[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.47,0:00:02.86,Default,,0000,0000,0000,,>> In this module we're talking\Nabout colligative properties,
Dialogue: 0,0:00:02.86,0:00:05.37,Default,,0000,0000,0000,,specifically boiling\Npoint elevation,
Dialogue: 0,0:00:05.37,0:00:07.86,Default,,0000,0000,0000,,freezing point depression,\Nand osmotic pressure.
Dialogue: 0,0:00:08.08,0:00:10.51,Default,,0000,0000,0000,,First of all, what a\Ncolligative property is.
Dialogue: 0,0:00:10.51,0:00:12.86,Default,,0000,0000,0000,,It's a property of a solution
Dialogue: 0,0:00:13.05,0:00:17.37,Default,,0000,0000,0000,,that depends only upon how\Nmany solute particles there are
Dialogue: 0,0:00:17.37,0:00:17.98,Default,,0000,0000,0000,,and not their identity.
Dialogue: 0,0:00:24.72,0:00:26.22,Default,,0000,0000,0000,,So, boiling point elevation.
Dialogue: 0,0:00:26.30,0:00:28.64,Default,,0000,0000,0000,,First of all, this is\Nthe important equation
Dialogue: 0,0:00:28.64,0:00:29.49,Default,,0000,0000,0000,,that you're going to memorize.
Dialogue: 0,0:00:29.62,0:00:31.86,Default,,0000,0000,0000,,It says that the change\Nin the boiling point
Dialogue: 0,0:00:32.18,0:00:34.29,Default,,0000,0000,0000,,of a solvent is equal
Dialogue: 0,0:00:34.29,0:00:37.24,Default,,0000,0000,0000,,to something called the van't\NHoff factor times the molality
Dialogue: 0,0:00:37.24,0:00:39.04,Default,,0000,0000,0000,,of that solution, times the --
Dialogue: 0,0:00:39.04,0:00:42.95,Default,,0000,0000,0000,,what's called the boiling\Npoint elevation constant.
Dialogue: 0,0:00:43.12,0:00:45.35,Default,,0000,0000,0000,,So, first of all, the\Nparts of this equation.
Dialogue: 0,0:00:45.67,0:00:48.15,Default,,0000,0000,0000,,The change in the boiling\Npoint is the boiling point
Dialogue: 0,0:00:48.15,0:00:50.73,Default,,0000,0000,0000,,of the solution minus\Nthe boiling point
Dialogue: 0,0:00:50.73,0:00:53.52,Default,,0000,0000,0000,,of the pure solvent,\Nbefore you add something in.
Dialogue: 0,0:00:53.74,0:00:57.90,Default,,0000,0000,0000,,Now, it ends up that\Nfor any liquid,
Dialogue: 0,0:00:57.99,0:01:01.52,Default,,0000,0000,0000,,if you add a non-volatile\Nsolute,
Dialogue: 0,0:01:01.99,0:01:03.66,Default,,0000,0000,0000,,you will increase\Nthe boiling point,
Dialogue: 0,0:01:03.75,0:01:05.46,Default,,0000,0000,0000,,that's what this is\Ndescribing of that --
Dialogue: 0,0:01:05.46,0:01:09.34,Default,,0000,0000,0000,,of the resulting solution\Ncompared to the pure solvent.
Dialogue: 0,0:01:09.34,0:01:11.58,Default,,0000,0000,0000,,Because, this number's\Nalways bigger than this,
Dialogue: 0,0:01:11.64,0:01:13.57,Default,,0000,0000,0000,,this will always be\Na positive number.
Dialogue: 0,0:01:14.30,0:01:17.87,Default,,0000,0000,0000,,This i, the van't Hoff\Nfactor, this is the definition,
Dialogue: 0,0:01:17.87,0:01:20.72,Default,,0000,0000,0000,,it's the mols of particles\Nin the solution divided
Dialogue: 0,0:01:20.72,0:01:22.21,Default,,0000,0000,0000,,by the mols of the solute.
Dialogue: 0,0:01:22.88,0:01:25.83,Default,,0000,0000,0000,,If it's a molecular\Ncompound, then i ,
Dialogue: 0,0:01:25.83,0:01:29.07,Default,,0000,0000,0000,,the van't Hoff factor is just\N1, because for every particle
Dialogue: 0,0:01:29.07,0:01:30.05,Default,,0000,0000,0000,,for that molecular --
Dialogue: 0,0:01:30.12,0:01:32.69,Default,,0000,0000,0000,,that molecule you put in\Nthere you just get, you know,
Dialogue: 0,0:01:32.69,0:01:34.46,Default,,0000,0000,0000,,it's the same particle inside.
Dialogue: 0,0:01:35.09,0:01:37.67,Default,,0000,0000,0000,,However, if it's\Nan ionic compound,
Dialogue: 0,0:01:38.17,0:01:40.81,Default,,0000,0000,0000,,then it can dissociate\Nin that solution.
Dialogue: 0,0:01:41.04,0:01:43.57,Default,,0000,0000,0000,,And when it does, you\Ncreate more particles.
Dialogue: 0,0:01:43.57,0:01:46.21,Default,,0000,0000,0000,,For example, if you put\Nsodium chloride into water,
Dialogue: 0,0:01:46.60,0:01:48.97,Default,,0000,0000,0000,,it dissociates into sodium\Nions and chloride ions.
Dialogue: 0,0:01:48.97,0:01:50.68,Default,,0000,0000,0000,,So, it would be about 2 over 1.
Dialogue: 0,0:01:50.68,0:01:52.60,Default,,0000,0000,0000,,So, 1 -- for every 1\Nmol of sodium chloride,
Dialogue: 0,0:01:52.88,0:01:54.90,Default,,0000,0000,0000,,we get about two mols\Nof particles roughly.
Dialogue: 0,0:01:56.19,0:01:58.09,Default,,0000,0000,0000,,Come back to that in a second.
Dialogue: 0,0:01:58.25,0:02:00.12,Default,,0000,0000,0000,,Molality is the same\Nas we saw before.
Dialogue: 0,0:02:00.12,0:02:01.81,Default,,0000,0000,0000,,Remember the definition\Nof molality,
Dialogue: 0,0:02:01.81,0:02:04.99,Default,,0000,0000,0000,,it's mols of solute divided\Nby kilograms of solvent.
Dialogue: 0,0:02:05.89,0:02:10.12,Default,,0000,0000,0000,,And this boiling point elevation\Nconstant, you look it up,
Dialogue: 0,0:02:10.12,0:02:11.80,Default,,0000,0000,0000,,you don't have to\Nmemorize any of them.
Dialogue: 0,0:02:12.26,0:02:14.45,Default,,0000,0000,0000,,Just, the only thing -- the\Nimportant thing to remember is
Dialogue: 0,0:02:14.50,0:02:19.44,Default,,0000,0000,0000,,that you look it up based upon\Nwhat the solvent is, that's all.
Dialogue: 0,0:02:19.86,0:02:21.32,Default,,0000,0000,0000,,So, this van't Hoff factor.
Dialogue: 0,0:02:22.06,0:02:24.70,Default,,0000,0000,0000,,So, like we were saying,\Nif it's sodium chloride,
Dialogue: 0,0:02:25.15,0:02:28.94,Default,,0000,0000,0000,,we'd expect one formula unit\Nof sodium chloride to break
Dialogue: 0,0:02:28.94,0:02:32.26,Default,,0000,0000,0000,,up into two particles, the\Nsodium ion and the chloride ion,
Dialogue: 0,0:02:32.37,0:02:35.73,Default,,0000,0000,0000,,we'd expect it to be -- the\Nvan't Hoff factor to be 2.0.
Dialogue: 0,0:02:36.18,0:02:38.18,Default,,0000,0000,0000,,In fact, it's about 1.9.
Dialogue: 0,0:02:38.55,0:02:41.57,Default,,0000,0000,0000,,The reason for this is\Nsomething called ion pairing.
Dialogue: 0,0:02:41.57,0:02:44.58,Default,,0000,0000,0000,,So, when we put that sodium\Nion in the water, okay,
Dialogue: 0,0:02:44.58,0:02:46.86,Default,,0000,0000,0000,,most of the sodium ions and\Nthe chloride ions separate
Dialogue: 0,0:02:46.86,0:02:49.04,Default,,0000,0000,0000,,from each other, and\Nthey're surrounded by a sea
Dialogue: 0,0:02:49.04,0:02:50.69,Default,,0000,0000,0000,,of solvation by water molecules.
Dialogue: 0,0:02:50.92,0:02:53.38,Default,,0000,0000,0000,,However, some of\Nthem still say --
Dialogue: 0,0:02:53.38,0:02:54.88,Default,,0000,0000,0000,,stay associated with each other.
Dialogue: 0,0:02:55.19,0:02:57.27,Default,,0000,0000,0000,,And that means instead\Nof getting two particles,
Dialogue: 0,0:02:57.36,0:03:01.27,Default,,0000,0000,0000,,this whole thing here only\Ncounts as one particle.
Dialogue: 0,0:03:01.80,0:03:04.26,Default,,0000,0000,0000,,Magnesium chloride, we'd\Nexpect to get three particles,
Dialogue: 0,0:03:04.44,0:03:05.88,Default,,0000,0000,0000,,two chloride and one magnesium.
Dialogue: 0,0:03:06.27,0:03:08.73,Default,,0000,0000,0000,,So, we'd expect that\Nvan't Hoff factor
Dialogue: 0,0:03:08.73,0:03:10.84,Default,,0000,0000,0000,,to be three, in fact it's 2.7.
Dialogue: 0,0:03:11.34,0:03:14.34,Default,,0000,0000,0000,,And here, just for\Ncomparison notice that glucose,
Dialogue: 0,0:03:14.34,0:03:17.54,Default,,0000,0000,0000,,which is a molecular\Ncompound does not dissociate.
Dialogue: 0,0:03:17.54,0:03:18.99,Default,,0000,0000,0000,,So, we'd expect it's\Nvan't Hoff factor
Dialogue: 0,0:03:18.99,0:03:21.51,Default,,0000,0000,0000,,to be 1, and it actually is.
Dialogue: 0,0:03:22.60,0:03:24.99,Default,,0000,0000,0000,,So, freezing point\Ndepression works
Dialogue: 0,0:03:24.99,0:03:29.15,Default,,0000,0000,0000,,like boiling point elevation,\Nexcept that the freezing point
Dialogue: 0,0:03:29.15,0:03:32.75,Default,,0000,0000,0000,,of a solution is always\Nlower than the freezing point
Dialogue: 0,0:03:32.75,0:03:37.70,Default,,0000,0000,0000,,of the pure solvent when you\Nadd a non-volatile solute to it.
Dialogue: 0,0:03:37.73,0:03:42.09,Default,,0000,0000,0000,,So, here delta T F, the change\Nin freezing point is equal
Dialogue: 0,0:03:42.09,0:03:43.62,Default,,0000,0000,0000,,to the freezing point\Nof the solution,
Dialogue: 0,0:03:43.62,0:03:46.27,Default,,0000,0000,0000,,minus the freezing point of\Nthe solvent -- pure solvent.
Dialogue: 0,0:03:46.88,0:03:48.61,Default,,0000,0000,0000,,Now, because this would\Nbe lower than this,
Dialogue: 0,0:03:48.65,0:03:50.60,Default,,0000,0000,0000,,this will be a negative\Nnumber delta T F.
Dialogue: 0,0:03:51.19,0:03:54.45,Default,,0000,0000,0000,,When you're putting it here\Nin this equation to use it,
Dialogue: 0,0:03:54.50,0:03:57.11,Default,,0000,0000,0000,,just make it positive, it'll\Nwork out better that way.
Dialogue: 0,0:03:57.48,0:03:59.52,Default,,0000,0000,0000,,Same van't Hoff factor,\Nsame volatility.
Dialogue: 0,0:03:59.83,0:04:01.40,Default,,0000,0000,0000,,The freezing point\Ndepression constant,
Dialogue: 0,0:04:01.40,0:04:04.29,Default,,0000,0000,0000,,again it depends only\Nupon what the solvent is.
Dialogue: 0,0:04:04.60,0:04:07.97,Default,,0000,0000,0000,,You look it up, you don't\Nhave to memorize any of them.
Dialogue: 0,0:04:08.31,0:04:09.47,Default,,0000,0000,0000,,So, why does this happen?
Dialogue: 0,0:04:09.73,0:04:12.15,Default,,0000,0000,0000,,Well this is a Phase\Ndiagram for water.
Dialogue: 0,0:04:13.06,0:04:15.25,Default,,0000,0000,0000,,Please pardon my attempt\Nto draw curves here.
Dialogue: 0,0:04:15.85,0:04:20.10,Default,,0000,0000,0000,,But, the idea is this, we\Nknow from the last module
Dialogue: 0,0:04:20.23,0:04:22.40,Default,,0000,0000,0000,,that by adding a solute
Dialogue: 0,0:04:22.72,0:04:25.12,Default,,0000,0000,0000,,to a solvent we lower\Nthe vapor pressure
Dialogue: 0,0:04:25.36,0:04:26.70,Default,,0000,0000,0000,,of the resulting solution.
Dialogue: 0,0:04:27.31,0:04:30.51,Default,,0000,0000,0000,,And so, in this Phase diagram,\Nremember it's solid over here,
Dialogue: 0,0:04:30.73,0:04:32.71,Default,,0000,0000,0000,,it's liquid here, and\Nits gas over here.
Dialogue: 0,0:04:33.33,0:04:35.82,Default,,0000,0000,0000,,So, this is pressure\Nversus temperature.
Dialogue: 0,0:04:36.24,0:04:37.47,Default,,0000,0000,0000,,As we go along this curve,
Dialogue: 0,0:04:37.56,0:04:40.89,Default,,0000,0000,0000,,the pressure along this curve\Nhere will be the vapor pressure
Dialogue: 0,0:04:41.35,0:04:43.57,Default,,0000,0000,0000,,of that liquid, in\Nthis case water.
Dialogue: 0,0:04:44.18,0:04:47.49,Default,,0000,0000,0000,,So, by adding -- so, we know\Nby adding the solute we go
Dialogue: 0,0:04:47.49,0:04:48.28,Default,,0000,0000,0000,,over the vapor pressure.
Dialogue: 0,0:04:48.28,0:04:52.02,Default,,0000,0000,0000,,That's why the green curve\Nwhich it the vapor pressure
Dialogue: 0,0:04:52.02,0:04:56.16,Default,,0000,0000,0000,,of the solution, is lower\Nthan the white curve,
Dialogue: 0,0:04:56.16,0:04:57.76,Default,,0000,0000,0000,,which is the vapor\Npressure of pure water.
Dialogue: 0,0:04:58.39,0:05:01.19,Default,,0000,0000,0000,,So, green is the vapor\Npressure of the solution,
Dialogue: 0,0:05:01.47,0:05:02.74,Default,,0000,0000,0000,,white is the vapor\Npressure of water.
Dialogue: 0,0:05:03.72,0:05:06.52,Default,,0000,0000,0000,,And so, if we look at, okay,\Nthis is one atmosphere,
Dialogue: 0,0:05:06.63,0:05:09.30,Default,,0000,0000,0000,,the pressure -- the\Nexternal pressure, right?
Dialogue: 0,0:05:09.64,0:05:15.25,Default,,0000,0000,0000,,Let's say that the atmospheric\Npressure was one atmosphere.
Dialogue: 0,0:05:15.71,0:05:18.30,Default,,0000,0000,0000,,The solution or the\Nsolvent will boil
Dialogue: 0,0:05:18.61,0:05:23.12,Default,,0000,0000,0000,,when its vapor pressure is\Nequal to the external pressure.
Dialogue: 0,0:05:23.68,0:05:25.93,Default,,0000,0000,0000,,So, when the vapor pressure\Nreaches one atmosphere,
Dialogue: 0,0:05:26.27,0:05:27.69,Default,,0000,0000,0000,,that temperature is\Nthe boiling point.
Dialogue: 0,0:05:28.17,0:05:31.31,Default,,0000,0000,0000,,Because this is -- this curve\Nis higher than this curve,
Dialogue: 0,0:05:31.77,0:05:35.28,Default,,0000,0000,0000,,we reach that point at a lower\Ntemperature for the pure solvent
Dialogue: 0,0:05:35.61,0:05:37.07,Default,,0000,0000,0000,,than we do for the solution.
Dialogue: 0,0:05:37.66,0:05:40.47,Default,,0000,0000,0000,,This difference here is the\Nchange in the boiling point --
Dialogue: 0,0:05:40.47,0:05:43.01,Default,,0000,0000,0000,,the boiling point elevation\Ngoing from here pure solvent
Dialogue: 0,0:05:43.25,0:05:45.86,Default,,0000,0000,0000,,to here , the solution\NSame thing works
Dialogue: 0,0:05:45.86,0:05:48.10,Default,,0000,0000,0000,,with the freezing point,\Nonly the opposite way.
Dialogue: 0,0:05:48.60,0:05:51.92,Default,,0000,0000,0000,,This is a temperature at which,\Nyou go from liquid to solid
Dialogue: 0,0:05:51.92,0:05:53.41,Default,,0000,0000,0000,,or solid to liquid,\Nthe freezing point.
Dialogue: 0,0:05:53.94,0:05:55.93,Default,,0000,0000,0000,,So, from pure solvent,\Nin this case water,
Dialogue: 0,0:05:55.93,0:05:57.93,Default,,0000,0000,0000,,it would be this\Ntemperature right here.
Dialogue: 0,0:05:58.66,0:06:02.88,Default,,0000,0000,0000,,The lowering of the vapor\Npressure of the solution means
Dialogue: 0,0:06:02.88,0:06:04.78,Default,,0000,0000,0000,,that it freezes at a\Nlower temperature here.
Dialogue: 0,0:06:05.37,0:06:07.17,Default,,0000,0000,0000,,And this difference\Nhere would be the change
Dialogue: 0,0:06:07.17,0:06:07.89,Default,,0000,0000,0000,,in the freezing point.
Dialogue: 0,0:06:10.71,0:06:14.13,Default,,0000,0000,0000,,Now, we -- remember\Nwhen we talked
Dialogue: 0,0:06:14.13,0:06:17.40,Default,,0000,0000,0000,,about vapor pressure lowering\Nby an addition of the solute.
Dialogue: 0,0:06:17.65,0:06:21.74,Default,,0000,0000,0000,,We -- and talked about\Nhow entropy is involved,
Dialogue: 0,0:06:21.92,0:06:23.83,Default,,0000,0000,0000,,it still applies here,\Nit's the same reasoning.
Dialogue: 0,0:06:23.83,0:06:24.57,Default,,0000,0000,0000,,All right.
Dialogue: 0,0:06:25.20,0:06:29.60,Default,,0000,0000,0000,,So, the other colligative\Nproperty is osmotic pressure.
Dialogue: 0,0:06:29.66,0:06:32.09,Default,,0000,0000,0000,,This is supposed to be a\Npi, we call it pi capital.
Dialogue: 0,0:06:32.80,0:06:34.57,Default,,0000,0000,0000,,And osmotic pressure is equal
Dialogue: 0,0:06:34.57,0:06:36.59,Default,,0000,0000,0000,,to the van't Hoff\Nfactor, same as before.
Dialogue: 0,0:06:37.02,0:06:39.57,Default,,0000,0000,0000,,Now this M is the\NMolarity not the molality,
Dialogue: 0,0:06:39.66,0:06:42.66,Default,,0000,0000,0000,,watch out it's a capital M,\Ntimes the gas law constant,
Dialogue: 0,0:06:42.66,0:06:44.69,Default,,0000,0000,0000,,which we know from before\Nfrom the gas factor.
Dialogue: 0,0:06:45.07,0:06:47.21,Default,,0000,0000,0000,,And T is the temperature\Nin kelvin.
Dialogue: 0,0:06:48.29,0:06:52.90,Default,,0000,0000,0000,,R, remember is .08206 liters\Natmospheres per kelvin mol.
Dialogue: 0,0:06:53.35,0:06:55.49,Default,,0000,0000,0000,,And that's only four sig figs.
Dialogue: 0,0:06:56.46,0:06:58.50,Default,,0000,0000,0000,,So, what is osmotic pressure?
Dialogue: 0,0:06:58.50,0:07:03.88,Default,,0000,0000,0000,,Well, it ends up if you put\Na solution and separate it
Dialogue: 0,0:07:03.88,0:07:05.37,Default,,0000,0000,0000,,from some of its solvent
Dialogue: 0,0:07:05.61,0:07:07.71,Default,,0000,0000,0000,,by what's called a\Nsemi-permeable membrane,
Dialogue: 0,0:07:09.12,0:07:14.05,Default,,0000,0000,0000,,the column of liquid on the\Nside of the solution will rise,
Dialogue: 0,0:07:14.51,0:07:17.70,Default,,0000,0000,0000,,and -- but you can\Nstop it from rising
Dialogue: 0,0:07:17.70,0:07:19.48,Default,,0000,0000,0000,,by exerting an --\Na pressure on it.
Dialogue: 0,0:07:19.94,0:07:24.26,Default,,0000,0000,0000,,And that pressure for that\Nsolution is the osmotic pressure
Dialogue: 0,0:07:24.45,0:07:25.13,Default,,0000,0000,0000,,of that solution.
Dialogue: 0,0:07:26.61,0:07:28.21,Default,,0000,0000,0000,,So, if we look at it, this --
Dialogue: 0,0:07:28.21,0:07:31.86,Default,,0000,0000,0000,,what this semi-permeable\Nsolution does is it allows
Dialogue: 0,0:07:31.86,0:07:33.58,Default,,0000,0000,0000,,solvent particles\Nto pass through
Dialogue: 0,0:07:33.58,0:07:35.14,Default,,0000,0000,0000,,but not solute particles.
Dialogue: 0,0:07:35.54,0:07:38.22,Default,,0000,0000,0000,,So, what the -- what'll here\Nis just left to its own,
Dialogue: 0,0:07:38.54,0:07:40.42,Default,,0000,0000,0000,,the pure solvent\Nside is going to try
Dialogue: 0,0:07:40.51,0:07:43.39,Default,,0000,0000,0000,,to dilute the solution side.
Dialogue: 0,0:07:43.39,0:07:45.55,Default,,0000,0000,0000,,And it's -- what\Nit's trying to do is
Dialogue: 0,0:07:45.55,0:07:48.05,Default,,0000,0000,0000,,to get the concentration the\Nsame on both sides, zero,
Dialogue: 0,0:07:48.05,0:07:48.73,Default,,0000,0000,0000,,it can never get there.
Dialogue: 0,0:07:48.73,0:07:51.65,Default,,0000,0000,0000,,But what happens is, as\Nthe solvent particles pass
Dialogue: 0,0:07:51.65,0:07:53.27,Default,,0000,0000,0000,,through the semi-permeable\Nmembrane,
Dialogue: 0,0:07:53.99,0:07:56.71,Default,,0000,0000,0000,,this column of liquid\Nstarts increasing in height.
Dialogue: 0,0:07:56.71,0:07:59.44,Default,,0000,0000,0000,,And just like with the\Nmercury barometer we talked
Dialogue: 0,0:07:59.44,0:08:01.89,Default,,0000,0000,0000,,about in the gas chapter,\Nas it increases --
Dialogue: 0,0:08:01.89,0:08:03.59,Default,,0000,0000,0000,,as it goes up higher and\Nhigher in this column,
Dialogue: 0,0:08:04.05,0:08:06.87,Default,,0000,0000,0000,,gravity is exerting\Na force down,
Dialogue: 0,0:08:07.32,0:08:10.81,Default,,0000,0000,0000,,and thus creating a\Npressure at this barrier.
Dialogue: 0,0:08:11.20,0:08:15.09,Default,,0000,0000,0000,,And when it reaches a height\Nsuch that the pressure exerted
Dialogue: 0,0:08:15.09,0:08:16.53,Default,,0000,0000,0000,,down balances the pressure
Dialogue: 0,0:08:16.53,0:08:18.67,Default,,0000,0000,0000,,of the solvent coming\Nthrough, it stops rising.
Dialogue: 0,0:08:19.08,0:08:22.18,Default,,0000,0000,0000,,That difference is a measure\Nof the osmotic pressure.
Dialogue: 0,0:08:22.40,0:08:25.72,Default,,0000,0000,0000,,So, let's do an example.
Dialogue: 0,0:08:25.72,0:08:27.98,Default,,0000,0000,0000,,This one happens to deal with\Nfreezing point depression,
Dialogue: 0,0:08:28.56,0:08:30.09,Default,,0000,0000,0000,,but boiling point elevation
Dialogue: 0,0:08:30.09,0:08:32.75,Default,,0000,0000,0000,,and osmotic pressure\Ncalculations are pretty similar,
Dialogue: 0,0:08:32.75,0:08:33.94,Default,,0000,0000,0000,,just remember those equations.
Dialogue: 0,0:08:34.78,0:08:37.83,Default,,0000,0000,0000,,So, we have 1 gram of quinine\N-- quinine, it's a molecule,
Dialogue: 0,0:08:37.83,0:08:39.95,Default,,0000,0000,0000,,a molecular compound,\Nis dissolved
Dialogue: 0,0:08:39.95,0:08:42.75,Default,,0000,0000,0000,,in 10 grams of cyclohexane.
Dialogue: 0,0:08:42.75,0:08:44.57,Default,,0000,0000,0000,,So, quinine is a\Nmolecular compound
Dialogue: 0,0:08:44.57,0:08:45.97,Default,,0000,0000,0000,,as is cyclohexane an enzyme.
Dialogue: 0,0:08:46.59,0:08:47.31,Default,,0000,0000,0000,,The freezing point
Dialogue: 0,0:08:47.31,0:08:50.52,Default,,0000,0000,0000,,of the resulting solution\Nis .24 degrees Celsius,
Dialogue: 0,0:08:50.52,0:08:51.76,Default,,0000,0000,0000,,almost 0 not quite.
Dialogue: 0,0:08:52.20,0:08:53.52,Default,,0000,0000,0000,,What's the normal\Nmass of quinine?
Dialogue: 0,0:08:53.66,0:08:57.29,Default,,0000,0000,0000,,If we have pure cyclohexane\Nit ends up that it freezes
Dialogue: 0,0:08:57.29,0:08:59.35,Default,,0000,0000,0000,,at 6.47 degrees Celsius.
Dialogue: 0,0:08:59.85,0:09:01.43,Default,,0000,0000,0000,,And the freezing point\Ndepression constant
Dialogue: 0,0:09:01.43,0:09:03.62,Default,,0000,0000,0000,,for the solvent which\Nis cyclohexane is this,
Dialogue: 0,0:09:03.62,0:09:05.90,Default,,0000,0000,0000,,20.2 kelvins per molal.
Dialogue: 0,0:09:06.33,0:09:08.55,Default,,0000,0000,0000,,So, here our approach\Nis going to be,
Dialogue: 0,0:09:08.55,0:09:10.65,Default,,0000,0000,0000,,we're going to use the freezing\Npoint depression equation.
Dialogue: 0,0:09:14.49,0:09:16.72,Default,,0000,0000,0000,,And we're going to solve\Nit for the molality.
Dialogue: 0,0:09:17.69,0:09:21.29,Default,,0000,0000,0000,,Because -- okay, so, the freeze\N-- first the freezing point,
Dialogue: 0,0:09:21.29,0:09:22.38,Default,,0000,0000,0000,,it's the freezing point
Dialogue: 0,0:09:22.78,0:09:24.74,Default,,0000,0000,0000,,of the solution minus\Nthe freezing point
Dialogue: 0,0:09:24.74,0:09:25.71,Default,,0000,0000,0000,,of the pure solid.
Dialogue: 0,0:09:26.10,0:09:28.57,Default,,0000,0000,0000,,Which is, .24 minus\N6.47 degrees,
Dialogue: 0,0:09:28.57,0:09:31.32,Default,,0000,0000,0000,,it's negative 6.23\Ndegrees Celsius.
Dialogue: 0,0:09:32.45,0:09:36.50,Default,,0000,0000,0000,,And because, the\Ndifference in the temperature
Dialogue: 0,0:09:36.65,0:09:39.56,Default,,0000,0000,0000,,in Celsius is the same as\Nthe difference in kelvin,
Dialogue: 0,0:09:39.85,0:09:42.07,Default,,0000,0000,0000,,this is also the difference\Nin the freezing point,
Dialogue: 0,0:09:42.07,0:09:45.87,Default,,0000,0000,0000,,the change in the freezing point\Nis also negative 6.23 Kelvin.
Dialogue: 0,0:09:45.97,0:09:47.38,Default,,0000,0000,0000,,Think about that if\Nyou're not sure why
Dialogue: 0,0:09:47.38,0:09:48.26,Default,,0000,0000,0000,,that is, but that's true.
Dialogue: 0,0:09:49.26,0:09:51.83,Default,,0000,0000,0000,,And so, now we have the freezing\Npoint depression equation,
Dialogue: 0,0:09:51.83,0:09:53.64,Default,,0000,0000,0000,,we're going to solve\Nfor the molality.
Dialogue: 0,0:09:54.12,0:09:59.10,Default,,0000,0000,0000,,Here, because quinine is a\Nmolecular compound, i is 1.
Dialogue: 0,0:09:59.10,0:09:59.93,Default,,0000,0000,0000,,So, we don't have\Nto worry about it.
Dialogue: 0,0:10:00.05,0:10:05.98,Default,,0000,0000,0000,,If we solve for the molality,\Njust divide the change
Dialogue: 0,0:10:05.98,0:10:08.90,Default,,0000,0000,0000,,in freezing point, we're going\Nto make it positive, by the,
Dialogue: 0,0:10:09.32,0:10:11.77,Default,,0000,0000,0000,,well i, the van't Hoff factor\Ntimes the freezing point
Dialogue: 0,0:10:11.77,0:10:13.54,Default,,0000,0000,0000,,depression constant, i is 1.
Dialogue: 0,0:10:14.59,0:10:17.60,Default,,0000,0000,0000,,We saw that k is\N-- k sub F is 20.2,
Dialogue: 0,0:10:17.60,0:10:19.09,Default,,0000,0000,0000,,I gave you that T\Nin the problem.
Dialogue: 0,0:10:19.44,0:10:22.26,Default,,0000,0000,0000,,Do the division, we get\N.308, three sig figs,
Dialogue: 0,0:10:22.32,0:10:24.44,Default,,0000,0000,0000,,first insignificant\Nfigure is a 4, molality.
Dialogue: 0,0:10:24.98,0:10:26.57,Default,,0000,0000,0000,,So, that's the molality\Nof the solution But,
Dialogue: 0,0:10:26.57,0:10:29.37,Default,,0000,0000,0000,,we want the molar\Nmass of the quinine.
Dialogue: 0,0:10:29.52,0:10:33.47,Default,,0000,0000,0000,,So, what we're going to do\Nis, because the definition
Dialogue: 0,0:10:33.47,0:10:36.48,Default,,0000,0000,0000,,of molality is mols of solute\Nover kilograms of solvent.
Dialogue: 0,0:10:37.02,0:10:38.83,Default,,0000,0000,0000,,And we know how many kilograms
Dialogue: 0,0:10:38.83,0:10:41.30,Default,,0000,0000,0000,,of solvent, which\Nis cyclohexane.
Dialogue: 0,0:10:41.44,0:10:46.29,Default,,0000,0000,0000,,We had 10.0 grams, which\Nis 0.1000 kilograms --
Dialogue: 0,0:10:46.36,0:10:47.94,Default,,0000,0000,0000,,two zeros after that.
Dialogue: 0,0:10:47.94,0:10:51.81,Default,,0000,0000,0000,,If we multiply the molality\Nthat we just figured
Dialogue: 0,0:10:51.81,0:10:55.37,Default,,0000,0000,0000,,out times the kilograms of\Nsolvent, we get mols of solute.
Dialogue: 0,0:10:55.50,0:10:58.59,Default,,0000,0000,0000,,So, this is mols of\Nquinine, the solute.
Dialogue: 0,0:10:58.73,0:11:03.94,Default,,0000,0000,0000,,And all we have to do now to get\Nthe molar mass is take the mass
Dialogue: 0,0:11:03.94,0:11:07.26,Default,,0000,0000,0000,,of the quinine in grams divided\Nby the mols, grams per mol.
Dialogue: 0,0:11:07.26,0:11:11.02,Default,,0000,0000,0000,,So, the molality was 1 gram\Ndivided by that many mols.
Dialogue: 0,0:11:11.02,0:11:13.84,Default,,0000,0000,0000,,When we do the division\Nwe get to three sig figs,
Dialogue: 0,0:11:13.84,0:11:16.49,Default,,0000,0000,0000,,and you are 3.24 grams per mol.
Dialogue: 0,0:11:16.63,0:11:17.26,Default,,0000,0000,0000,,And there you go.