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Okay, so let's start. Okay. Solutions.
Multiple choice, okay. Is a homogeneous mixture of two of more substances.
Okay? So solution is not water and sand. Okay?
That's not- the two of them aren't
in the same phase, okay? A solution is a mixture of two or more substances.
So look here. Gasses. What am I breathing?
What is the percentage of nitrogen in this room right now?
Who can—- who's got a— what's that?
78%! That's right. So we've got 78% of nitrogen, we've got oxygen, about 20%. Okay?
So that's a mixture. Okay? Gas or liquid.
Right here. A soda. Okay? It's got CO2 in it, and it's got a liquid.
Gas solid. You don't need to know that. That's kind of a weird one.
Liquid liquid. Ethanol and water. Why would they tell students who are 18
about mixing water- like, scotch and water? These people are nuts.
Solid liquid, okay? Salt in water. And then solid, solid, brass. Okay.
So anyway, there's several different types of ways to mix things up
and we're going to learn about mixing them, and in science
we have to say something about being able to quantify
what these things are. So the solution, the solute, is the substance
present in the smaller amounts And the solvent
is the one in the bigger amount. So if I take
one cup of ethanol and I add it to two cups
of water then the ethanol is the solute,
and the water is the solvent. Okay? If I like my drinks a little stronger,
and I have two cups of ethanol and one cup of water,
then ethanol is the solvent. Okay? You need to know the difference.
It's a nice, little quick question on an exam. The difference between solvent
and solute. Okay. Now this is just
once again a saturated solution is something that's got-
it's in equilibrium. Okay? It has got as much
stuff dissolved in it as possible. Okay? How many remember me talking about having
gone to the dead sea and swam in the dead sea? Remember that story? Okay.
There's about this much salt on the bottom of the dead sea.
Okay? So it is completely saturated. You cannot get any more salt in the water.
So, so much salt that it just falls out. So this is saturated.
Unsaturated is if there had been no salt on the bottom. Like the ocean.
Ocean's got saltwater, but it's not saturated. Okay? If it were saturated, we'd have a problem.
And super saturated, you can kind of get this by having something that's saturated, and
then cooling it down. Now that is a lead-in to what we're going
to be talking about today in terms of how to make things dissolve in water or other
things by changing the temperature. So once again, this is the kind of thing that
you can get by sort of tricking things a little bit.
And this is what they look like, okay? So- well, I guess this is what it looks like.
This is if you have a super saturated, they drop a little tiny piece of something in there,
BOOM. It just *snaps fingers* just like that. It
just forms up and makes a solid and a precipitate. So that would be a super saturated solution.
I tried to do this when I was a kid. Because I'm a scientist.
And I couldn't do it. You know? Too much crap floating around the house.
You get a little bit of lint or something, a piece falls in that little thing of salt,
it's got to be very, very clean for this experiment to work.
All right! So you're book does it a little differently,
this is a Hess's law thing. Okay? If I have a solvent, which is just water,
we know water has hydrogen bonding, we know water is happy to be just water, and then
you have a piece of salt, let's say. Okay? Or something. This could be salt, this
could be ethanol.
So you can have this in two steps or three steps; this breaks into pieces, this breaks
into pieces, and then it forms this, or you could just
draw a line from here to here and here to here. Okay?
The way it's drawn here, you have a certain amount of energy that you have to put into
this in order to separate it certain amount of energy that has to go into
here to separate it, and then you add these together and boom,
it all falls out. This is going to be exothermic. And so the overall equation looks like that.
Okay? And your book—-you can miss a step or two
what I really care about, folks, is that you can take this,
and take this, and make this. Okay? Sort of important.
Now this is something that you will have heard of before, but will particularly when you
get to organic chemistry, at least when I went to organic chemistry,
but this is a very important thing. "Like dissolves like."
Okay? Things that have similar chemical characteristics tend to dissolve in each other. Okay?
So once again, if you have something, let's say that we have
decane. Okay? That is C10H22. You don't need to write that
down. If I take some of that and then I have some
dodecane, which is C12H26, If I mix one with the other, *pshhht*
they mix completely. Okay? If instead I take decane, which is C10H22,
and I add water to it, they separate completely. Okay?
Because one is total dispersion, which is the carbon-hydrogen based one,
the other has dispersion, it's got hydrogen bonding, it's got dipole-dipole moment stuff.
And so water and oil, you've heard that before, water and oil don't mix. Okay?
They don't mix; they don't have the same or similar intermolecular forces.
So what this means, folks, is that if you had something,
if I had a stain right here, if I happened to drop
some grease or something right here on the floor,
water's not going to get it up. Okay? Water is not going to get it up.
I need to pour some gasoline or something on that thing.
Or some oil-based something so that the— all this dispersion stuff can go together
and and they grab on to each other and you sort
of wipe it off. Okay? So like dissolves like—it's huge in organic
chemistry. At least that's where I learned it.
So let me just put these things up here. So once again, it's just—
you have things that are similar, or that these have positive charges, negative
charges, this has a positive and negative end, so it's
going to dissolve in this. This is ethanol and water, and this is something
that has not got any hydrogen bonding, and so something like this which has dispersion
and this which has dispersion are going to be
we call them missable, okay. Or they are going to be able to be mixed together.
So you could be given a question on an exam that says:
Which of these are—you know—which of these would dissolve
this? Or in which would this be a solution? And you say, "okay, not water, not this, not
this, it would have to be something that was non-polar and
like that." Soap
Soap, folks, does this for us. Okay? So soap has one end that is greasy, and one
end that's like water. Okay? So you end up with this.
This end down here, you've got this positive Na,
we know that sodium chloride, NaCl will dissolve in water,
so what happens is we've got this long hydrocarbon tail.
Huge—this is bigger than decane. So this is good for grease. Okay?
You've got grease on your hands, you take some soap, you put it on there, and it's this
end, THIS END, that grabs on to that.
And then you have water washing over your hands,
and the water likes this. So this part sticks to the water, the water
rushes over your hand, and you clean up a bit. Okay?
Now I was telling some of the students who were in office hours that when,
because I grew up on a farm, that when we had to do all of our,
we had to fix all of the motors in the tractors and the lawnmowers, everything else,
and we would end up, my dad and I would end up with just—probably
just almost black grease, you know, from about here to here,
and when we would finish, my father would take a can of gasoline
and pour it on me. He called it a "gasoline bath."
No wonder he died of cancer and I had cancer, you know.
That's only half funny. And so you've got to wonder.
I was doing this, like, you know, once or twice a week, it seemed, like.
So anyway, but you know what, it was amazing. I had these horribly greasy arms and hands
and stuff, and my dad would pour some gas on a rag, and
we would do this, and it just came off like magic. Okay?
And that was because of this. Okay? Because the grease was hydrocarbon based,
and the gasoline was hydrocarbon based. So like dissolves like, and that's how soap
works, okay? Look at that. Grease, see how these little ends here that
are the greasy, the fatty ones, the hydrocarbon ends do this,
here's the ones that have the positive charge, and then the water just grabs right on to
those guys, and washes it away. Pretty cool.
Okay! So, solubility.
This is what solubility is. Okay? Is grease
soluble in water? No. Okay? It's not.
And in fact if I had some gasoline, If I had half water, half gasoline in here,
and I shook it and shook it and shook it, it wouldn't take long at all before the water
was on the bottom, because the density of water at room temperature is about 1,
the density of gasoline is about .8 So the .8 stuff would float.
So I could shake it up and I could make it look a little like it was mixed up, but it
would just be bubbles. It wouldn't take long before all the gasoline
was on top and very little— I mean, if we got rid of that, and then we
smelled the water, it would smell a little bit like gasoline. But there's very little
bit dissolved in there. Okay. So that's what we're calling solubility.
So if we look at this, generally speaking, these are all solids. So like a salt. So if
you have salts, and you put them in water, look at this one. Potassium nitrate. At 0
degrees, and this is degrees C, not K, okay, so 0 degrees C,
it's not very soluble. In fact, it's the least soluble of all of these. Okay? At 0 degrees.
But look what happens to it at 100 degrees. Holy smoke. It goes—
it's increase in solubility is like a factor of 20 or so. Okay?
Almost all of them increase as you warm things up. Okay?
Almost everything you can make more soluble as you warm up. Not everything, okay, look
at this. This one decreases as you get warmer, and
this one decreases. But rule of thumb,
rule of thumb is that things, the solubility of something, salts, solids, that type of
thing, increase as you warm them up. Okay? I cannot, on an exam, say "do they," but I
can say "do they usually." And the answer is yes. Okay? If you've got
something that you want to put into solution, warm it up. Okay?
You can only put so much -- I don't know if I told you this --
my wife loves to put, like, a ton of sugar in her coffee. And I try to explain to her
that it doesn't do it any good. Because it's solid sugar sitting on the bottom
of the cup. You cannot get any more -- she has it completely
saturated. Coffee's warm. Okay? So look at this. Even
at a warmer temperature, maybe she's drinking it somewhere in here,
which for some of these things wouldn't matter, but if she put potassium nitrate in there,
the solubility, if this were sugar, it would go up quite a
bit. I'm not suggesting that, okay. Don't tell
my wife. But as her coffee cools, guess what happens?
She has even a bigger pile of sugar in her coffee. Okay? Because of this, you move back
this way. If your KCl right here— I like this one. Sodium nitrate.
If this is where you're at 100, look at that. That's about—
there's 75, that's about 150— this is twice as soluble going from 0 to 100,
you get twice as much sodium nitrate in solution, in the water,
than you would at 0 degrees. Okay? So what that means is if I can get everything
dissolved in a cup of water, hot water, here, somewhere between here and
0 degrees, I've got to start having stuff fall out like it's rain.
It's going to go from the liquid phase into forming solid sodium nitrate
in the bottom of the pot. Okay? So you generally warm things up, solubility increases. There
are some outliers where that is not in fact the case.
So this is just a -- they've taken two examples from that last
one. This is sodium chloride, which doesn't seem to change much. Okay? The amount of sodium
chloride that you can get in at so we can get 34.2 grams per 100 grams of
water at 0 degrees, and it only goes to 38. So it's only, like, 10% more. 10% more soluble
as you warm it up here. This one is huge. Look at this. It goes up
by a factor of 9. And this has to do with lattice energy, solvation
energy, and so on as to why that actually happens.
And I say that here. Okay, well that's all stuff that you can get
off of the off of this itself.
Unfortunately, it's just the opposite for gases.
So all your life you've been able to do products minus reactants and
and then that enthalpies thing always said it was reactants minus products, kind of screwed
you up. Well this is just the opposite. Look at this.
This is the solubility of oxygen at 0 degrees, so it drops by more than a factor of 2 going
from 0 to 100. Now folks this,
even though it's not intuitive, it should be a little bit not a big surprise, okay,
if I have a soda, and I would have to do this, I would have to pop it first and let,
because when they ship it to you, and we'll have a problem on this,
when they ship it to you, there's a lot of carbon dioxide on the top.
Okay, but let's say we pop it like this, and let's say that we're sitting at, you know,
1 degree Celsius. Okay? So right above freezing. Pop it like this, it's going to make a sound
when I do this. And then say I take another one and pop it, okay. So now they both have
been vented. And now I put one of them back in the refrigerator,
and I put the other one right here. Okay. So now one of them still sits right here,
the other goes to here. Look at the difference. The CO2 is similar.
Look at the difference in the amount of CO2 that the water itself and the soda can maintain.
Okay? It can maintain this much, once again, this is oxygen, not CO2.
From here to there, that's a factor of like, 40% or so.
What that means is that if I were to then very carefully, in a well-done experiment,
do this, put this right next to this and go *twists cap off*
nothing. That was the cold one. Here's the warm one. *experiment fails* Dang.
Okay. It wasn't controlled enough.
The point is is that if we did have the cold and the warm,
you'd hear a bigger hiss coming out of the warm one because of this.
Okay? So once again for gases, folks, it's just the opposite.
So why do you think, why do you think this worries me?
Why would this plot—I showed you a plot earlier that was density versus temperature.
Okay? Density of water versus temperature and it looked like this. It maximized at 4
degrees. Anywhere on either side of 4 degrees, if you
went toward 4, you became more dense. If you moved away from 4, you became less dense.
Okay? Why does this scare Don? Yes. *student answers question*
Okay. Well he's -- okay. So in this case, that's a very good thing.
Well, it's a bad thing. But it's a good observation.
Um, yes. Since this is oxygen, what he's saying is that if we warmed up the oceans, the oceans
could not maintain as much oxygen as they have.
Okay. This could be a problem. Okay. I'm more worried about—let's say this
were CO2. The more CO2 would come out of the ocean.
CO2 is a greenhouse gas, and so if just by the temperature rising,
we force some CO2 that is right now happily in the water,
we force it out because we warm up, you know, 1 degree here,
then we kick out more CO2, CO2's a greenhouse gas,
warms things up even more. Okay? So it's a big concern and something
that we need to be worried about. But this is the general trend for gases in
solution. And it shouldn't—
so how many have ever made this observation? That if you let a soda sit open for a while,
as it warms up, it has less fizz. You ever notice that?
Okay. Well that's what's happening. Okay? We'll do some calculations. It will be more
fun than this part of the lecture. Okay. In your notes, it just says c is equal
to kP. I've actually added to this, because in your book they use s. Okay? It doesn't
matter. I prefer c because it's the concentration
in moles per liter. Okay? The solubility of a gas in liquid is proportional
to the pressure of the gas over the solution.
So folks, what that's saying is that if Don has a soda right here,
and I put a certain amount of CO2, I can re-fizz this thing, folks.
I can actually re-make my soda. Okay? That has lost the fizz by just putting a certain
pressure of CO2 in here. Okay? So you've got Henry's law constants,
and this is something that is temperature dependent, obviously, we saw
the last plot. So it's absolutely temperature dependent,
so you have to, I have to specifically say "at this temperature,
the Henry's law constant is x." Okay? So lots of Henry's law constants have
been calculated or determined, probably not the Henry's law constant for
diet pepsi. You know? So we can do some experiments to figure that
out. But this is the equation. Okay? It's a very
simple equation, it will be on the exam.
Okay? I promise, because I love Henry's law. So once again: c or s is the concentration
in moles per liter of dissolved gas.
Moles per liter of -- molarity. It's perfect. Okay?
P is the pressure of the gas over the solution, and k is a constant for each gas. Okay? That
depends only on the temperature. So look at -- look at this picture right here.
Look at the one on the left, we've got one, two, three, four, five, six,
seven, we've got eight gas molecules inside this,
and we have four, then, down here. If I increase the pressure up here,
like I have over here, I haven't counted those, but maybe there's
16, I don't know, then you end up with more in here.
So the higher the pressure here, the more you're actually going to be able
to cram into here. And that is totally dependent on the Henry's law constant. Okay?
If the Henry's law constant were twice as big,
then if we had 8 here, we'd have 8 here. Okay? But as it is right now for this particular
one, the Henry's law constant—'cause look at the concentration.
Okay. The concentration is directly proportional to the Henry's law constant.
So if I make the Henry's law constant twice as big, the concentration gets bigger.
If I don't change the Henry's law constant, but I have more pressure,
I can make this. So I can dial a concentration just by determining how much of a particular
gas I put up here. You better write that down.
Sometimes it is just c = kP, I put the h in there just because we have a lot of k's in
science, and so this is the Henry's law. Hmm? Hmm? How many of you have ever done that?
Yeah. Do this.
So read this, I think I'm nice enough to give you—
yes. So read that, it says how many grams of carbon dioxide gas is dissolved in a 1
Liter bottle of carbonated water if the manufacturer uses a pressure of 2.4 atmospheres in the
bottling process at 25 degrees C? Given that the Henry's law constant for CO2
is 0.034 moles per liter atmosphere at 25 C. Okay? So I'm going to go to the next page,
which has then got the equation, and it's got all the information you need.
Here we are. So concentration is equal to kP.
Okay, take those calculators out, folks. Don't cheat yourself.
You're paying for this education—at least, somebody is.
So folks, there are a lot of very simple equations that we're going to be doing.
Okay? There are going to be four that I will show you later,
maybe today, maybe not, and they are all this straightforward and
simple. Okay? So
this kind of a problem on an exam is probably, like, 4 points.
I mean, it is a plug and chug. There's absolutely nothing here that's exciting.
You've got k, you've got P. And look at the look at the units. If you don't know what
to do, look at the units. Okay? The reason this is in moles per liter atmosphere
is because you're going to—is that— because this is going to be a pressure in
atmospheres. Okay?
So what do we get? What's the answer? look at this.
This is equal to that times that equal to that.
Pretty simple? Okay, now I'm going to give it a little twist.
Okay? Here we go. Same everything, except this time I'm going
to say that the pressure is 1,000 torr.
Now do this problem. 1,000 torr is P.
1,000 torr. How many need a little help?
So everybody knows that there's 760 torr in an atmosphere.
So 1,000 torr is about 1 and 1/3 atmospheres. So I'm predicting here that the number is
going to be about, like, 0.05
47! Okay. So folks, you must convert
if you're given two different kinds of units. You can either change this
to moles per liter torr, or you can turn my value that I gave you here,
instead of P is equal to 2.4 atmospheres, or what did I say, 1. or 1,000 torr,
you can convert torr to atmospheres. There's either way, okay?
All right, so let's move on. Don't forget, that was not the whole question.
So this is the moles per liter, or the molarity. Okay?
So remember, we had 1 liter. We had 1 liter of soda.
Okay? So the question was how many grams of CO2?
Do it. CO2. Remember? Carbon is 12, oxygen is 16.
So there it is. So this is a multi-step question, but it's
quite simple. Okay? The question is how many grams were in that
drink, how many grams of CO2,
look at the units, once again, if you can get something in moles
per liter, folks, you're home. Okay? So you look at the units and you go, okay, well this
was in atmospheres, this is in liter atmospheres, I can make this into molarity,
and then you have 1 liter, so 1 liter, it says somewhere here, and then
you just do your simple little math here, and you come up with about 3 and a half grams.
Okay? Straightforward and simple.
Okay. Do this one. It says estimating solubility.
Verify that the concentration of oxygen in lake water is normally adequate to sustain
aquatic life, which requires oxygen concentrations of at least 0.13 millimoles per liter.
Okay, the partial pressure of oxygen, so why do we say the partial pressure of oxygen
is 0.21? Partial pressure means what is the percentage
of that particular gas. Okay? Because the fish don't care about nitrogen.
The fish want oxygen. Well, we have 21% oxygen in the air. So that means that the amount
in one atmosphere, the .21 of it, is oxygen. So that's the P that you're going to use.
Okay? So all you've got to do is to calculate what
is the actual concentration of oxygen in this
lake water, and is it greater than 0.13? If it is, then the fish are happy.
If not, the fish die. Or at least are unhappy.
Okay, this is the simple part, so I'm going to move on to the next page.
There's the answer. Everybody get that? Once again, simple.
Henry's law constant times the pressure, this many millimoles. Okay?
So 2.7 is certainly greater than 1.3 So don't forget, look at the units.
This is 2.7*10^(-4) This is also.
This is 1.3*10^(4). Millimoles means 10^(-3). Okay?
So this is they're kind of playing with you a little
bit, they just want you to be able to convert this
millimoles, Once again, milli meaning a thousandth, so
a thousandth of a mole, .13 thousandths of a mole, or .13 millimoles,
or millimolar, and then this they didn't bother, they just
put it in scientific notation. So we clearly have more than the—
we have twice as much oxygen. Okay? Any questions? Pretty simple? Okay.
So we have different ways we have to, in science, have a way to
for one scientist to another to say, "there's this much of this in this."
Okay? And grams per something or whatever doesn't always help.
So we tend to go with either percent by mass, which I'm not really big on,
but certainly mole fraction. The mole fraction, folks,
is where it's at. I mean, this is simple. Okay?
If I say that I've got 50 grams of sodium chloride in 100 grams of water,
then the percent by mass of sodium chloride is .3333---
or 33.33%. Okay, that's like 6th grade math. Okay? But it doesn't tell us a whole lot.
Because we work in a world where moles go together. You need one one mole of this plus
two moles of this to make one mole of that. It's not that you need 6 pounds of this and
12 pounds of this. Okay?
How many of you have ever baked a cake? Raise your hand.
Okay. Let me give you a scenario and you tell me if you think—those bakers
in this room— if this sort of makes sense.
So you need 1 lb of flour am I doing okay?
One pound of flour, 1 cup of milk,
and 3 eggs. I don't know what I'm going to make with that,
but I'm going to make something, okay? So I hope that sounds like something you might
read in a recipe book. What you don't get is
1 lb of flour, 1lb of milk, and 1 lb of eggs. Eggs don't come in pounds, right?
It's a molar thing. You want 3 eggs. 3 eggs per cake. Okay?
So it's not like you're adding masses together. You actually need an egg, okay?
So that's where moles are important. So that's why this mole fraction is important.
It is the number of moles of something over the total number of all the moles.
Okay? So there are 454 seats in this room, and there's at least 50 of them that are empty.
So I'm going to assume that there are 400 people in this room right now. Okay?
I know there's at least one person in this room named Don. Okay?
Is anybody else's name Don in this room? Is there a Don?
What is wrong with this generation? What's wrong with your parents?
I mean, Don was like, you know, when I was a little kid in the 50's
in class, there might be two of us in each classroom
of, like, 30 kids. Don and John and Bob and Bill,
or my sisters. Nancy, Judy, and Pattie. You know?
And now, I don't know. Okay, so, what is the mole fraction of Dons
in this room right now? Do it!
1 over 400. Yeah. Okay? So that's maybe important to know.
Okay? That's how you do mole fractions. Okay? What is the mole fraction of Dons and Ralphs
in the room? There's probably no one else in this room
called Ralph. Although, I don't know. Besides you.
Good looking, yes, yes. Is anybody named Ralph in this room?
Besides you, okay. So there's two of us.
So the mole fraction is 2 over 100. And there's probably only one Josette.
Probably. But there's probably more than one Brittney.
and one Jasmine. and one... let me come up with a guy's name.
Andy! How many Andy's? Yeah, look at this. One, two, three, four,
five Andy's. Okay? I knew Andy Effenberger, that was it.
He was the only Andy in my school, I think. The point is that sometimes you need mole
fractions. The mole fraction is quite easy, and it is the mole fraction, folks, that is
going to determine a lot. So I'm hoping that you understand mole fraction because if you
don't, there's a big truck coming right at you.
Okay, molarity. You learned this I hope in high school, I hope that it's something that's
easy, it is moles of solute, okay, over a liter of solution.
Okay? What that means is that if this were a liter bottle,
and I put one mole of something into it, and I fill this liter bottle up to exactly
one liter, after I had added this stuff, that is one
molar. Okay? Per liter. So molarity is an easy one, I prefer
molarity when I'm doing problems, but... and then there's molality. Which is moles
of solute, same thing on top, except this is mass of solvent. Okay? So that means you
start with one liter, or let's say 1,000 grams, and then you add
stuff to it. Okay, so you don't end up with a liter, you
have a certain value, and it's in kilograms. Okay?
Now what's important for you to understand, because of my obsession with the density of
water, okay, is that if I have 1 liter of water,
if I have one liter of water, guess what, if it fills right to the top of the 1 liter,
you know what I can make it do? I can make it overflow. By warming it up.
Okay? because the density at 60 degrees is maybe
.98 I can get rid of 20 cc's of water by changing
the temperature. Okay? So the molarity can actually change
by warming it up. The number of moles doesn't change here, and
the number of molecules doesn't change, but the volume changes.
In this case it doesn't. A kilogram stays a kilogram.
Okay? So this is a more— a non-dynamic value.
And we use it in pretty much all of the calculations and the equations that you're going to be
going over not just today, but next week.
Don't forget there's an exam a week from today. And it's really long.
It's getting shorter though, right? We're going to work on it.
Okay. So this is what your book gives me, folks. You paid for it, I've got to give it
to you. You know? I would love to not do this.
I'm not sure...okay. So in this case, folks,
mole solute and a kilogram of solvent. What are we doing here? Molality or molarity?
Molality, right? Because we've got a kilogram of solvent.
Okay? This is a mole fraction easy to do
as long as we know how many moles of this we have
and how many moles of this we have we can come up with a mole fraction. Not a
problem at all. If instead we have some of this, we add it
to this, and it comes up to be exactly 1 liter, now we've got moles per liter.
Not a liter here, in the end, a liter. Okay? So it's after you've done the adding you have
one liter then that is the molarity.
Okay? That's the difference, okay? So a clue for the most part is going to be
if you're given something in kilograms, particularly solvent-wise, you're going to
end up with molality if you've got something, and you were given
this many moles per liter, then obviously you're talking about molarity.
Okay. Shown is the molar concentration of sodium
chloride in water as a function of temperature. Which is the corresponding plot for molality?
Okay, so once again as we increase the temperature
on the liquid the density decreases, therefore the volume
increases therefore the molarity drops.
Okay? Look, there's the molarity right there, dropping with temperature.
Which one is the molality? Right there.
Because a kilogram warmed up is still a kilogram. Okay? So this is the molality, this is the
molarity. Ta-da.
That is why I am a distinguished professor. Okay? Right there. That is—that is it.
I know molality and molarity. Do this one!
So I'll read it, and then I'm going to go to the next page where I think I'm giving
you some actual information, and try to be nice.
What is the molality of the sugar fructose, that's it right there, in a solution prepared
by dissolving 90.5 grams of fructose in 250. grams of water? Now folks—
if you see this on an exam, I never do this. Point. Okay? That point means that you've
got three significant figures. Okay? If I just said 250, you'd have two significant
figures, because the zeroes don't count unless there's a decimal point. Okay?
So that has three sig figs there, that has three sig figs here, so that means you darn
well come up with three sig figs for fructose. Okay?
Let's go to the next page. So you are going to do the molality.
I guess I didn't—son of a gun, I could've sworn I did that.
So, figure out yourself. Carbon, so maybe we have to revert to two
sig figs, 12, 1, 16. Figure the molecular weight out
with that. Okay! Sorry. I got carried away here.
So folks! Once you've figured out what that number is,
which I think it's like 180, wouldn't you say it's about half of that?
Or twice that? So you've got about half a mole.
Then you look at this and go, hmmm, I have a quarter of a kilogram. The answer's got
to be close to 2. So you should be able to go boom, boom, boom
just to see if you got an answer that makes sense
Okay? So once again, 90.5 divided by 180
over, this is a quarter, so you end up with two moles per kilogram
which means the molality is 2.01 or 2 if you didn't have all these extra sig figs
from this, okay? Is that simple? Simple and boring.
Any questions? Okay.
Do this one. So let me read it. What is the molality of
benzene, C6H6, dissolved in toluene, C6H5C— C7H8, Okay?
in a solution for which the mole fraction of benzene is .15?
So folks, you've got two things, okay? If you've got two things,
and they tell you what the mole fraction of one of them is,
what's the other got to be? Whatever's left.
Okay? So if the mole fraction of benzene is .15, then the mole fraction of toluene has
to be .85 They have to add up to be 100%
So there's a lot of what's going on on this next page that you're not going to need. But
basically, if you only got two things—now if you got
three, it's not so easy
So you're going to figure out the molecular weights
and then do this. I mean this is, we already did this.
So we'll go to the next one. Okay. This is how you do it folks, okay.
So once again, this part right here, one minus that
is obvious, okay we're talking mole fraction, you've got two
things It's got to be, if you're given what x is,
then the other is just 1-x, okay? So, for toluene, this is saying you've got
0.85 moles times, this is the molecular weight of toluene,
times 100—
1 kilogram per 1,000 kilograms, per thousand grams, you have grams here,
so here's your .85, here's your toluene, and there is the amount of weight of toluene.
I want to be very straightforward so we've got—this is the amount of toluene,
in this mixture, okay, so it's not quite a mole
okay, a mole would be 92.13 grams so this is a smaller amount than, than that
So the molality then, we already know from what we were given that the number of
moles or the mole fraction for benzene was .15
so it would be .15, this whole thing again is what we got from the page before
so the molality is equal to about 2 moles per kilogram. Or the molality is about point—is
about 2. So folks, this would be probably, I mean,
it's probably an eight point question on an exam
ones like these you do have to know. Folks, you've got to know
particularly when we get to Raoult's law you've got to understand the 1-x part
okay? Okay now this, folks, to me, these are the
most difficult. This is the most difficult question I can ask. Okay?
Because this involves density, and all So, you're going to have to practice these
if you don't get it right now, don't be disappointed, okay, or just discouraged
Maybe you've done this before in some other class, and that's wonderful, but I find these
to be the longest ones between starting off and ending. Okay?
So practice this. Don't forget, 9.7, folks, do it. One more thing.
Everybody pay attention. I got an email the other day from a student,
I have no idea from whom, or what class. And they said that it would be useful, Professor
Blake, if you posted the answers to the discussion. Because sometimes in discussion, the T.A.'s
don't get through all the questions. What you have to understand, folks, is that
almost, maybe not all, but I try to make it so that every question that is asked, is an
even one in the book. You have all the odd answers in your solution
manual. Okay? And so, whereas, maybe problem 9.22
is this one, 9.21 or 9.23 would have the same thing except
the density would be 1.08 and this might be 1.14.
Okay? It's the same exact approach. So I'm not trying to be stingy, I mean, we
could certainly post those, the problem is that then even fewer students would go to
discussion. Okay? So I don't know if that's—I don't
think it solves the problem. The problem is that
you need to be looking at that solution manual. You need to be doing a lot of problems. I'm
just giving you problems that are ones that I think are important.
Doesn't mean there are others given to you in the back of the chapter
that you shouldn't be doing All right! So
let's do this problem. So the first step, and it's not a bad idea,
folks, I mean I'm a kind of do-things-my-own-way
but I find on these, that if I don't follow these steps, I end up screwing something up
Okay, so maybe you're just smarter than I am
most of you are, so if you've got a way of solving these problems,
great. So: step 1. Find the mass of exactly 1 Liter.
Okay? Of the solution, and you know what the density
is, okay? So look at this. This many grams per mL, so that if you want
one liter, you multiply that by 1,000. So you've got
exactly 1.14 times 10^(3) grams. That's how many—
that is how many grams you've got in this mix. Okay?
So that's step 1. Figure out how many grams you've got.
Step 2: Find the mass of solute in exactly 1 liter of solution from
solute minus solvent, blah blah blah. Okay? So
we had sucrose, okay? The molecular weight: 342
that's how many grams per mole, okay? So here we have moles here, this is the molarity,
right? 1.06 moles per liter So that's an easy one
you've got 1 liter here, so you have 1.06 moles of that.
So the number's going to be about 350. 360. Okay?
What have I got? 363. Okay. So once again, you are given the molarity—
molarities are easy, people. If I say the molarity is 2, and I say you've
got a half a liter, then it's going to be 2 moles per liter times
½ a liter is equal to 1 mole.
This is a simple thing. So coming up then with how many grams of sucrose
because, folks, on the other one we came up with how many grams
of the total stuff. The total stuff had a density of 1.14
So we said, okay, well in 1 liter you've got 1.4—
1.140*10^(3) grams so now we have to pull the thing apart.
How much of it is the sugar, how much of it is the solvent?
So this is then how much sucrose we have. So we have 363 grams of sucrose,
we had this much total, the whole mixture wighed this much per liter,
we had that many per liter, in 1 liter we had this many grams of
the sucrose So how much water was there? Well, it's whatever
is left! Whatever's left, okay?
So it's 1140 minus 363, you've got 780 grams.
So you have 780 grams of water So now, folks, you've got the number of moles
of— or the weight, the molecular weight, or I
should say the total mass of the sucrose and now you've got this many grams
now you can do molality because any time you've got this in grams,
that is .78 kilograms So now you go 1.06 moles over point—
078 kilograms the molarity is equal to 1.4
So folks, it's four steps okay? It's not an easy problem
you have to understand density, you have to be able to manipulate molecular weights,
you've got to be able to subtract. a lot, a lot of things could go wrong on this,
okay? This would be at least a ten point question
on an exam okay?
Which means in a sixty-five minute exam, you should spend five or eight minutes on this
problem So, folks, I went over it a little fast,
you need to do this. This is posted, it's in your book.
There's even more information in your book on how to do this particular example.
I kind of cut and paste stuff. All right? So, please go back and do this and do this
and do this. I don't know how to not follow the steps and
get it right. I still know how to not follow the steps and
get it wrong. All right!
So we have something that we call colligative properties
and it's interesting because they only depend on the number of solute particles in the solution
Okay, and not on the nature of the particles So it doesn't matter sort of what it is, it
just is how many you've got in there Okay?
So look at vapor pressure lowering Okay? Same type of little 3-step problem
pressure is going to equal the mole fraction times the pressure of the pure solvent. Okay?
Mole fraction times the pressure of the solvent. So this is sort of like what is
if the mole fraction, and this is the mole fraction. folks, of the
liquid, okay, so let's say this is water, let's just play like this is water
at room temperature, the vapor pressure of water is about 20 torr.
Okay? So this value would be 20. This is saying that at whatever condition,
pure water at 20 degrees is going to have a vapor pressure of 20 torr.
Okay? So if I have pure water There's a 1, right? The mole fraction is 1.
It's only water. So if I have a mole fraction of 1, and pure
water has a vapor pressure at 20 degrees of 20 torr,
what is the vapor pressure of water in this situation?
When you have a mole fraction of 1 for water? It's 2o torr.
Okay? It's 20 torr. This is the—what it would be if it were
pure, and it turns out if that's a one, it is pure.
Okay? So now let me ask you this: What if I take some salt and I pour it in
there, and I make the mole fraction of water be 0.5?
At 20 degrees? What is the pressure- the vapor pressure of
water going to be? it's going to be 10.
This was 20 on its own, and whatever this value is, this is 20 torr times the mole fraction
if it turns out that it's .5, then it's going to be .5 times 20 torr, so it's going to be
10 torr is going to be the pressure we can lower the vapor pressure of something
by adding stuff to it Okay? So once again, this is the mole fraction,
it's not weight, it's the mole fraction So you've got to really, really be comfortable
throwing around, adding up, and dividing, because you could be given,
I could say 16 grams of this, 46 grams of this, and 27 grams of this.
Okay? Let's do that! Okay. Nancy, write this down so I can remember
it. 100 grams of water
50 grams of sodium chloride and 50 grams of glucose
No- wait wait wait wait!!! Get rid of that sodium chloride one.
Let's just say—sorry everybody the fructose, okay
So one was sucrose, one was fructose, I think, wasn't it?
maybe I said glucose. Okay, something that's got
100 grams of water. We're okay on that. Okay. So let's just say 100 grams of water,
100 grams of glucose at 180 grams per mole and 180 grams of sucrose at 342.
They are all 100, I've decided. They are all 100, so I can remember it
and I think—so let's just say that the molecular weight of one of them is 180
and then what molecular weight of another one is 342?
Since they all have 100 grams, what is the vapor pressure of water?
At 20 degrees? Do it!
I think you're going to be surprised. You look like you're not going to get it right
anyway, but... I just mean, like, nobody's working.
So ware we confused or are we just, like, talking to each other?
Yes? I can't hear what you're saying. Loud!
I just said, 180 and 342. Yes.
But they all have 100 grams. So you've got 100 grams of water, 100 grams of this, and
100 grams of this. I want to know what is P1?
Okay, who's got an answer? Besides this guy? What'd you get?
17. So, look. It didn't change much. It went from 20 to 17. Why?
Because we had 10—no, we had 100 grams of water
at 18 grams per mole, that's like 5 moles. Everything else was tiny, okay?
You had 100 grams of something that's 180.9 moles
no-- .5 moles. Okay? And then you had another one that had
even a bigger molecular weight So that was like a quarter of a mole.
So you had a quarter of a mole plus a half a mole plus five moles.
So what was the actual mole fraction of water? 0.86. So it was really the dominant one.
So you just add everything up Everything
and then you have the the water, because that's the thing with the
vapor pressure on top. But look at how we could actually lower the vapor pressure by
putting something in the pure water. Okay? So what that means folks, is probably
the the vapor pressure of water in my soda
is probably lower than I've been thinking all these years
I've been thinking the vapor pressure of water at 20 torr in the room
that my that the vapor pressure of my soda was going
to be 20, but it's probably less than 20 because the mole fraction of water is probably not
100% here It's got some pepsi and some goo in it
and some coloring, okay? That is how you lower it. You absolutely have
to be comfortable, folks, calculating mole fractions
Okay? Because it just gets tougher after this So, once again, simple stuff, Raoult's law,
we just did it. Okay? So this is just a cartoon
it shows this at work This is the mole fraction going this direction
this is the vapor pressure going this way. So here's let's say water
at pure water right there. So this at room temperature would be 20
as we reduce the mole fraction by adding more and more stuff to it,
we can bring it down to 0. In theory.
Okay? That's the mole fraction. And you can have a lot of things
we had sucrose, we had glucose, we can add salt and whatever else we want
in there Okay? This is for a non-electrolyte
Okay? What I mean by that folks, is that sugar just dissolves in water but it does
not split apart Okay, if I put some glucose in my coffee,
it is a glucose molecule. The whole thing is a glucose molecule floating
around in my coffee If I put salt in my coffee,
I've got a sodium and a chloride The two are separated. And in that case folks,
guess what ooh! That's two things
Two particles! So we have to worry about how many things—
how things break in to other things, okay. We'll be talking about that later. But anyway,
this is once again just sort of a little cartoon of mole fraction.
Now! Okay, we just did this I thought I got rid of this picture.
This is more what I wanted to show you Okay? Look at this.
You have pure solvent, you have this many. I've got a pressure of 3-something up here.
You add some stuff to it, some nonvolatile solute
and you reduce the actual number of water molecules that are there. And this is how
it works. This is pure, this is once you've added something.
And we can make this pressure drop, drop drop drop drop by adding more and more stuff into
here Okay. Now
we can do this with liquids too okay? Not just something that is a nonvolatile
solute I can put liquid this, liquid this together
and we can then calculate what the actual pressure is going to be
and I like those a lot. There will be one of those on the exam.
I promise. Okay, so I'm not sure...
what have I done here? I'm not sure what this is...I'm going to
for some reason this didn't Okay
Let's do this. We've already gone and done it.
I'm so in tune with things that we've already done it, but let's just do it real quick.
So now I'll read it to you, and now you don't have to listen to me mumble and mumble.
Calculate the vapor pressure of water at 20 degrees in a solution prepared by dissolving
10 grams of the nonelectrolyte sucrose in 100 grams of water.
Okay? And I've given you this up here. look, I've even
look at how nice I am. I gave you this, and the vapor pressure of water at 20 degrees
is 17.56 torr. Okay? So let's just do it.
Should be quick Once again folks, the more you do these things,
as bored as you are, the better off you're going to be.
So everybody stop. Everybody stop look at the question, okay?
100 grams of water we already figured out is like 5 moles
okay you've only got 10 grams of something that
weighs a lot okay, so you need to go into this thinking,
you know, geeze, water has to be like, 90, 95%, so
if I've got 17.56 torr I've got to go in to this thinking the pressure's
not going to drop more than a torr or something like this.
So once again, you want to get just a feeling of not what your answer's going to be, but
in the range of what your answer is going to be
and then you go through it, and if you get something that's
you know, 15, 16, or 17, you go, "okay, I'm probably right."
You get something where you've done something incorrectly,
and you get a vapor pressure where this drops to 6
Then you've probably done something wrong. So go ahead and keep going.
(asks student) What'd you get? (student answers: 17.46) Perfect.
So the answer is 17.4 something. Okay, I'll go to the next slide which actually
shows you how to do it, but hopefully you've already beat me to that
Okay so this is just the number of moles this is figuring out mole fraction
and here you are. Okay? So, once again, we knew
we knew that this was going to be about five moles
and this is a tiny number of moles So the mole fraction—unfortunately they
don't on the one before did they actually give you
what the mole fraction was? It doesn't look like they do
okay, yeah the mole fraction is is .995. Okay. So you know that it's not going
to change the vapor pressure by very much and in fact 17.54 to 17.45. A tiny little
change. But still!
Folks, what that tells us is that we can add stuff
and we can actually adjust a vapor pressure or something.
And it turns out, that's quite important. Let's quit.
Okay, everybody have a safe weekend. Study effectively! Folks, don't spend a lot
of time on sapling. Okay? It's there to help you, but it's not
there to distract you.