we discussed the liquid vapor equilibrium variety of solutions."},{"Start":"00:05.280 ","End":"00:09.870","Text":"In this video, we\u0027ll apply the results to fractional distillation."},{"Start":"00:09.870 ","End":"00:14.100","Text":"We\u0027re going to talk about fractional distillation."},{"Start":"00:14.100 ","End":"00:19.950","Text":"Fractional distillation is a method of separating volatile liquids."},{"Start":"00:19.950 ","End":"00:22.785","Text":"For example, alcohol and water,"},{"Start":"00:22.785 ","End":"00:26.760","Text":"or different volatile components of petroleum."},{"Start":"00:26.760 ","End":"00:32.189","Text":"I\u0027m going to plot the normal boiling temperature versus a mole fraction."},{"Start":"00:32.189 ","End":"00:37.030","Text":"For example, for a solution of benzene and toluene."},{"Start":"00:37.100 ","End":"00:42.230","Text":"Here\u0027s the normal boiling point versus the mole fraction of benzene."},{"Start":"00:42.230 ","End":"00:45.930","Text":"The red curve is for the solution."},{"Start":"00:46.420 ","End":"00:50.360","Text":"The green curve is for the vapor."},{"Start":"00:50.360 ","End":"00:55.460","Text":"Now, note that the vapor curve is above the solution curve."},{"Start":"00:55.460 ","End":"01:03.710","Text":"That\u0027s because the vapor has a higher fraction of benzene in it than does the solution."},{"Start":"01:03.710 ","End":"01:10.334","Text":"Remember, when we had 0.5 mole fraction of benzene in the solution,"},{"Start":"01:10.334 ","End":"01:14.630","Text":"in the vapor, we had 0.77."},{"Start":"01:15.800 ","End":"01:21.580","Text":"If we have a solution with this particular mole fraction of benzene,"},{"Start":"01:21.580 ","End":"01:23.905","Text":"is 0.3 in this case,"},{"Start":"01:23.905 ","End":"01:26.105","Text":"and we heat it up,"},{"Start":"01:26.105 ","End":"01:32.795","Text":"then the vapor will have a higher mole fraction of benzene."},{"Start":"01:32.795 ","End":"01:38.255","Text":"Then we can cool it and get a solution with a higher mole fraction of benzene,"},{"Start":"01:38.255 ","End":"01:40.025","Text":"and if again,"},{"Start":"01:40.025 ","End":"01:42.290","Text":"we take off the vapor heated,"},{"Start":"01:42.290 ","End":"01:43.865","Text":"so the vapor comes off,"},{"Start":"01:43.865 ","End":"01:48.515","Text":"the vapor will have an even higher mole fraction of benzene."},{"Start":"01:48.515 ","End":"01:56.180","Text":"This is a way, we could distill solution of benzene and toluene and get pure benzene."},{"Start":"01:56.180 ","End":"02:00.540","Text":"So we\u0027re going to end up with pure benzene."},{"Start":"02:01.910 ","End":"02:04.860","Text":"This is not to scale."},{"Start":"02:04.860 ","End":"02:06.560","Text":"This is just a summary."},{"Start":"02:06.560 ","End":"02:07.760","Text":"We heat the solution,"},{"Start":"02:07.760 ","End":"02:09.679","Text":"the vapor is rich in benzene,"},{"Start":"02:09.679 ","End":"02:13.519","Text":"we cool and heat again until pure benzene is obtained."},{"Start":"02:13.519 ","End":"02:16.445","Text":"Now, if we have several volatile liquids,"},{"Start":"02:16.445 ","End":"02:20.270","Text":"the one with the lowest boiling part will vaporize first,"},{"Start":"02:20.270 ","End":"02:24.590","Text":"and then we can do it again and get off the second one and so on."},{"Start":"02:24.590 ","End":"02:30.590","Text":"Here\u0027s a very approximate picture of what\u0027s going on."},{"Start":"02:30.590 ","End":"02:37.170","Text":"Here, we have the solution which is being heated,"},{"Start":"02:37.170 ","End":"02:38.410","Text":"then we have a column,"},{"Start":"02:38.410 ","End":"02:40.490","Text":"it\u0027s called a fractionating column."},{"Start":"02:40.490 ","End":"02:44.510","Text":"It usually has glass beads or something like that inside,"},{"Start":"02:44.510 ","End":"02:45.770","Text":"and it can be made very,"},{"Start":"02:45.770 ","End":"02:52.100","Text":"very high depending on what we\u0027re trying to separate."},{"Start":"02:52.100 ","End":"02:54.970","Text":"We\u0027re heating the solution,"},{"Start":"02:54.970 ","End":"02:57.030","Text":"forming a vapor,"},{"Start":"02:57.030 ","End":"03:00.500","Text":"the vapor then goes through this tube,"},{"Start":"03:00.500 ","End":"03:02.390","Text":"which is called a condenser,"},{"Start":"03:02.390 ","End":"03:07.250","Text":"and we flow water or something like that through it so that it\u0027s cooled."},{"Start":"03:07.250 ","End":"03:10.790","Text":"Then the vapor becomes a solution again"},{"Start":"03:10.790 ","End":"03:14.870","Text":"or becomes a liquid again and drops into a vessel."},{"Start":"03:14.870 ","End":"03:18.515","Text":"Now, up to now, we\u0027ve talked about ideal solutions."},{"Start":"03:18.515 ","End":"03:22.490","Text":"Now, what happens when we have a non-ideal solution?"},{"Start":"03:22.490 ","End":"03:24.560","Text":"Now, in a non-ideal solution,"},{"Start":"03:24.560 ","End":"03:29.425","Text":"the vapor pressure can be higher or lower than expected for an ideal solution."},{"Start":"03:29.425 ","End":"03:32.825","Text":"For example, acetone chloroform solution,"},{"Start":"03:32.825 ","End":"03:34.790","Text":"the attraction between the different molecules is"},{"Start":"03:34.790 ","End":"03:37.585","Text":"greater than the attraction between the same molecules,"},{"Start":"03:37.585 ","End":"03:42.664","Text":"so the Delta H of solution is negative, it\u0027s exothermic process."},{"Start":"03:42.664 ","End":"03:45.050","Text":"The vapor pressure is lower,"},{"Start":"03:45.050 ","End":"03:49.430","Text":"and the boiling point higher than an ideal solution."},{"Start":"03:49.430 ","End":"03:54.650","Text":"The opposite is true of acetone carbon disulfide solution."},{"Start":"03:54.650 ","End":"03:57.350","Text":"Because here, the attraction between the different molecules"},{"Start":"03:57.350 ","End":"04:00.050","Text":"is less than the attraction between the same molecules,"},{"Start":"04:00.050 ","End":"04:02.750","Text":"so Delta H of solution is positive,"},{"Start":"04:02.750 ","End":"04:05.690","Text":"it\u0027s an endothermic process."},{"Start":"04:05.690 ","End":"04:08.160","Text":"The vapor pressure is higher,"},{"Start":"04:08.160 ","End":"04:12.170","Text":"and the boiling point lower than a non-ideal solution."},{"Start":"04:12.170 ","End":"04:15.770","Text":"Now, we\u0027re going to talk about an azeotrope."},{"Start":"04:15.770 ","End":"04:18.440","Text":"Sometimes, there\u0027s a maximum or minimum,"},{"Start":"04:18.440 ","End":"04:22.490","Text":"the liquid vapor equilibrium that we saw in a previous video."},{"Start":"04:22.490 ","End":"04:29.215","Text":"That can lead to a minimum or maximum the boiling point versus composition graph."},{"Start":"04:29.215 ","End":"04:33.605","Text":"Examples are water and propanol or water and ethanol."},{"Start":"04:33.605 ","End":"04:38.120","Text":"Supposing we have propanol in a propanol-water solution."},{"Start":"04:38.120 ","End":"04:44.660","Text":"Then the minimum is at 71.7% by mass."},{"Start":"04:44.660 ","End":"04:47.560","Text":"That means at this minimum,"},{"Start":"04:47.560 ","End":"04:53.525","Text":"the composition of the solution and the vapor are identical."},{"Start":"04:53.525 ","End":"04:58.730","Text":"We can\u0027t make it more pure than this point by distillation."},{"Start":"04:58.730 ","End":"05:02.270","Text":"If this were ethanol-water solution,"},{"Start":"05:02.270 ","End":"05:05.015","Text":"this would be 96%."},{"Start":"05:05.015 ","End":"05:07.654","Text":"That\u0027s why when you buy alcohol,"},{"Start":"05:07.654 ","End":"05:11.960","Text":"you don\u0027t get better than 96% unless it\u0027s performed by"},{"Start":"05:11.960 ","End":"05:17.780","Text":"some other method for laboratory purposes or medicinal purposes."},{"Start":"05:17.780 ","End":"05:20.420","Text":"But normally, just by distillation,"},{"Start":"05:20.420 ","End":"05:23.090","Text":"you can\u0027t get better than 96%."},{"Start":"05:23.090 ","End":"05:27.840","Text":"In this video, we learned about fractional distillation."}],"ID":30939},{"Watched":false,"Name":"Freezing-Point Depression and Boiling-Point Elevation","Duration":"7m 41s","ChapterTopicVideoID":25629,"CourseChapterTopicPlaylistID":245377,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.310","Text":"In the previous video,"},{"Start":"00:02.310 ","End":"00:06.420","Text":"we talked about lowering of the vapor pressure of the solvent and solutions."},{"Start":"00:06.420 ","End":"00:08.640","Text":"In this video, we\u0027ll talk about properties that"},{"Start":"00:08.640 ","End":"00:12.400","Text":"result from this lowering of the vapor pressure."},{"Start":"00:12.400 ","End":"00:17.930","Text":"We\u0027re going to talk about Freezing Point Depression and Boiling Point Elevation,"},{"Start":"00:17.930 ","End":"00:21.540","Text":"these are 2 of the colligative properties."},{"Start":"00:25.870 ","End":"00:29.555","Text":"Here we have a phase diagram."},{"Start":"00:29.555 ","End":"00:34.830","Text":"The red curve is the phase diagram for a pure solvent."},{"Start":"00:36.230 ","End":"00:43.970","Text":"The blue for a solution is solute dissolved in a solvent,"},{"Start":"00:43.970 ","End":"00:46.380","Text":"this is a solution."},{"Start":"00:47.770 ","End":"00:52.265","Text":"If you recall, here we have a solid,"},{"Start":"00:52.265 ","End":"00:57.810","Text":"liquid, and vapor."},{"Start":"00:57.960 ","End":"01:00.400","Text":"Now in the previous video,"},{"Start":"01:00.400 ","End":"01:05.750","Text":"we saw that the vapor pressure of the solution is lower than that of the solvent,"},{"Start":"01:05.750 ","End":"01:09.865","Text":"so this blue curve is lower than the red curve."},{"Start":"01:09.865 ","End":"01:15.425","Text":"Then it meets the sublimation curve at a lower point,"},{"Start":"01:15.425 ","End":"01:21.339","Text":"so the solid liquid curve is the same as it is for the pure solvent,"},{"Start":"01:21.339 ","End":"01:23.905","Text":"but it\u0027s at lower temperatures."},{"Start":"01:23.905 ","End":"01:26.410","Text":"We have blue and we have red,"},{"Start":"01:26.410 ","End":"01:27.760","Text":"red for the solvent,"},{"Start":"01:27.760 ","End":"01:29.770","Text":"blue for the solution."},{"Start":"01:29.770 ","End":"01:34.780","Text":"Now in order to get the boiling point and the freezing point,"},{"Start":"01:34.780 ","End":"01:38.045","Text":"we draw a line at 1 atmosphere."},{"Start":"01:38.045 ","End":"01:42.050","Text":"We see that as far as the freezing point is concerned,"},{"Start":"01:42.050 ","End":"01:46.370","Text":"it beats the blue curve for the solution at"},{"Start":"01:46.370 ","End":"01:51.320","Text":"a lower temperature than it meets the red curve for the solvent."},{"Start":"01:51.320 ","End":"01:53.950","Text":"This is depression,"},{"Start":"01:53.950 ","End":"01:56.255","Text":"blue is lower than red,"},{"Start":"01:56.255 ","End":"02:01.505","Text":"so the freezing point is depressed, it\u0027s lowered."},{"Start":"02:01.505 ","End":"02:04.250","Text":"Then if we look at the boiling points,"},{"Start":"02:04.250 ","End":"02:07.160","Text":"we see that the boiling point of the solution"},{"Start":"02:07.160 ","End":"02:10.430","Text":"is higher than the boiling point of the solvent,"},{"Start":"02:10.430 ","End":"02:13.115","Text":"so this boiling point elevation."},{"Start":"02:13.115 ","End":"02:19.900","Text":"This is elevation and this is depression."},{"Start":"02:20.080 ","End":"02:24.725","Text":"Now, this is all summarized in the next few sentences."},{"Start":"02:24.725 ","End":"02:28.205","Text":"We\u0027re at the phase diagram for the pure solvent in red,"},{"Start":"02:28.205 ","End":"02:30.860","Text":"and for the solution in blue."},{"Start":"02:30.860 ","End":"02:36.695","Text":"Now, the vapor pressure in solution is lower than that of the solvent."},{"Start":"02:36.695 ","End":"02:39.215","Text":"In the fusion curve,"},{"Start":"02:39.215 ","End":"02:43.745","Text":"the solid to liquid curve moves to lower temperature for the solution,"},{"Start":"02:43.745 ","End":"02:45.620","Text":"here we have it as lower."},{"Start":"02:45.620 ","End":"02:49.415","Text":"Now we have the normal freezing point, that\u0027s the freezing point."},{"Start":"02:49.415 ","End":"02:57.035","Text":"1 atmosphere is lowered or depressed and the normal boiling point is raised or elevated."},{"Start":"02:57.035 ","End":"03:01.460","Text":"Now both of these are colligative properties we said before."},{"Start":"03:01.460 ","End":"03:03.845","Text":"Now for very dilute solutions,"},{"Start":"03:03.845 ","End":"03:05.300","Text":"freezing point depression and"},{"Start":"03:05.300 ","End":"03:09.470","Text":"boiling point elevation are written by the following equations."},{"Start":"03:09.470 ","End":"03:15.530","Text":"Delta T_f, that\u0027s the depression of freezing point is equal to minus K_f,"},{"Start":"03:15.530 ","End":"03:19.715","Text":"which is a constant depending on the solvent,"},{"Start":"03:19.715 ","End":"03:24.630","Text":"times m, the molality of the solution."},{"Start":"03:24.630 ","End":"03:29.730","Text":"Delta T_b the elevation of the boiling point is K_b,"},{"Start":"03:29.730 ","End":"03:37.090","Text":"and other constant for the boiling point for a particular solvent times m the molality."},{"Start":"03:37.090 ","End":"03:39.630","Text":"K_f and K_b are constants,"},{"Start":"03:39.630 ","End":"03:41.310","Text":"and here are some typical values."},{"Start":"03:41.310 ","End":"03:43.815","Text":"K_f is 1.86,"},{"Start":"03:43.815 ","End":"03:46.190","Text":"and K_b is 0.512,"},{"Start":"03:46.190 ","End":"03:51.405","Text":"and the units are degrees Celsius per mole times kilogram."},{"Start":"03:51.405 ","End":"03:54.045","Text":"These values are for water."},{"Start":"03:54.045 ","End":"03:56.835","Text":"m is the molality."},{"Start":"03:56.835 ","End":"04:00.650","Text":"Now, we can use these changes in boiling and freezing points"},{"Start":"04:00.650 ","End":"04:04.310","Text":"that we can measure in the lab to calculate the molality,"},{"Start":"04:04.310 ","End":"04:07.025","Text":"and from the molality, the molar mass."},{"Start":"04:07.025 ","End":"04:09.950","Text":"It turns out the freezing is better than boiling"},{"Start":"04:09.950 ","End":"04:14.945","Text":"because the K_f is larger typically than K_b,"},{"Start":"04:14.945 ","End":"04:18.335","Text":"so freezing is better than boiling."},{"Start":"04:18.335 ","End":"04:23.105","Text":"It\u0027s a bit difficult to do because we have to measure"},{"Start":"04:23.105 ","End":"04:27.695","Text":"the difference in temperature for the solution and solvent very accurately."},{"Start":"04:27.695 ","End":"04:31.745","Text":"That changes with the atmospheric pressure, which also varies."},{"Start":"04:31.745 ","End":"04:36.410","Text":"It\u0027s not as accurate as another method we\u0027ll learn about in the next video,"},{"Start":"04:36.410 ","End":"04:40.460","Text":"which is osmotic pressure."},{"Start":"04:40.460 ","End":"04:47.000","Text":"Now, these expressions are only good for very dilute solutions of non-electrolytes,"},{"Start":"04:47.000 ","End":"04:51.005","Text":"and we\u0027ll correct them in a later video for electrolytes."},{"Start":"04:51.005 ","End":"04:56.130","Text":"It\u0027s best to use a solvent with a very high volume of K_f."},{"Start":"04:56.630 ","End":"04:59.380","Text":"Here\u0027s an example."},{"Start":"04:59.380 ","End":"05:04.025","Text":"Calculate the molality of a solution of an unknown solute in water,"},{"Start":"05:04.025 ","End":"05:07.885","Text":"if the solution freezes at -2 degrees Celsius."},{"Start":"05:07.885 ","End":"05:14.000","Text":"If 2 grams of the solute is dissolved in 0.05 kilograms of water,"},{"Start":"05:14.000 ","End":"05:16.550","Text":"what is its molar mass?"},{"Start":"05:16.550 ","End":"05:19.340","Text":"Delta T_f, we have to work out."},{"Start":"05:19.340 ","End":"05:24.470","Text":"We saw that it freezes at -2 degrees Celsius for the solution,"},{"Start":"05:24.470 ","End":"05:30.090","Text":"so this is solution minus solvent."},{"Start":"05:31.160 ","End":"05:34.035","Text":"It\u0027s -2 - 0,"},{"Start":"05:34.035 ","End":"05:38.025","Text":"that gives us -2 degrees Celsius."},{"Start":"05:38.025 ","End":"05:47.700","Text":"Now, delta T_f is equal to -K_f times m. m is equal to minus delta T_f divided by K_f."},{"Start":"05:47.700 ","End":"05:50.150","Text":"Delta T_f has a minus sign."},{"Start":"05:50.150 ","End":"05:51.850","Text":"There\u0027s another minus sign here,"},{"Start":"05:51.850 ","End":"05:54.545","Text":"so altogether it\u0027s only plus."},{"Start":"05:54.545 ","End":"06:03.095","Text":"We have 2 degrees Celsius for Delta T_f divided by 1.86 degrees Celsius per mole."},{"Start":"06:03.095 ","End":"06:06.575","Text":"This is multiplied by kilogram of water."},{"Start":"06:06.575 ","End":"06:08.540","Text":"That when we divide it,"},{"Start":"06:08.540 ","End":"06:10.280","Text":"the degrees Celsius goes,"},{"Start":"06:10.280 ","End":"06:16.670","Text":"we get 1.08 moles of the solute per kilogram of water."},{"Start":"06:16.670 ","End":"06:20.720","Text":"Often people don\u0027t write moles of solute per kilograms of water."},{"Start":"06:20.720 ","End":"06:22.610","Text":"They don\u0027t write the solute in the water,"},{"Start":"06:22.610 ","End":"06:23.780","Text":"but here I\u0027ve written it,"},{"Start":"06:23.780 ","End":"06:25.430","Text":"so this is perfectly clear."},{"Start":"06:25.430 ","End":"06:33.455","Text":"Now, if we have the molality and that\u0027s 1.08 moles of solute per kilogram of water,"},{"Start":"06:33.455 ","End":"06:39.290","Text":"we can calculate the number of moles because the molality is number of"},{"Start":"06:39.290 ","End":"06:45.970","Text":"moles of solute divided by the mass in kilograms of the solvent."},{"Start":"06:45.970 ","End":"06:51.520","Text":"If we notice, this is 1.08 and it\u0027s n divided by 0.05."},{"Start":"06:51.520 ","End":"06:59.030","Text":"We can work out n. n is equal to 0.054 moles of solute."},{"Start":"06:59.030 ","End":"07:01.955","Text":"Once we have the number of moles,"},{"Start":"07:01.955 ","End":"07:04.450","Text":"that\u0027s 0.05 moles of solute,"},{"Start":"07:04.450 ","End":"07:09.055","Text":"we know that\u0027s the mass of the solute divided by the molar mass."},{"Start":"07:09.055 ","End":"07:16.775","Text":"The molar mass is the mass of the solute divided by the number of moles."},{"Start":"07:16.775 ","End":"07:22.380","Text":"The mass is 2 grams and the number of moles is 0.054."},{"Start":"07:22.380 ","End":"07:28.690","Text":"We divide that, we get 37.0 grams per mole."},{"Start":"07:28.690 ","End":"07:33.830","Text":"The molar mass is 37 grams per mole."},{"Start":"07:33.830 ","End":"07:41.160","Text":"In this video, we learned about freezing point depression and boiling point elevation."}],"ID":30940},{"Watched":false,"Name":"Liquid-Vapor Equilibrium in Ideal Solutions","Duration":"6m 27s","ChapterTopicVideoID":25630,"CourseChapterTopicPlaylistID":245377,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.470","Text":"In the previous video,"},{"Start":"00:01.470 ","End":"00:05.670","Text":"we discussed the lowering of the vapor pressure of a solvent in a solution."},{"Start":"00:05.670 ","End":"00:11.880","Text":"In this video, we\u0027ll discuss the liquid-vapor equilibrium of ideal solutions."},{"Start":"00:11.880 ","End":"00:15.995","Text":"Let\u0027s first look at the vapor pressure of an ideal solution."},{"Start":"00:15.995 ","End":"00:21.215","Text":"Remember an ideal solution is one the Delta H solution equal to 0."},{"Start":"00:21.215 ","End":"00:24.800","Text":"Take the example of benzene and toluene."},{"Start":"00:24.800 ","End":"00:29.030","Text":"What we saw in the previous video was that"},{"Start":"00:29.030 ","End":"00:30.950","Text":"if the mole fraction of benzene is"},{"Start":"00:30.950 ","End":"00:34.415","Text":"equal to the mole fraction of toluene, they\u0027re both 0.5."},{"Start":"00:34.415 ","End":"00:39.960","Text":"We saw that the vapor pressure of the vapor above"},{"Start":"00:39.960 ","End":"00:48.419","Text":"the solution of benzene and toluene is 47.6 millimeters of mercury."},{"Start":"00:49.270 ","End":"00:58.655","Text":"The vapor pressure of toluene above the solution is 14.2 millimeters of mercury."},{"Start":"00:58.655 ","End":"01:05.179","Text":"The total is 61.8 millimeters of mercury."},{"Start":"01:05.340 ","End":"01:07.540","Text":"Now before we continue,"},{"Start":"01:07.540 ","End":"01:11.155","Text":"we need to note that benzene is more volatile than toluene."},{"Start":"01:11.155 ","End":"01:13.825","Text":"It is a higher vapor pressure."},{"Start":"01:13.825 ","End":"01:17.755","Text":"Let us look at the composition of the vapor."},{"Start":"01:17.755 ","End":"01:21.340","Text":"To do this, we need to record Dalton\u0027s law of"},{"Start":"01:21.340 ","End":"01:25.810","Text":"partial pressures we met when we talked about gases."},{"Start":"01:25.810 ","End":"01:31.330","Text":"The pressure of benzene is equal to the vapor pressure of benzene."},{"Start":"01:31.330 ","End":"01:37.265","Text":"We\u0027re talking about vapor pressure of benzene in the vapor above the solution."},{"Start":"01:37.265 ","End":"01:45.940","Text":"The vapor is equal to Chi benzene times P_total vapor pressure."},{"Start":"01:45.940 ","End":"01:47.930","Text":"We can work out from that,"},{"Start":"01:47.930 ","End":"01:53.250","Text":"that Chi of benzene is equal to P of benzene,"},{"Start":"01:53.250 ","End":"01:55.546","Text":"vapor pressure of the benzene,"},{"Start":"01:55.546 ","End":"01:57.740","Text":"divided by the total vapor pressure."},{"Start":"01:57.740 ","End":"02:00.800","Text":"That\u0027s 47.6 millimeters of mercury,"},{"Start":"02:00.800 ","End":"02:02.495","Text":"that we calculated before,"},{"Start":"02:02.495 ","End":"02:06.088","Text":"divide by 61.8 millimeters of mercury,"},{"Start":"02:06.088 ","End":"02:10.195","Text":"and that gives us 0.77."},{"Start":"02:10.195 ","End":"02:15.050","Text":"Instead of 0.5 inside the solution,"},{"Start":"02:15.050 ","End":"02:20.685","Text":"we go up to 0.77 in the vapor above the solution."},{"Start":"02:20.685 ","End":"02:22.975","Text":"Now let\u0027s look at toluene."},{"Start":"02:22.975 ","End":"02:24.541","Text":"P of toluene,"},{"Start":"02:24.541 ","End":"02:26.330","Text":"the pressure of toluene,"},{"Start":"02:26.330 ","End":"02:28.455","Text":"is equal to Chi of toluene."},{"Start":"02:28.455 ","End":"02:32.870","Text":"Now we\u0027re talking about the amount that\u0027s in the vapor,"},{"Start":"02:32.870 ","End":"02:37.105","Text":"the mole fraction in the vapor times P_total,"},{"Start":"02:37.105 ","End":"02:41.380","Text":"so Chi toluene is equal to P toluene divided by P_total."},{"Start":"02:41.380 ","End":"02:44.630","Text":"That\u0027s 14.2, that we calculated before,"},{"Start":"02:44.630 ","End":"02:45.875","Text":"millimeters of mercury,"},{"Start":"02:45.875 ","End":"02:49.475","Text":"divided by 61.8 millimeters of mercury."},{"Start":"02:49.475 ","End":"02:55.450","Text":"If we work that out, we get 0.23."},{"Start":"02:55.450 ","End":"02:59.210","Text":"Of course, we could simply have subtracted"},{"Start":"02:59.210 ","End":"03:03.965","Text":"0.77 for the benzene from the total, which is 1."},{"Start":"03:03.965 ","End":"03:06.240","Text":"1 minus 0.77,"},{"Start":"03:06.240 ","End":"03:08.950","Text":"of course, is just 0.23."},{"Start":"03:08.950 ","End":"03:17.810","Text":"We\u0027ve gone from 0.5 in the solution to 0.23 in the vapor above the solution."},{"Start":"03:17.810 ","End":"03:25.595","Text":"Now we can see that the vapor is richer in benzene than it is in toluene."},{"Start":"03:25.595 ","End":"03:31.630","Text":"Vapor is richer in the more volatile component, which is benzene."},{"Start":"03:31.630 ","End":"03:39.695","Text":"We\u0027ll see in a video after this that this is the basis of fractional distillation."},{"Start":"03:39.695 ","End":"03:47.035","Text":"This is a basis which I can use to distill a mixture of benzene and toluene."},{"Start":"03:47.035 ","End":"03:54.658","Text":"Now we\u0027re going to draw a figure which explains all this or what we\u0027ve calculated above,"},{"Start":"03:54.658 ","End":"03:58.460","Text":"and we call it the liquid-vapor equilibrium."},{"Start":"03:58.460 ","End":"04:04.175","Text":"We\u0027re plotting the vapor pressure versus the mole fraction of benzene."},{"Start":"04:04.175 ","End":"04:07.055","Text":"When the mole fraction of benzene is 0,"},{"Start":"04:07.055 ","End":"04:10.800","Text":"that means the solution is toluene."},{"Start":"04:10.870 ","End":"04:14.180","Text":"When the mole fraction of benzene is 1,"},{"Start":"04:14.180 ","End":"04:17.960","Text":"it means that it\u0027s all benzene."},{"Start":"04:17.960 ","End":"04:21.400","Text":"The first curve, which is in blue,"},{"Start":"04:21.400 ","End":"04:24.985","Text":"is the vapor pressure of benzene in solution."},{"Start":"04:24.985 ","End":"04:29.894","Text":"Here we have the vapor pressure versus the mole fraction of benzene."},{"Start":"04:29.894 ","End":"04:34.300","Text":"Starts at 0, goes up to the volume for pure benzene,"},{"Start":"04:34.300 ","End":"04:38.500","Text":"which is 95.1 millimeters of mercury."},{"Start":"04:38.500 ","End":"04:42.310","Text":"Then we can do the same sort of graph for toluene."},{"Start":"04:42.310 ","End":"04:48.310","Text":"It starts at 0 here and goes up to the volume for pure toluene,"},{"Start":"04:48.310 ","End":"04:52.075","Text":"which is 28.4 millimeters of mercury."},{"Start":"04:52.075 ","End":"04:56.245","Text":"The blue curve is for benzene,"},{"Start":"04:56.245 ","End":"05:00.600","Text":"straight line, and the red one for toluene."},{"Start":"05:00.600 ","End":"05:03.320","Text":"Now these are in solution."},{"Start":"05:03.320 ","End":"05:05.935","Text":"Now we can work out the total,"},{"Start":"05:05.935 ","End":"05:08.664","Text":"and here\u0027s that the green line is the total,"},{"Start":"05:08.664 ","End":"05:11.490","Text":"the total of benzene and toluene."},{"Start":"05:11.490 ","End":"05:14.050","Text":"Here\u0027s the green line."},{"Start":"05:14.100 ","End":"05:18.760","Text":"Now let\u0027s look at the point we calculated at the beginning."},{"Start":"05:18.760 ","End":"05:22.690","Text":"Chi is equal to 0.5 for both benzene and toluene."},{"Start":"05:22.690 ","End":"05:25.795","Text":"Here\u0027s this five-pointed star."},{"Start":"05:25.795 ","End":"05:30.325","Text":"Now let\u0027s look at the vapor above the solution."},{"Start":"05:30.325 ","End":"05:33.160","Text":"Now with the same pressure,"},{"Start":"05:33.160 ","End":"05:41.130","Text":"so we can draw a line along that\u0027s called a tie-line."},{"Start":"05:41.130 ","End":"05:45.920","Text":"Here we\u0027re drawing the line to the composition of the vapor."},{"Start":"05:45.920 ","End":"05:55.355","Text":"Remember the composition of vapor was 0.77 mole fraction of benzene and 0.23 of toluene."},{"Start":"05:55.355 ","End":"05:58.360","Text":"Here is the tie-line."},{"Start":"05:58.360 ","End":"06:04.580","Text":"This orange curve connects all the points for the vapor."},{"Start":"06:04.580 ","End":"06:09.680","Text":"Each time I can calculate for different amount of toluene and benzene,"},{"Start":"06:09.680 ","End":"06:11.402","Text":"gets a particularly value,"},{"Start":"06:11.402 ","End":"06:16.395","Text":"draw across another tie-line to the composition of the vapor,"},{"Start":"06:16.395 ","End":"06:18.320","Text":"and I get this curve."},{"Start":"06:18.320 ","End":"06:23.765","Text":"In this video, we discussed the liquid-vapor equilibrium of ideal solutions."},{"Start":"06:23.765 ","End":"06:28.060","Text":"We\u0027ll use this for distillation."}],"ID":30941},{"Watched":false,"Name":"Osmosis and Osmotic Pressure","Duration":"5m 55s","ChapterTopicVideoID":25631,"CourseChapterTopicPlaylistID":245377,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"In the previous videos,"},{"Start":"00:01.950 ","End":"00:05.445","Text":"we learned about 3 of the 4 colligative properties."},{"Start":"00:05.445 ","End":"00:10.980","Text":"In this video, we\u0027ll talk about the fourth property, osmotic pressure."},{"Start":"00:10.980 ","End":"00:14.310","Text":"First we\u0027re going to talk about the fact that water"},{"Start":"00:14.310 ","End":"00:17.865","Text":"flows from a dilute to a more concentrated solution."},{"Start":"00:17.865 ","End":"00:23.830","Text":"This principle underlines all that we\u0027ll learn about osmosis."},{"Start":"00:23.830 ","End":"00:28.190","Text":"We\u0027re going to consider a solution with a non-volatile solute such as"},{"Start":"00:28.190 ","End":"00:33.095","Text":"glucose or urea in a volatile solvent such as water."},{"Start":"00:33.095 ","End":"00:35.059","Text":"We\u0027re going to have 2 solutions."},{"Start":"00:35.059 ","End":"00:36.320","Text":"One, A,"},{"Start":"00:36.320 ","End":"00:41.255","Text":"that\u0027s more dilute than B. Here\u0027s a picture."},{"Start":"00:41.255 ","End":"00:43.190","Text":"This is A,"},{"Start":"00:43.190 ","End":"00:47.785","Text":"this is B, and A is more dilute than B."},{"Start":"00:47.785 ","End":"00:51.110","Text":"Now, the fact that it\u0027s more dilute means that"},{"Start":"00:51.110 ","End":"00:55.085","Text":"the mole fraction of water in A is higher than in B."},{"Start":"00:55.085 ","End":"00:59.465","Text":"The vapor pressure is higher in A than in B,"},{"Start":"00:59.465 ","End":"01:02.840","Text":"vapor pressure is higher in this one."},{"Start":"01:02.840 ","End":"01:12.495","Text":"What will happen is that water will vaporize from A and condense in B."},{"Start":"01:12.495 ","End":"01:14.540","Text":"It will go from A to B,"},{"Start":"01:14.540 ","End":"01:19.040","Text":"and this will continue until A and B have the same concentration."},{"Start":"01:19.040 ","End":"01:22.460","Text":"That means it\u0027s the same mole fraction of water."},{"Start":"01:22.460 ","End":"01:26.740","Text":"We\u0027re going to have A and B with the same concentration."},{"Start":"01:26.740 ","End":"01:31.615","Text":"Now we need to know what a semipermeable membrane is."},{"Start":"01:31.615 ","End":"01:34.435","Text":"Here\u0027s an illustration."},{"Start":"01:34.435 ","End":"01:37.815","Text":"In osmosis and osmotic pressure,"},{"Start":"01:37.815 ","End":"01:43.655","Text":"a semipermeable membrane is a material with submicroscopic holes."},{"Start":"01:43.655 ","End":"01:45.260","Text":"These are supposed to be"},{"Start":"01:45.260 ","End":"01:51.680","Text":"submicroscopic holes that permit the solvent molecules to pass through the holes,"},{"Start":"01:51.680 ","End":"01:52.820","Text":"but not the solute."},{"Start":"01:52.820 ","End":"02:00.920","Text":"These blue circles are supposed to be solvent molecules and the red solute."},{"Start":"02:00.920 ","End":"02:04.445","Text":"The solid molecules are larger than the solvent molecules."},{"Start":"02:04.445 ","End":"02:10.895","Text":"The solvent molecules can move through the holes backward and forward,"},{"Start":"02:10.895 ","End":"02:14.870","Text":"but the solid molecules cannot go through the holes."},{"Start":"02:14.870 ","End":"02:19.290","Text":"Now we can talk about osmosis and osmotic pressure."},{"Start":"02:19.290 ","End":"02:23.480","Text":"Osmosis is the tendency of a solvent to flow through"},{"Start":"02:23.480 ","End":"02:28.550","Text":"a semipermeable membrane into a more concentrated solution."},{"Start":"02:28.550 ","End":"02:32.065","Text":"That\u0027s the definition of osmosis."},{"Start":"02:32.065 ","End":"02:36.420","Text":"Here we have a picture on U-tube."},{"Start":"02:36.420 ","End":"02:39.510","Text":"It\u0027s not exactly a U-tube, but near enough."},{"Start":"02:39.510 ","End":"02:43.250","Text":"On the left-hand side, we have solvent."},{"Start":"02:43.250 ","End":"02:46.445","Text":"The right-hand side solution."},{"Start":"02:46.445 ","End":"02:50.905","Text":"This is supposed to be the semipermeable membrane."},{"Start":"02:50.905 ","End":"02:59.420","Text":"Initially, solvent molecules move through the semipermeable membrane into the solution."},{"Start":"02:59.420 ","End":"03:03.600","Text":"The water molecules are going from left to right."},{"Start":"03:03.850 ","End":"03:11.565","Text":"At equilibrium, the flow in each direction is equal so there\u0027s no net flow of solvent."},{"Start":"03:11.565 ","End":"03:15.965","Text":"The water molecules or solvent molecules are going from left to right"},{"Start":"03:15.965 ","End":"03:20.775","Text":"and right to left and so there\u0027s no net flow."},{"Start":"03:20.775 ","End":"03:25.100","Text":"Now, because the solvent has moved from left to right,"},{"Start":"03:25.100 ","End":"03:29.845","Text":"it\u0027s now higher on the right-hand side than on the left-hand side."},{"Start":"03:29.845 ","End":"03:35.939","Text":"The difference in height between the 2 arms is called the osmotic pressure."},{"Start":"03:35.950 ","End":"03:38.630","Text":"This is the osmotic pressure,"},{"Start":"03:38.630 ","End":"03:43.360","Text":"the difference in height between the right-hand side and the left-hand side."},{"Start":"03:43.360 ","End":"03:53.885","Text":"Now, supposing we apply a pressure on the right-hand side and push the solvent down,"},{"Start":"03:53.885 ","End":"03:56.510","Text":"it rises on the left-hand side."},{"Start":"03:56.510 ","End":"03:59.270","Text":"We do this until they\u0027re equal."},{"Start":"03:59.270 ","End":"04:04.100","Text":"The pressure we have to apply is equal to the osmotic pressure."},{"Start":"04:04.100 ","End":"04:07.715","Text":"If you apply a pressure equal to the osmotic pressure,"},{"Start":"04:07.715 ","End":"04:12.280","Text":"the 2 arms become the same height."},{"Start":"04:12.280 ","End":"04:16.580","Text":"The osmotic pressure is often defined as a pressure required to stop"},{"Start":"04:16.580 ","End":"04:21.145","Text":"the flow of solvent through a semipermeable membrane."},{"Start":"04:21.145 ","End":"04:26.105","Text":"Now, the osmotic pressure is based on laws similar to the ideal gas law."},{"Start":"04:26.105 ","End":"04:30.445","Text":"Remember the ideal gas law, PV equals nRT."},{"Start":"04:30.445 ","End":"04:34.280","Text":"Instead of P, we have Pi for those multi-pressure,"},{"Start":"04:34.280 ","End":"04:38.615","Text":"and it\u0027s Pi V equal to nRT."},{"Start":"04:38.615 ","End":"04:43.650","Text":"That means that Pi is equal to n divided by VRT,"},{"Start":"04:43.650 ","End":"04:44.960","Text":"n is the number of moles,"},{"Start":"04:44.960 ","End":"04:46.475","Text":"V is the volume,"},{"Start":"04:46.475 ","End":"04:53.130","Text":"and that ratio is the same as the molarity M. We have Pi,"},{"Start":"04:53.130 ","End":"04:55.390","Text":"is equal to MRT."},{"Start":"04:55.390 ","End":"05:00.500","Text":"R of course is the gas constant and T is the temperature in Kelvin."},{"Start":"05:00.500 ","End":"05:09.205","Text":"Now, you\u0027ve probably heard of reverse osmosis used to desalinate seawater."},{"Start":"05:09.205 ","End":"05:12.710","Text":"Now, if we apply external pressures greater than"},{"Start":"05:12.710 ","End":"05:17.460","Text":"the osmotic pressure of the solution, here\u0027s a diagram."},{"Start":"05:17.460 ","End":"05:23.270","Text":"We\u0027re applying pressure that\u0027s greater than the osmotic pressure and that\u0027s going to push"},{"Start":"05:23.270 ","End":"05:30.290","Text":"the water down the right-hand side and up on the left-hand side."},{"Start":"05:30.290 ","End":"05:37.105","Text":"Water is flowing from the seawater into the freshwater."},{"Start":"05:37.105 ","End":"05:40.250","Text":"Because the flow is from the right-hand side to"},{"Start":"05:40.250 ","End":"05:43.340","Text":"the left-hand side instead of what we had before,"},{"Start":"05:43.340 ","End":"05:45.155","Text":"which was from the left to the right,"},{"Start":"05:45.155 ","End":"05:49.440","Text":"it\u0027s called reverse osmosis."},{"Start":"05:49.510 ","End":"05:54.840","Text":"In this video, we learned about osmotic pressure."}],"ID":30942},{"Watched":false,"Name":"Using Osmotic Pressure to Find Molar Mass","Duration":"3m 41s","ChapterTopicVideoID":25632,"CourseChapterTopicPlaylistID":245377,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.620","Text":"In the previous video,"},{"Start":"00:01.620 ","End":"00:03.615","Text":"we talked about osmotic pressure."},{"Start":"00:03.615 ","End":"00:09.015","Text":"In this video, we\u0027ll use osmotic pressure to calculate the molar mass."},{"Start":"00:09.015 ","End":"00:14.535","Text":"We can use the osmotic pressure to find the molar mass of the solute."},{"Start":"00:14.535 ","End":"00:18.210","Text":"Here\u0027s the expression for osmotic pressure."},{"Start":"00:18.210 ","End":"00:20.760","Text":"Pie V equals nRT."},{"Start":"00:20.760 ","End":"00:22.650","Text":"Pi is osmotic pressure V,"},{"Start":"00:22.650 ","End":"00:24.584","Text":"the volume of the solution,"},{"Start":"00:24.584 ","End":"00:25.935","Text":"n is the number of moles,"},{"Start":"00:25.935 ","End":"00:28.965","Text":"R is the gas constant and T is the temperature in Kelvin."},{"Start":"00:28.965 ","End":"00:34.170","Text":"Now we know that the number of moles is equal to the mass divided by the molar mass."},{"Start":"00:34.170 ","End":"00:39.210","Text":"We have Pi V is equal to mRT divided by Mw."},{"Start":"00:39.210 ","End":"00:46.920","Text":"We can isolate molar mass and find that molar mass is equal to mRT"},{"Start":"00:46.920 ","End":"00:55.250","Text":"divided by Pi V. This is a very similar expression to that we found using the gas law."},{"Start":"00:55.250 ","End":"00:59.765","Text":"There we found the molar mass using the gas law."},{"Start":"00:59.765 ","End":"01:05.240","Text":"This is a much more sensitive methods than freezing point depression because"},{"Start":"01:05.240 ","End":"01:10.900","Text":"it\u0027s easy to measure the pressure in millimeters mercury."},{"Start":"01:10.900 ","End":"01:12.914","Text":"Here\u0027s an example."},{"Start":"01:12.914 ","End":"01:18.955","Text":"A 50 milliliters aqueous solution contains 1.34 grams of hemoglobin."},{"Start":"01:18.955 ","End":"01:25.085","Text":"The osmotic precious 7.25 millimeters of mercury at 25 degrees Celsius."},{"Start":"01:25.085 ","End":"01:27.860","Text":"What is the molar mass of hemoglobin?"},{"Start":"01:27.860 ","End":"01:33.550","Text":"The first thing to do is to collect the data and make sure it\u0027s in the correct units."},{"Start":"01:33.550 ","End":"01:36.995","Text":"Pi 7.25 millimeters of mercury,"},{"Start":"01:36.995 ","End":"01:40.180","Text":"we need to convert that to atmospheres."},{"Start":"01:40.180 ","End":"01:45.165","Text":"We know that 1 atmosphere is 760 millimeters of mercury."},{"Start":"01:45.165 ","End":"01:51.710","Text":"7.25 millimeters of mercury divided by 760 millimeters of"},{"Start":"01:51.710 ","End":"01:58.435","Text":"mercury gives us 9.54 times 10^-3 atmospheres."},{"Start":"01:58.435 ","End":"02:01.445","Text":"Now we have the pressure in atmospheres."},{"Start":"02:01.445 ","End":"02:03.964","Text":"We need the volume in liters,"},{"Start":"02:03.964 ","End":"02:06.545","Text":"and that\u0027s 0.05 liters."},{"Start":"02:06.545 ","End":"02:07.880","Text":"The mass in grams,"},{"Start":"02:07.880 ","End":"02:10.685","Text":"mass is equal to 1.34 grams,"},{"Start":"02:10.685 ","End":"02:17.450","Text":"are the gas constant in atmospheres times liters per mole per K,"},{"Start":"02:17.450 ","End":"02:24.305","Text":"and that\u0027s 0.08206, exactly as we used in the gas law."},{"Start":"02:24.305 ","End":"02:27.775","Text":"T in Kelvin is 298 Kelvin."},{"Start":"02:27.775 ","End":"02:33.460","Text":"Now we can substitute in the expression we had for the molar mass."},{"Start":"02:33.460 ","End":"02:39.220","Text":"Here we have the mass of the solute, 1.34 grams,"},{"Start":"02:39.220 ","End":"02:43.780","Text":"the gas constant 0.0826"},{"Start":"02:43.780 ","End":"02:49.270","Text":"atmospheres times liters per mole per K. Then we have the temperature in Kelvin,"},{"Start":"02:49.270 ","End":"02:58.330","Text":"298 Kelvin divided by Pi and atmosphere is 9.54 times 10^-3 atmospheres and V,"},{"Start":"02:58.330 ","End":"03:02.485","Text":"the volume in liters, 0.05 liters."},{"Start":"03:02.485 ","End":"03:04.270","Text":"Let\u0027s look at the units first."},{"Start":"03:04.270 ","End":"03:06.430","Text":"Atmosphere cancels with atmosphere,"},{"Start":"03:06.430 ","End":"03:08.410","Text":"liter cancels with liter."},{"Start":"03:08.410 ","End":"03:12.270","Text":"Kelvin cancels with Kelvin to the power of minus 1,"},{"Start":"03:12.270 ","End":"03:17.800","Text":"and we\u0027re left with grams per mole."},{"Start":"03:18.230 ","End":"03:20.540","Text":"So we do the arithmetic."},{"Start":"03:20.540 ","End":"03:25.625","Text":"We get 6.87 times 10^4 grams per mole."},{"Start":"03:25.625 ","End":"03:29.300","Text":"Now we have the molar mass of hemoglobin,"},{"Start":"03:29.300 ","End":"03:35.105","Text":"which is a large biological molecule."},{"Start":"03:35.105 ","End":"03:40.920","Text":"In this video, we calculated the molar mass from the osmotic pressure."}],"ID":30943},{"Watched":false,"Name":"Vant Hoff Factor","Duration":"3m 32s","ChapterTopicVideoID":25633,"CourseChapterTopicPlaylistID":245377,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.340","Text":"In previous videos, we talked about the colligative properties for non-electrolytes."},{"Start":"00:05.340 ","End":"00:09.105","Text":"In this video, we will discuss them for electrolytes."},{"Start":"00:09.105 ","End":"00:14.100","Text":"We\u0027re talking about anomalous colligative properties."},{"Start":"00:14.100 ","End":"00:16.185","Text":"In the 19th century,"},{"Start":"00:16.185 ","End":"00:22.770","Text":"van\u0027t Hoff observed that certain solutes have higher colligative effects than expected."},{"Start":"00:22.770 ","End":"00:25.830","Text":"For example, NaCl."},{"Start":"00:25.830 ","End":"00:30.270","Text":"For a 0.01 molal solution,"},{"Start":"00:30.270 ","End":"00:32.310","Text":"we expect that Delta Tf,"},{"Start":"00:32.310 ","End":"00:34.845","Text":"that\u0027s the depression of the freezing point,"},{"Start":"00:34.845 ","End":"00:38.855","Text":"that\u0027s equal to minus Kf times m will be equal to"},{"Start":"00:38.855 ","End":"00:44.030","Text":"minus 1.86 degrees Celsius per mole times kilogram,"},{"Start":"00:44.030 ","End":"00:46.935","Text":"that\u0027s Kf for water,"},{"Start":"00:46.935 ","End":"00:53.089","Text":"times 0.01 moles per kilogram."},{"Start":"00:53.089 ","End":"00:54.790","Text":"That\u0027s the molality."},{"Start":"00:54.790 ","End":"00:56.460","Text":"When we multiply that out,"},{"Start":"00:56.460 ","End":"01:01.110","Text":"we get is it\u0027s equal to minus 0.0186"},{"Start":"01:01.110 ","End":"01:06.225","Text":"degrees Celsius because moles^power minus 1 goes with moles,"},{"Start":"01:06.225 ","End":"01:09.450","Text":"kilogram with kilograms^power minus 1."},{"Start":"01:09.450 ","End":"01:11.000","Text":"Now for NaCl,"},{"Start":"01:11.000 ","End":"01:14.765","Text":"the measured effect is approximately twice as large."},{"Start":"01:14.765 ","End":"01:20.720","Text":"We define the van\u0027t Hoff factor i to be the measured Delta Tf,"},{"Start":"01:20.720 ","End":"01:24.110","Text":"divided by the expected value of Delta Tf,"},{"Start":"01:24.110 ","End":"01:27.370","Text":"and that is approximately 2."},{"Start":"01:27.370 ","End":"01:33.835","Text":"Now Arrhenius\u0027s theory of electrolytic dissociation can explain these results."},{"Start":"01:33.835 ","End":"01:37.780","Text":"NaCl is a strong electrolyte that gives 2 moles of"},{"Start":"01:37.780 ","End":"01:41.530","Text":"ions in solution so that i is equal to 2,"},{"Start":"01:41.530 ","End":"01:50.515","Text":"we get Na plus aqueous and Cl minus aqueous in solution."},{"Start":"01:50.515 ","End":"01:52.810","Text":"Now if we had a non-electrolyte,"},{"Start":"01:52.810 ","End":"01:54.115","Text":"i would just be 1."},{"Start":"01:54.115 ","End":"01:56.995","Text":"The equation stays the same as before."},{"Start":"01:56.995 ","End":"02:01.785","Text":"Now, we have a weak electrolyte that only partially ionizes in water,"},{"Start":"02:01.785 ","End":"02:04.715","Text":"i is slightly greater than 1."},{"Start":"02:04.715 ","End":"02:09.880","Text":"For example, we have acetic acid CH3CO2H."},{"Start":"02:09.880 ","End":"02:13.690","Text":"It will ionize slightly to give acetate,"},{"Start":"02:13.690 ","End":"02:15.780","Text":"which is solvated,"},{"Start":"02:15.780 ","End":"02:19.500","Text":"and H plus, which is also solvated."},{"Start":"02:19.500 ","End":"02:23.070","Text":"It\u0027s a very small proportion of these so that\u0027s"},{"Start":"02:23.070 ","End":"02:27.505","Text":"the reason that i is only slightly greater than 1."},{"Start":"02:27.505 ","End":"02:32.155","Text":"Now we can modify the equations for the colligative properties."},{"Start":"02:32.155 ","End":"02:33.990","Text":"Here are the new equations,"},{"Start":"02:33.990 ","End":"02:40.710","Text":"Delta Tf is equal to minus i times Kf times m. Delta Tb is i"},{"Start":"02:40.710 ","End":"02:48.100","Text":"times Kb times m and Pi is equal to i times MRT."},{"Start":"02:48.100 ","End":"02:52.085","Text":"The first one is for the depression of the freezing point."},{"Start":"02:52.085 ","End":"03:02.570","Text":"The second one for the elevation of the boiling point and the third one is for osmosis."},{"Start":"03:02.570 ","End":"03:06.150","Text":"That\u0027s the osmotic pressure is Pi."},{"Start":"03:06.550 ","End":"03:10.970","Text":"You see here that we\u0027ve got the same equation as before,"},{"Start":"03:10.970 ","End":"03:18.860","Text":"except the i is included and i equal to 1 returns us to the previous equation,"},{"Start":"03:18.860 ","End":"03:20.675","Text":"so i is equal to 1,"},{"Start":"03:20.675 ","End":"03:25.020","Text":"we go back to the equations for non-electrolytes."},{"Start":"03:26.830 ","End":"03:31.860","Text":"In this video, we discussed the van\u0027t Hoff factor."}],"ID":30944},{"Watched":false,"Name":"Vapor Pressure of Solutions","Duration":"6m 15s","ChapterTopicVideoID":25634,"CourseChapterTopicPlaylistID":245377,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"In the previous video,"},{"Start":"00:01.800 ","End":"00:05.655","Text":"we learned about solutions of gases and liquids and in this video,"},{"Start":"00:05.655 ","End":"00:10.365","Text":"we\u0027ll discuss lowering the vapor pressure of the solvent in a solution."},{"Start":"00:10.365 ","End":"00:14.910","Text":"Lowering the vapor pressure of a solvent in the solution is one of"},{"Start":"00:14.910 ","End":"00:19.740","Text":"the colligative properties and we\u0027ll talk about these just now."},{"Start":"00:19.740 ","End":"00:24.060","Text":"Now, colligative properties are properties of solutions that"},{"Start":"00:24.060 ","End":"00:28.875","Text":"depend on the relative numbers of solute and solvent molecules."},{"Start":"00:28.875 ","End":"00:33.020","Text":"They don\u0027t depend on the chemical identity of the solute or solvent,"},{"Start":"00:33.020 ","End":"00:37.745","Text":"just on the relative numbers of molecules or moles."},{"Start":"00:37.745 ","End":"00:42.829","Text":"Now, there are four colligative properties lowering the vapor pressure of the solvent."},{"Start":"00:42.829 ","End":"00:44.840","Text":"We\u0027ll talk about that in this video."},{"Start":"00:44.840 ","End":"00:47.840","Text":"Raising its boiling point and lowering its freezing point."},{"Start":"00:47.840 ","End":"00:50.510","Text":"We\u0027ll talk about that in another video and finally,"},{"Start":"00:50.510 ","End":"00:53.110","Text":"we\u0027ll talk about osmosis."},{"Start":"00:53.110 ","End":"00:56.210","Text":"I\u0027m going to talk first about vapor pressure of"},{"Start":"00:56.210 ","End":"01:00.490","Text":"solutions and the law we\u0027re talking about is Raoult\u0027s law."},{"Start":"01:00.490 ","End":"01:04.870","Text":"François-Marie Raoult in the 1880s,"},{"Start":"01:04.870 ","End":"01:09.815","Text":"found experimentally that a non-volatile solute"},{"Start":"01:09.815 ","End":"01:14.950","Text":"is a solute that doesn\u0027t go easily into the vapor phase."},{"Start":"01:14.950 ","End":"01:18.980","Text":"He found when a non-volatile solute is dissolved in a solvent,"},{"Start":"01:18.980 ","End":"01:22.165","Text":"it lowers the vapor pressure of the solvent."},{"Start":"01:22.165 ","End":"01:25.843","Text":"Here\u0027s his law, Raoult\u0027s law,"},{"Start":"01:25.843 ","End":"01:29.765","Text":"for a solution of a non-volatile solute in a solvent."},{"Start":"01:29.765 ","End":"01:35.420","Text":"Raoult\u0027s law says that the vapor pressure of the solvent is equal to"},{"Start":"01:35.420 ","End":"01:43.355","Text":"its mole fraction multiplied by the vapor pressure of the pure solvent."},{"Start":"01:43.355 ","End":"01:47.480","Text":"Here it\u0027s written out Chi_solvent is the mole fraction of solvent,"},{"Start":"01:47.480 ","End":"01:50.045","Text":"P_solvent is vapor pressure of solution,"},{"Start":"01:50.045 ","End":"01:54.970","Text":"and P_solvent 0 is the vapor pressure of the pure solvent."},{"Start":"01:54.970 ","End":"02:01.174","Text":"Now we know that the mole fraction of the solvent is either 1 when it\u0027s pure."},{"Start":"02:01.174 ","End":"02:07.600","Text":"1 is pure, less than 1 is in the solution."},{"Start":"02:07.600 ","End":"02:10.690","Text":"Chi_solvent is less or equal to 1,"},{"Start":"02:10.690 ","End":"02:15.710","Text":"which means that the vapor pressure of the solvent in the solution is"},{"Start":"02:15.710 ","End":"02:21.265","Text":"less or equal to the vapor pressure of the solvent when it\u0027s pure."},{"Start":"02:21.265 ","End":"02:24.555","Text":"Now this law applies to ideal solutions."},{"Start":"02:24.555 ","End":"02:27.665","Text":"We discussed ideal solutions in a previous video."},{"Start":"02:27.665 ","End":"02:33.455","Text":"That means the Delta H of solution is 0."},{"Start":"02:33.455 ","End":"02:37.010","Text":"However, even in non-ideal solutions,"},{"Start":"02:37.010 ","End":"02:41.780","Text":"where Delta H solution can be exothermic or slightly endothermic,"},{"Start":"02:41.780 ","End":"02:44.655","Text":"it works well for the solvent in"},{"Start":"02:44.655 ","End":"02:50.075","Text":"dilute solutions and becomes more exact as the concentration approaches 0."},{"Start":"02:50.075 ","End":"02:52.445","Text":"This is called a limiting law,"},{"Start":"02:52.445 ","End":"02:57.005","Text":"when the law becomes better when the concentration approaches 0."},{"Start":"02:57.005 ","End":"03:00.485","Text":"Now there\u0027s a simple thermodynamic explanation for this."},{"Start":"03:00.485 ","End":"03:07.100","Text":"There are many different explanations ranging from the simple to the more complicated."},{"Start":"03:07.100 ","End":"03:10.940","Text":"Now, if we recall the vapor pressure is the pressure"},{"Start":"03:10.940 ","End":"03:15.100","Text":"exerted by the vapor in dynamic equilibrium with its liquid."},{"Start":"03:15.100 ","End":"03:20.075","Text":"We have the vapor above the liquid in a closed vessel."},{"Start":"03:20.075 ","End":"03:23.165","Text":"There\u0027s an equilibrium between the liquid and the vapor."},{"Start":"03:23.165 ","End":"03:25.625","Text":"Delta G is equal to 0."},{"Start":"03:25.625 ","End":"03:30.199","Text":"Now the entropy of a solution is greater than the entropy of a solvent."},{"Start":"03:30.199 ","End":"03:34.354","Text":"A solution is more disordered than a solvent."},{"Start":"03:34.354 ","End":"03:37.070","Text":"The enthalpy is the same for both the solution than"},{"Start":"03:37.070 ","End":"03:41.330","Text":"the solvent so that Gibbs free energy of the solvent is lower,"},{"Start":"03:41.330 ","End":"03:45.970","Text":"more negative for a solution than for a pure solvent."},{"Start":"03:45.970 ","End":"03:48.770","Text":"If S is greater,"},{"Start":"03:48.770 ","End":"03:51.665","Text":"G will be lower."},{"Start":"03:51.665 ","End":"03:55.879","Text":"Now we want to maintain equilibrium between the solution and its vapor"},{"Start":"03:55.879 ","End":"04:00.655","Text":"so the Gibbs free energy of vapor above the solution must also be lower."},{"Start":"04:00.655 ","End":"04:05.465","Text":"It can be shown that the Gibbs free energy depends on the vapor pressure."},{"Start":"04:05.465 ","End":"04:08.360","Text":"Vapor pressure must also be lower."},{"Start":"04:08.360 ","End":"04:15.545","Text":"Now before we talked about a non-volatile liquid in a volatile solvent."},{"Start":"04:15.545 ","End":"04:19.100","Text":"Now, to consider the case of two volatile liquids."},{"Start":"04:19.100 ","End":"04:20.780","Text":"We just call them A and B."},{"Start":"04:20.780 ","End":"04:25.010","Text":"Now it\u0027s irrelevant which is the solvent which is the solute."},{"Start":"04:25.010 ","End":"04:30.560","Text":"Now Raoult\u0027s law reads P_A for component A of the solution is"},{"Start":"04:30.560 ","End":"04:36.710","Text":"equal to Chi_A P_A 0 and P_B is equal to Chi_B P_B 0."},{"Start":"04:36.710 ","End":"04:40.114","Text":"Here\u0027s an example, a solution of benzene and toluene."},{"Start":"04:40.114 ","End":"04:46.339","Text":"This is an ideal solution because benzene and toluene are very similar."},{"Start":"04:46.339 ","End":"04:48.405","Text":"Now here\u0027s the example."},{"Start":"04:48.405 ","End":"04:51.515","Text":"For a solution of benzene and for benzene,"},{"Start":"04:51.515 ","End":"04:56.300","Text":"the vapor pressure of pure benzene is 95.1 millimeters of mercury and"},{"Start":"04:56.300 ","End":"05:01.625","Text":"toluene and it\u0027s vapor pressure is 28.4 millimeters of mercury."},{"Start":"05:01.625 ","End":"05:05.300","Text":"Calculate the vapor pressures of benzene and toluene and"},{"Start":"05:05.300 ","End":"05:08.540","Text":"the total vapor pressure for the case where"},{"Start":"05:08.540 ","End":"05:13.310","Text":"Chi_benzene is equal to Chi_toluene is equal to 0.5."},{"Start":"05:13.310 ","End":"05:16.650","Text":"Same number of moles of benzene and toluene."},{"Start":"05:16.650 ","End":"05:18.475","Text":"Now from Raoult\u0027s law,"},{"Start":"05:18.475 ","End":"05:25.600","Text":"we see that the vapor pressure of benzene is equal to Chi_benzene times P_benzene 0."},{"Start":"05:25.600 ","End":"05:32.449","Text":"So that\u0027s 0.5 for the mole fraction times 95.1 millimeters of mercury,"},{"Start":"05:32.449 ","End":"05:37.460","Text":"that\u0027s the vapor pressure of pure benzene and we multiply that out."},{"Start":"05:37.460 ","End":"05:40.655","Text":"It\u0027s 47.6 millimeters of mercury."},{"Start":"05:40.655 ","End":"05:42.830","Text":"Do the same thing for toluene."},{"Start":"05:42.830 ","End":"05:46.710","Text":"P_toluene is equal to Chi_toluene P0_toluene that\u0027s"},{"Start":"05:46.710 ","End":"05:50.540","Text":"0.5 times 28.4 millimeters of mercury,"},{"Start":"05:50.540 ","End":"05:54.845","Text":"and that\u0027s equal to 14.2 millimeters of mercury."},{"Start":"05:54.845 ","End":"05:59.255","Text":"The total is just the sum of these two quantities,"},{"Start":"05:59.255 ","End":"06:02.855","Text":"47.6 plus 14.2 millimeters of mercury,"},{"Start":"06:02.855 ","End":"06:06.370","Text":"which is 61.8 millimeters of mercury."},{"Start":"06:06.370 ","End":"06:08.520","Text":"In this video,"},{"Start":"06:08.520 ","End":"06:14.220","Text":"we discussed the lowering of the vapor pressure of the solvent in a solution."}],"ID":30945},{"Watched":false,"Name":"Applications of Colligative Properties","Duration":"3m 1s","ChapterTopicVideoID":25635,"CourseChapterTopicPlaylistID":245377,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"In the previous videos,"},{"Start":"00:01.980 ","End":"00:04.715","Text":"we learned about the 4 colligative properties."},{"Start":"00:04.715 ","End":"00:08.850","Text":"In this video, we\u0027ll talk about some of their applications."},{"Start":"00:08.850 ","End":"00:13.185","Text":"We\u0027ve already discussed some of the applications."},{"Start":"00:13.185 ","End":"00:16.830","Text":"For example, we discussed fractional distillation,"},{"Start":"00:16.830 ","End":"00:20.445","Text":"which is a consequence of lowering of vapor pressure."},{"Start":"00:20.445 ","End":"00:23.720","Text":"We showed how to calculate the molar mass."},{"Start":"00:23.720 ","End":"00:30.035","Text":"This can be done either using freezing point depression or osmotic pressure."},{"Start":"00:30.035 ","End":"00:35.420","Text":"We discussed desalination, which is an example of osmosis,"},{"Start":"00:35.420 ","End":"00:38.360","Text":"in this case, reverse osmosis."},{"Start":"00:38.360 ","End":"00:42.394","Text":"Now, there are other applications of freezing point depression."},{"Start":"00:42.394 ","End":"00:47.405","Text":"For example, use of antifreeze in automobiles or cars."},{"Start":"00:47.405 ","End":"00:51.320","Text":"Now, ethylene glycol is an antifreeze that"},{"Start":"00:51.320 ","End":"00:55.729","Text":"lowers the freezing point of water and prevents it from freezing."},{"Start":"00:55.729 ","End":"00:58.025","Text":"If it\u0027s left in the summer,"},{"Start":"00:58.025 ","End":"01:03.305","Text":"it raises the boiling point of water and can prevent it from boiling over."},{"Start":"01:03.305 ","End":"01:08.360","Text":"Another application is the de-icing of roads in winter."},{"Start":"01:08.360 ","End":"01:14.915","Text":"Now, salt can melt ice at temperatures as low as minus 21 degrees Celsius."},{"Start":"01:14.915 ","End":"01:18.710","Text":"It\u0027s used for melting ice and snow."},{"Start":"01:18.710 ","End":"01:22.640","Text":"We can also use it to prepare ice cream."},{"Start":"01:22.640 ","End":"01:26.720","Text":"We can add salt ice to lower its temperature."},{"Start":"01:26.720 ","End":"01:29.600","Text":"This is often called a slush bath."},{"Start":"01:29.600 ","End":"01:33.500","Text":"It helps to freeze the ice cream as we mix it."},{"Start":"01:33.500 ","End":"01:37.210","Text":"There are also other applications of osmosis."},{"Start":"01:37.210 ","End":"01:41.340","Text":"For example, intravenous fluids."},{"Start":"01:41.340 ","End":"01:45.350","Text":"Now, the osmotic pressure inside the red blood cell is equivalent to"},{"Start":"01:45.350 ","End":"01:50.060","Text":"0.92% mass per volume of sodium chloride,"},{"Start":"01:50.060 ","End":"01:52.595","Text":"when aqueous solution of sodium chloride."},{"Start":"01:52.595 ","End":"01:56.030","Text":"This is often called a saline solution."},{"Start":"01:56.030 ","End":"02:01.070","Text":"Now, if we want to use an intravenous fluid to counter dehydration,"},{"Start":"02:01.070 ","End":"02:06.700","Text":"it must have this concentration and then it\u0027s said to be isotonic."},{"Start":"02:06.700 ","End":"02:09.305","Text":"Here\u0027s the red blood cell."},{"Start":"02:09.305 ","End":"02:15.785","Text":"If the saline solution has this concentration 0.92%,"},{"Start":"02:15.785 ","End":"02:19.070","Text":"then the red blood cells will stay the same."},{"Start":"02:19.070 ","End":"02:20.965","Text":"There\u0027ll be unaffected."},{"Start":"02:20.965 ","End":"02:24.210","Text":"This is called isotonic."},{"Start":"02:24.210 ","End":"02:27.620","Text":"If the saline solution has a lower concentration,"},{"Start":"02:27.620 ","End":"02:29.360","Text":"then it\u0027s called hypotonic,"},{"Start":"02:29.360 ","End":"02:32.930","Text":"and water will flow into the cell."},{"Start":"02:32.930 ","End":"02:37.915","Text":"The cell can get larger and may even burst."},{"Start":"02:37.915 ","End":"02:41.645","Text":"If the saline solution has a higher concentration,"},{"Start":"02:41.645 ","End":"02:47.660","Text":"water will flow out of the cell and this is called a hypertonic solution."},{"Start":"02:47.660 ","End":"02:50.975","Text":"This is hypertonic."},{"Start":"02:50.975 ","End":"02:54.220","Text":"This is hypotonic."},{"Start":"02:54.220 ","End":"03:01.200","Text":"In this video, we talked about some of the applications of the colligative properties."}],"ID":30946}],"Thumbnail":null,"ID":245377},{"Name":"Colloidal Mixtures","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Colloids","Duration":"6m 9s","ChapterTopicVideoID":25636,"CourseChapterTopicPlaylistID":245378,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"In previous videos,"},{"Start":"00:01.770 ","End":"00:04.020","Text":"we talked about many solutions,"},{"Start":"00:04.020 ","End":"00:07.005","Text":"and in this video we\u0027ll talk about colloids."},{"Start":"00:07.005 ","End":"00:11.370","Text":"Let\u0027s recall what we know about solutions and mixtures."},{"Start":"00:11.370 ","End":"00:13.905","Text":"In a true homogeneous solution,"},{"Start":"00:13.905 ","End":"00:18.045","Text":"the solid molecules are small and stay in solution indefinitely."},{"Start":"00:18.045 ","End":"00:22.125","Text":"For example, glucose dissolved in water or salt and water."},{"Start":"00:22.125 ","End":"00:25.845","Text":"Some heterogeneous mixtures appear to be solutions,"},{"Start":"00:25.845 ","End":"00:26.970","Text":"but after some time,"},{"Start":"00:26.970 ","End":"00:30.075","Text":"the components separate out due to gravity."},{"Start":"00:30.075 ","End":"00:33.930","Text":"For example, fine clay particles dispersed in water."},{"Start":"00:33.930 ","End":"00:36.405","Text":"These clay particles are, of course,"},{"Start":"00:36.405 ","End":"00:40.240","Text":"larger than the solute particles."},{"Start":"00:40.240 ","End":"00:45.785","Text":"Now let\u0027s talk about a colloidal dispersion or colloid for short."},{"Start":"00:45.785 ","End":"00:48.529","Text":"A colloidal dispersion, or colloid,"},{"Start":"00:48.529 ","End":"00:51.815","Text":"is intermediate between these 2 extremes."},{"Start":"00:51.815 ","End":"00:54.680","Text":"It\u0027s larger than the solute molecules and"},{"Start":"00:54.680 ","End":"00:59.705","Text":"smaller than the particles in the heterogeneous mixtures."},{"Start":"00:59.705 ","End":"01:04.325","Text":"Colloid particles have diameters between 5 and 1,000 nanometers,"},{"Start":"01:04.325 ","End":"01:08.510","Text":"where solute particles are smaller than 5 nanometers."},{"Start":"01:08.510 ","End":"01:12.995","Text":"The colloids can be gases, liquids, or solids."},{"Start":"01:12.995 ","End":"01:16.250","Text":"Many common substances are colloids,"},{"Start":"01:16.250 ","End":"01:24.570","Text":"like paints or ink, or even marshmallows."},{"Start":"01:24.850 ","End":"01:28.910","Text":"How can we classify these colloids?"},{"Start":"01:28.910 ","End":"01:34.325","Text":"Colloids are classified according to the phases of their components."},{"Start":"01:34.325 ","End":"01:39.785","Text":"The equivalent of the solvent is called the dispersing substance."},{"Start":"01:39.785 ","End":"01:45.480","Text":"The equivalent of the solute is called the dispersed substance."},{"Start":"01:45.610 ","End":"01:50.285","Text":"Let\u0027s look at several types of colloids."},{"Start":"01:50.285 ","End":"01:52.759","Text":"We have the dispersing substance."},{"Start":"01:52.759 ","End":"01:54.920","Text":"It can be gas, liquid, or solid,"},{"Start":"01:54.920 ","End":"01:57.980","Text":"and then the phase of the colloid itself will be gas,"},{"Start":"01:57.980 ","End":"02:00.160","Text":"liquid, or solid."},{"Start":"02:00.160 ","End":"02:04.297","Text":"Supposing we have a dispersing substance that is a gas,"},{"Start":"02:04.297 ","End":"02:07.955","Text":"then the dispersed substance can be a liquid or a solid."},{"Start":"02:07.955 ","End":"02:09.200","Text":"If it\u0027s a liquid,"},{"Start":"02:09.200 ","End":"02:11.012","Text":"it\u0027s called an aerosol,"},{"Start":"02:11.012 ","End":"02:12.695","Text":"and the examples are fog,"},{"Start":"02:12.695 ","End":"02:14.195","Text":"mist, or hairspray."},{"Start":"02:14.195 ","End":"02:17.765","Text":"If it\u0027s a solid dispersed in a gas,"},{"Start":"02:17.765 ","End":"02:19.564","Text":"it\u0027s also called an aerosol,"},{"Start":"02:19.564 ","End":"02:23.165","Text":"and examples are smoke or smog."},{"Start":"02:23.165 ","End":"02:26.825","Text":"Now if the dispersing substance is a liquid,"},{"Start":"02:26.825 ","End":"02:32.990","Text":"then the phase of the colloid will also be liquid and the dispersed substance can be gas,"},{"Start":"02:32.990 ","End":"02:34.295","Text":"liquid, or solid."},{"Start":"02:34.295 ","End":"02:37.430","Text":"If it\u0027s a gas dispersed to the liquid,"},{"Start":"02:37.430 ","End":"02:43.155","Text":"it\u0027s a foam, and the examples are fire extinguisher foam."},{"Start":"02:43.155 ","End":"02:46.315","Text":"If it\u0027s a liquid dispersed in a liquid,"},{"Start":"02:46.315 ","End":"02:48.010","Text":"it\u0027s called an emulsion,"},{"Start":"02:48.010 ","End":"02:51.715","Text":"and examples of these are milk or mayonnaise."},{"Start":"02:51.715 ","End":"02:54.309","Text":"If it\u0027s a solid dispersed in a liquid,"},{"Start":"02:54.309 ","End":"02:56.620","Text":"it\u0027s called a sol or gel."},{"Start":"02:56.620 ","End":"02:59.845","Text":"The examples are paint or printing ink."},{"Start":"02:59.845 ","End":"03:03.895","Text":"Now, if the dispersing substance is a solid,"},{"Start":"03:03.895 ","End":"03:07.523","Text":"then the phase of the colloid will be solid,"},{"Start":"03:07.523 ","End":"03:10.870","Text":"and dispersed substance can be gas, liquid, or solid."},{"Start":"03:10.870 ","End":"03:13.240","Text":"If it\u0027s a gas dispersed in a solid,"},{"Start":"03:13.240 ","End":"03:15.035","Text":"it is called solid foam,"},{"Start":"03:15.035 ","End":"03:18.415","Text":"and examples are insulating foam or marshmallow."},{"Start":"03:18.415 ","End":"03:21.160","Text":"If it\u0027s a liquid dispersed in a solid,"},{"Start":"03:21.160 ","End":"03:23.050","Text":"it is called solid emulsion."},{"Start":"03:23.050 ","End":"03:24.445","Text":"Examples are butter,"},{"Start":"03:24.445 ","End":"03:26.620","Text":"ice cream, or gelatin."},{"Start":"03:26.620 ","End":"03:28.455","Text":"Gelatin is also called a gel."},{"Start":"03:28.455 ","End":"03:31.650","Text":"If it\u0027s a solid dispersed in a solid,"},{"Start":"03:31.650 ","End":"03:33.920","Text":"it is called a solid dispersion."},{"Start":"03:33.920 ","End":"03:36.350","Text":"An example is ruby glass,"},{"Start":"03:36.350 ","End":"03:41.780","Text":"in which gold chloride is dispersed in glass."},{"Start":"03:41.780 ","End":"03:45.050","Text":"Now I\u0027m going to talk about the Tyndall effect,"},{"Start":"03:45.050 ","End":"03:51.740","Text":"which is a way of checking or examining colloids."},{"Start":"03:51.740 ","End":"03:55.580","Text":"Now, colloid particles may be so small that the dispersion"},{"Start":"03:55.580 ","End":"03:59.150","Text":"appears to be homogeneous, even under microscope."},{"Start":"03:59.150 ","End":"04:02.030","Text":"If we take homogenized milk,"},{"Start":"04:02.030 ","End":"04:05.810","Text":"then the particles of fat in"},{"Start":"04:05.810 ","End":"04:10.280","Text":"the milk are so small that you probably can\u0027t see them under microscope."},{"Start":"04:10.280 ","End":"04:14.525","Text":"However, these particles are large enough to scatter light,"},{"Start":"04:14.525 ","End":"04:19.130","Text":"and visible light can be seen as it passes through a colloidal dispersion."},{"Start":"04:19.130 ","End":"04:24.990","Text":"This effect is called the Tyndall effect after John Tyndall who discovered it."},{"Start":"04:25.520 ","End":"04:28.500","Text":"Supposing we have red light,"},{"Start":"04:28.500 ","End":"04:32.165","Text":"and here we have a solution."},{"Start":"04:32.165 ","End":"04:34.790","Text":"The red light will pass through the solution,"},{"Start":"04:34.790 ","End":"04:38.225","Text":"but you won\u0027t be able to see its path through the solution."},{"Start":"04:38.225 ","End":"04:41.615","Text":"However, when it goes through the colloidal dispersion,"},{"Start":"04:41.615 ","End":"04:44.975","Text":"you should be able to see its path because"},{"Start":"04:44.975 ","End":"04:49.825","Text":"the particles inside the colloid will scatter the light."},{"Start":"04:49.825 ","End":"04:56.375","Text":"Now light scattering is proportional to the frequency to the power 4."},{"Start":"04:56.375 ","End":"05:05.555","Text":"Nu is a frequency and that\u0027s inversely proportional to the wavelength, the power 4."},{"Start":"05:05.555 ","End":"05:07.550","Text":"That means that blue light,"},{"Start":"05:07.550 ","End":"05:09.425","Text":"which has a higher frequency,"},{"Start":"05:09.425 ","End":"05:11.570","Text":"is scattered more than red light,"},{"Start":"05:11.570 ","End":"05:13.340","Text":"which has a lower frequency."},{"Start":"05:13.340 ","End":"05:17.120","Text":"For example, flour suspended in water appears blue,"},{"Start":"05:17.120 ","End":"05:22.910","Text":"and blue eyes result from the iris scattering blue light more than red."},{"Start":"05:22.910 ","End":"05:27.090","Text":"Now a word or 2 about aqueous colloids."},{"Start":"05:27.620 ","End":"05:31.710","Text":"Aqueous colloids can either be hydrophilic,"},{"Start":"05:31.710 ","End":"05:36.095","Text":"attracting water, or hydrophobic, repelling water."},{"Start":"05:36.095 ","End":"05:39.800","Text":"Now, examples of hydrophobic colloids are milk,"},{"Start":"05:39.800 ","End":"05:42.680","Text":"which is a suspension of fat and water, or mayonnaise,"},{"Start":"05:42.680 ","End":"05:46.135","Text":"which is a suspension of water and fat."},{"Start":"05:46.135 ","End":"05:50.735","Text":"Examples of hydrophilic colloids are gels and puddings."},{"Start":"05:50.735 ","End":"05:54.140","Text":"The macromolecular protein in gelatin and the starch in"},{"Start":"05:54.140 ","End":"05:58.910","Text":"pudding have many hydrophilic groups that attract water."},{"Start":"05:58.910 ","End":"06:02.075","Text":"In this video, we talked about colloids."},{"Start":"06:02.075 ","End":"06:03.950","Text":"Now this is an enormous subject,"},{"Start":"06:03.950 ","End":"06:09.180","Text":"and we\u0027ve only touched on a few of the aspects of it."}],"ID":26461}],"Thumbnail":null,"ID":245378}]

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Name: Solution Concentration: Video + Workbook | Proprep
Uploaded: 2021-06-10T06:58:15.4730000
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Description: Physical Properties of Solutions - Solution Concentration. Watch the video made by an expert in the field. Download the workbook and maximize your learning.

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