0:00:00.000,0:00:00.790


0:00:00.790,0:00:04.310
Let's think about what might[br]happen to the boiling point or

0:00:04.310,0:00:08.550
the freezing point of any[br]solution if we start adding

0:00:08.550,0:00:10.860
particles, or we start[br]adding solute to it.

0:00:10.860,0:00:14.470
For our visualization, let's[br]just think about water again.

0:00:14.470,0:00:15.550
It doesn't have to be water.

0:00:15.550,0:00:17.880
It can be any solvent, but let's[br]just think about water

0:00:17.880,0:00:19.640
in its liquid state.

0:00:19.640,0:00:24.100
The particles are reasonably[br]disorganized because of their

0:00:24.100,0:00:27.150
kinetic energy, but they still[br]have that hydrogen bonds that

0:00:27.150,0:00:30.340
wants to make them be[br]near each other.

0:00:30.340,0:00:32.240
So this is in the liquid[br]state, and they have a

0:00:32.240,0:00:34.740
reasonable amount of[br]kinetic energy.

0:00:34.740,0:00:37.120
You know, each of these[br]particles is moving in some

0:00:37.120,0:00:39.460
direction, rubbing against each[br]other, bouncing off of

0:00:39.460,0:00:40.490
each other.

0:00:40.490,0:00:44.910
Now, to move it into the solid[br]state, or to freeze it, what

0:00:44.910,0:00:46.250
has to happen?

0:00:46.250,0:00:48.620
The ice has to enter kind of[br]a crystalline structure.

0:00:48.620,0:00:51.710
It has to get pretty organized,[br]so let's say it has

0:00:51.710,0:00:54.530
to look something like this.

0:00:54.530,0:00:56.610
The water molecules are going[br]to have a regular structure

0:00:56.610,0:01:01.460
where the hydrogen bonds[br]dominate any kind of kinetic

0:01:01.460,0:01:03.340
movement they want to do, and[br]all the kinetic movement,

0:01:03.340,0:01:04.980
they're just vibrating[br]in place.

0:01:04.980,0:01:07.840
So you have to get a[br]little bit orderly

0:01:07.840,0:01:10.060
right there, right?

0:01:10.060,0:01:12.310
And then, obviously, this[br]lattice structure goes on and

0:01:12.310,0:01:17.150
on with a gazillion[br]water molecules.

0:01:17.150,0:01:19.470
But the interesting[br]thing is that this

0:01:19.470,0:01:20.880
somehow has to get organized.

0:01:20.880,0:01:25.540
And what happens if we start[br]introducing molecules into

0:01:25.540,0:01:26.250
this water?

0:01:26.250,0:01:30.130
Let's say the example of[br]sodium-- actually, I won't do

0:01:30.130,0:01:30.750
any example.

0:01:30.750,0:01:32.770
Let's just say some arbitrary[br]molecule, if I were to

0:01:32.770,0:01:35.070
introduce it there, if I[br]were to put something--

0:01:35.070,0:01:35.930
let me draw it again.

0:01:35.930,0:01:40.350
So now I'll just use that same--[br]I'll introduce some

0:01:40.350,0:01:43.300
molecules, and let's say they're[br]pretty large, so they

0:01:43.300,0:01:46.060
push all of these water[br]molecules out of the way.

0:01:46.060,0:01:48.900
So the water molecules are now[br]on the outside of that, and

0:01:48.900,0:01:52.890
let's have another one that's[br]over here, some relatively

0:01:52.890,0:01:55.860
large molecules of solute[br]relative to water, and this is

0:01:55.860,0:01:58.890
because a water molecule[br]really isn't that big.

0:01:58.890,0:02:02.310
Now, do you think it's going[br]to be easier or harder to

0:02:02.310,0:02:03.220
freeze this?

0:02:03.220,0:02:06.040
Are you going to have to remove[br]more or less energy to

0:02:06.040,0:02:08.199
get to a frozen state?

0:02:08.199,0:02:10.979
Well, because these molecules,[br]they're not going to be part

0:02:10.979,0:02:13.690
of this lattice structure[br]because frankly, they wouldn't

0:02:13.690,0:02:14.780
even fit into it.

0:02:14.780,0:02:16.990
They're actually going to make[br]it harder for these water

0:02:16.990,0:02:20.495
molecules to get organized[br]because to get organized, they

0:02:20.495,0:02:22.500
have to get at the right[br]distance for the hydrogen

0:02:22.500,0:02:23.330
bonds to form.

0:02:23.330,0:02:26.600
But in this case, even as you[br]start removing heat from the

0:02:26.600,0:02:33.000
system, maybe the ones that[br]aren't near the solute

0:02:33.000,0:02:35.440
particles, they'll start to[br]organize with each other.

0:02:35.440,0:02:39.020


0:02:39.020,0:02:41.500
But then when you introduce a[br]solute particle, let's say a

0:02:41.500,0:02:44.740
solute particle is sitting[br]right here.

0:02:44.740,0:02:47.340
It's going to be very hard for[br]someone to organize with this

0:02:47.340,0:02:50.240
guy, to get near enough for[br]the hydrogen bond to start

0:02:50.240,0:02:52.470
taking hold.

0:02:52.470,0:02:54.280
This distance would make[br]it very difficult.

0:02:54.280,0:02:57.220
And so the way I think about[br]it is that these solute

0:02:57.220,0:03:00.170
particles make the structure[br]irregular, or they add more

0:03:00.170,0:03:01.710
disorder, and we'll eventually[br]talk about

0:03:01.710,0:03:03.000
entropy and all of that.

0:03:03.000,0:03:05.500
But they make it more irregular,[br]and it's making it

0:03:05.500,0:03:08.130
harder to get into[br]a regular form.

0:03:08.130,0:03:12.100
And so the intuition is is that[br]this should lower the

0:03:12.100,0:03:13.960
boiling point or make[br]it-- oh, sorry,

0:03:13.960,0:03:14.780
lower the melting point.

0:03:14.780,0:03:27.440
So solute particles make you[br]have a lower boiling point.

0:03:27.440,0:03:30.300
Let's say if we're talking[br]about water at standard

0:03:30.300,0:03:33.870
temperature and pressure or at[br]one atmosphere then instead of

0:03:33.870,0:03:36.390
going to 0 degrees, you might[br]have to go to negative 1 or

0:03:36.390,0:03:38.330
negative 2 degrees, and we're[br]going to talk a little bit

0:03:38.330,0:03:39.800
about what that is.

0:03:39.800,0:03:43.400
Now, what's the intuition of[br]what this will do when you

0:03:43.400,0:03:45.080
want to go into a gaseous[br]state, when you

0:03:45.080,0:03:45.850
want to boil it?

0:03:45.850,0:03:49.320
So my initial gut was, hey, I'm[br]already in a disordered

0:03:49.320,0:03:53.390
state, which is closer to what[br]a gas is, so wouldn't that

0:03:53.390,0:03:54.620
make it easier to boil?

0:03:54.620,0:03:57.040
But it turns out it also makes[br]it harder to boil, and this is

0:03:57.040,0:03:58.170
how I think about it.

0:03:58.170,0:04:00.780
Remember, everything with[br]boiling deals with what's

0:04:00.780,0:04:04.020
happening at the surface, and[br]we talked about that in our

0:04:04.020,0:04:05.160
vapor pressure.

0:04:05.160,0:04:08.980
So at the surface, we said if[br]I have a bunch of water

0:04:08.980,0:04:15.390
molecules in the liquid state,[br]we knew that although the

0:04:15.390,0:04:19.000
average temperature might not[br]be high enough for the water

0:04:19.000,0:04:21.680
molecules to evaporate, that[br]there's a distribution of

0:04:21.680,0:04:22.480
kinetic energies.

0:04:22.480,0:04:24.580
And some of these water[br]molecules on the surface

0:04:24.580,0:04:27.070
because the surface ones[br]might be going

0:04:27.070,0:04:29.030
fast enough to escape.

0:04:29.030,0:04:33.340
And when they escape into vapor,[br]then they create a

0:04:33.340,0:04:34.510
vapor pressure above here.

0:04:34.510,0:04:37.550
And if that vapor pressure is[br]high enough, you can almost

0:04:37.550,0:04:40.910
view them as linemen blocking[br]the way for more molecules to

0:04:40.910,0:04:43.530
kind of run behind them as they[br]block all of the other

0:04:43.530,0:04:46.930
ambient air pressure[br]above them.

0:04:46.930,0:04:49.980
So if there's enough of them and[br]they have enough energy,

0:04:49.980,0:04:54.180
they can start to push back or[br]to push outward is the way I

0:04:54.180,0:04:57.410
think about it, so that more[br]guys can come in behind them.

0:04:57.410,0:05:00.910
So I hope that lineman analogy[br]doesn't completely lose you.

0:05:00.910,0:05:04.210
Now, what happens if you were[br]to introduce solute into it?

0:05:04.210,0:05:06.210
Some of the solute particle[br]might be down here.

0:05:06.210,0:05:09.170
It probably doesn't have much[br]of an effect down here, but

0:05:09.170,0:05:12.290
some of it's going to be[br]bouncing on the surface, so

0:05:12.290,0:05:15.020
they're going to be taking up[br]some of the surface area.

0:05:15.020,0:05:17.310
And because, and this is at[br]least how I think of it, since

0:05:17.310,0:05:19.590
they're going to be taking up[br]some of the surface area,

0:05:19.590,0:05:22.300
you're going to have less[br]surface area exposed to the

0:05:22.300,0:05:25.200
solvent particle or to the[br]solution or the stuff that'll

0:05:25.200,0:05:26.080
actually vaporize.

0:05:26.080,0:05:27.820
You're going to have a[br]lower vapor pressure.

0:05:27.820,0:05:33.610


0:05:33.610,0:05:36.250
And remember, your boiling[br]point is when the vapor

0:05:36.250,0:05:39.020
pressure, when you have enough[br]particles with enough kinetic

0:05:39.020,0:05:41.820
energy out here to start[br]pushing against the

0:05:41.820,0:05:44.340
atmospheric pressure, when the[br]vapor pressure is equal to the

0:05:44.340,0:05:45.800
atmospheric pressure,[br]you start boiling.

0:05:45.800,0:05:49.100
But because of these guys, I[br]have a lower vapor pressure.

0:05:49.100,0:05:51.170
So I'm going to have to add even[br]more kinetic energy, more

0:05:51.170,0:05:54.030
heat to the system in order to[br]get enough vapor pressure up

0:05:54.030,0:05:57.130
here to start pushing back[br]the atmospheric pressure.

0:05:57.130,0:06:02.440
So solute also raises[br]the boiling point.

0:06:02.440,0:06:05.450


0:06:05.450,0:06:08.230
So the way that you can think[br]about it is solute, when you

0:06:08.230,0:06:11.780
add something to a solution,[br]it's going to make it want to

0:06:11.780,0:06:13.830
be in the liquid state more.

0:06:13.830,0:06:15.520
Whether you lower the[br]temperature, it's going to

0:06:15.520,0:06:17.580
want to stay in liquid as[br]opposed to ice, and if you

0:06:17.580,0:06:19.000
raise the temperature, it's[br]going to want to stay in

0:06:19.000,0:06:20.530
liquid as opposed to gas.

0:06:20.530,0:06:22.460
I found this neat-- hopefully,[br]it shows up

0:06:22.460,0:06:23.850
well on this video.

0:06:23.850,0:06:26.205
I have to give due credit, this[br]is from chem.purdue.edu/

0:06:26.205,0:06:34.820
gchelp/solutions/eboil.html, but[br]I thought it was a pretty

0:06:34.820,0:06:37.690
neat graphic, or at least[br]a visualization.

0:06:37.690,0:06:39.970
This is just the surface of[br]water molecules, and it gives

0:06:39.970,0:06:43.060
you a sense of just how things[br]vaporize as well.

0:06:43.060,0:06:46.090
There's some things on the[br]surface that just bounce off.

0:06:46.090,0:06:48.970
And here's an example where[br]they visualized sodium

0:06:48.970,0:06:50.390
chloride at the surface.

0:06:50.390,0:06:52.910
And because the sodium chloride[br]is kind of bouncing

0:06:52.910,0:06:56.120
around on the surface with the[br]water molecules, fewer of

0:06:56.120,0:06:59.880
those water molecules kind of[br]have the room to escape, so

0:06:59.880,0:07:03.050
the boiling point[br]gets elevated.

0:07:03.050,0:07:06.080
Now, the question is by how[br]much does it get elevated?

0:07:06.080,0:07:08.900
And this is one of the neat[br]things in life is that the

0:07:08.900,0:07:11.730
answer is actually[br]quite simple.

0:07:11.730,0:07:17.860
The change in boiling or[br]freezing point, so the change

0:07:17.860,0:07:24.970
in temperature of vaporization,[br]is equal to some

0:07:24.970,0:07:31.120
constant times the number of[br]moles, or at least the mole

0:07:31.120,0:07:37.590
concentration, the molality,[br]times the molality of the

0:07:37.590,0:07:41.200
solute that you're putting[br]into your solution.

0:07:41.200,0:07:51.230
So, for example, let's say I[br]have 1 kilogram of-- so let's

0:07:51.230,0:07:55.790
say my solvent is water.

0:07:55.790,0:07:57.840
I'll switch colors.

0:07:57.840,0:08:00.110
And I have 1 kilogram of water,[br]and let's say we're

0:08:00.110,0:08:02.020
just at atmospheric pressure.

0:08:02.020,0:08:08.040
And let's say I have some[br]sodium chloride, NaCl.

0:08:08.040,0:08:14.170
And let's say I have[br]2 moles of NaCl.

0:08:14.170,0:08:15.420
I'll have 2 moles.

0:08:15.420,0:08:17.670


0:08:17.670,0:08:23.480
The question is how much will[br]this raise the boiling point

0:08:23.480,0:08:24.610
of this water?

0:08:24.610,0:08:31.070
So first of all, you just have[br]to figure out the molality,

0:08:31.070,0:08:36.299
which is just equal to the[br]number of moles of solute,

0:08:36.299,0:08:42.159
this 2 moles, divided[br]by the number of

0:08:42.159,0:08:43.570
kilograms of solvent.

0:08:43.570,0:08:51.610
So let's say we have 1[br]kilogram of solvent.

0:08:51.610,0:08:54.090
This was, of course, moles.

0:08:54.090,0:08:56.820
So our molality is 2[br]moles per kilogram.

0:08:56.820,0:08:59.840
So we just have to figure out[br]what this constant is, and

0:08:59.840,0:09:01.225
then we'll know the temperature[br]elevation.

0:09:01.225,0:09:03.430
And actually, that same[br]Purdue site, they

0:09:03.430,0:09:04.510
gave a list of tables.

0:09:04.510,0:09:07.146
I haven't run the experiments[br]myself.

0:09:07.146,0:09:08.760
They have some neat[br]charts here.

0:09:08.760,0:09:11.640
But they say, OK water, normal[br]boiling point is 100 degrees

0:09:11.640,0:09:15.200
Celsius at standard atmospheric[br]pressure.

0:09:15.200,0:09:19.740
And then they say that the[br]constant is 0.512 Celsius

0:09:19.740,0:09:21.360
degrees per mole.

0:09:21.360,0:09:24.070
So let's just say 0.5.

0:09:24.070,0:09:25.320
So it equals 0.5.

0:09:25.320,0:09:27.700


0:09:27.700,0:09:28.950
So k is equal to 0.5.

0:09:28.950,0:09:32.430


0:09:32.430,0:09:35.670
And I want to be very clear here[br]because this is a very--

0:09:35.670,0:09:38.040
I won't say a subtle point, but[br]it's an interesting point.

0:09:38.040,0:09:41.350
So I said that there's 2-- the[br]molality of-- I just realized

0:09:41.350,0:09:42.030
I made a mistake.

0:09:42.030,0:09:44.710
I said the molality of[br]sodium chloride is 2.

0:09:44.710,0:09:47.620
2 moles per kilograms. But[br]that would be if sodium

0:09:47.620,0:09:51.940
chloride stayed in this[br]molecular state, if it stayed

0:09:51.940,0:09:53.840
together, right?

0:09:53.840,0:09:56.150
But what happens is that the[br]sodium chloride actually

0:09:56.150,0:09:57.790
disassociates, and we learned[br]all about it in

0:09:57.790,0:09:59.430
that previous video.

0:09:59.430,0:10:04.750
Each molecule or each sodium[br]chloride pair disassociates

0:10:04.750,0:10:08.810
into two molecules,[br]into a sodium ion

0:10:08.810,0:10:12.430
and a chlorine anion.

0:10:12.430,0:10:15.820
And because of that, because[br]this disassociates into two,

0:10:15.820,0:10:18.950
the molality is actually going[br]to be two times the number of

0:10:18.950,0:10:22.670
moles of sodium chloride I have.[br]So it's going to be two

0:10:22.670,0:10:23.730
times this.

0:10:23.730,0:10:26.040
So my molality will[br]actually be 4.

0:10:26.040,0:10:26.890
And this is an interesting[br]point.

0:10:26.890,0:10:30.700
If I was dealing with--[br]and I wrote it here.

0:10:30.700,0:10:36.400
So this right here is glucose,[br]and this is sodium chloride,

0:10:36.400,0:10:38.370
or at least sodium chloride[br]in its crystal form.

0:10:38.370,0:10:43.080
One molecule, I guess you can[br]view it, or one salt of it.

0:10:43.080,0:10:45.310
I guess you could just view it[br]as one of these little pairs

0:10:45.310,0:10:46.320
right here.

0:10:46.320,0:10:48.780
But the interesting thing is[br]is you could have the same

0:10:48.780,0:10:51.320
number of moles of sodium[br]chloride when you view it as a

0:10:51.320,0:10:53.230
compound and glucose.

0:10:53.230,0:10:56.730
But glucose, when it goes into[br]water, it just stays as one

0:10:56.730,0:10:58.460
molecule of glucose.

0:10:58.460,0:11:01.990
So a mole of glucose will[br]disassociate into a mole of

0:11:01.990,0:11:03.170
glucose in water.

0:11:03.170,0:11:04.950
Well, I guess it won't[br]disassociate.

0:11:04.950,0:11:07.510
It'll just stay as one mole,[br]while a mole of sodium

0:11:07.510,0:11:10.490
chloride will turn into[br]two moles because it

0:11:10.490,0:11:11.480
disassociates.

0:11:11.480,0:11:13.890
It turns into two separate[br]particles.

0:11:13.890,0:11:18.780
So in my example, when I start[br]with a mole of this, I end

0:11:18.780,0:11:26.910
up-- actually, once I dissolve[br]it in water, I ended up with 4

0:11:26.910,0:11:28.890
moles per kilogram of molality,[br]because this turns

0:11:28.890,0:11:30.280
into two particles.

0:11:30.280,0:11:34.600
So given that the molality[br]is 4 moles.

0:11:34.600,0:11:38.240
2 moles of sodium, 2 moles[br]of chloride per kilogram.

0:11:38.240,0:11:42.270
So I just use that constant that[br]I just got from Purdue.

0:11:42.270,0:11:46.100
And I get the change in[br]temperature is equal to that

0:11:46.100,0:11:54.910
constant, 0.5, times 4, which[br]is equal to 2 degrees.

0:11:54.910,0:11:59.440
So my boiling point will be[br]elevated by 2 degrees.

0:11:59.440,0:12:03.970
Now, if I had the same number of[br]moles, if I had 2 moles of

0:12:03.970,0:12:08.460
glucose dissolved into my water,[br]I'd only get half as

0:12:08.460,0:12:10.610
much, half as much[br]of an increase.

0:12:10.610,0:12:13.150
Because the molality would[br]be half as much.

0:12:13.150,0:12:15.210
Because it doesn't turn[br]into two particles.

0:12:15.210,0:12:16.720
In some textbooks, you'll[br]actually see it

0:12:16.720,0:12:18.090
written like this.

0:12:18.090,0:12:22.990
You'll actually see the same[br]formula written like change in

0:12:22.990,0:12:25.860
boiling temperature, or vapor[br]temperature, or whatever you

0:12:25.860,0:12:30.250
want to think, is equal to k[br]times m times i, where they'll

0:12:30.250,0:12:34.270
say this is the molality[br]of the compound

0:12:34.270,0:12:35.110
you're talking about.

0:12:35.110,0:12:39.880
In this case, this number[br]would be 2, and i is the

0:12:39.880,0:12:43.330
number of molecules or the[br]number of things that it

0:12:43.330,0:12:44.640
disassociates into.

0:12:44.640,0:12:46.910
So in this case, this[br]would have been 2.

0:12:46.910,0:12:49.520
And that's where we would have[br]gotten 4 times k, which is

0:12:49.520,0:12:51.020
0.5, which is 2.

0:12:51.020,0:12:53.180
In the case of water, this would[br]be-- oh, sorry, in the

0:12:53.180,0:12:55.390
case of the glucose, this[br]would still be 2.

0:12:55.390,0:12:57.400
But it only turns into one[br]particle when it goes in the

0:12:57.400,0:12:58.640
water, so that would be 1.

0:12:58.640,0:13:02.370
So you would only have a 1[br]degree increase in the boiling

0:13:02.370,0:13:03.370
point of water.

0:13:03.370,0:13:06.460
Now, freezing point[br]is the same thing.

0:13:06.460,0:13:10.500
Change in freezing[br]point is also

0:13:10.500,0:13:13.180
proportional to the molality.

0:13:13.180,0:13:15.800
And you can either say the[br]molality of the original

0:13:15.800,0:13:18.690
non-in-water compound times[br]the number of compounds it

0:13:18.690,0:13:22.290
disassociates into, although[br]this k is going to be

0:13:22.290,0:13:26.440
different for freezing than[br]it is for boiling.

0:13:26.440,0:13:29.660
Of course, this k changes at[br]different pressures and for

0:13:29.660,0:13:30.370
different elements.

0:13:30.370,0:13:33.950
But the really big takeaway is[br]just to realize that even if

0:13:33.950,0:13:37.160
you have a mole of this and a[br]mole of that, and they're

0:13:37.160,0:13:39.810
going to be dissolved into the[br]same amount of water, because

0:13:39.810,0:13:42.160
this dissociates into two[br]particles and this

0:13:42.160,0:13:45.170
disassociates into only one[br]for every-- or this

0:13:45.170,0:13:47.620
disassociates into two moles for[br]every mole of the crystal

0:13:47.620,0:13:50.610
you have-- this doesn't[br]disassociate; it just stays as

0:13:50.610,0:13:53.810
one-- this'll have twice as[br]large of an effect on the

0:13:53.810,0:13:57.300
freezing point change or on the[br]boiling point elevation

0:13:57.300,0:13:59.320
than the glucose will.

0:13:59.320,0:13:59.737