Lecture 12. Physical Equilibrium, Part II
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Blake, Donald R.
UCI Chem 1B is the second quarter of General Chemistry and covers the following topics: properties of gases, liquids, solids; changes of state; properties of solutions; stoichiometry; thermochemistry; and thermodynamics. 
 So let's start. Okay, solutions. Multiple choice, okay? It's a homogeneous mixture of two or 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, gases. What am I breathing? What is the percentage of nitrogen in this room right now? Who's got a...what's up? 78%, that's right. So we have 78% of nitrogen. We got oxygen about 20%, okay? So that's a mixture, okay? Gas of liquid. Right here, a soda. Okay, it's got CO2 in it and it's got 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? People are nuts. Solid liquid, okay? Salt in water and then solid, solid brass, okay? So we have several different types of way to mix things up and we're going to learn about mixing them and in science we have to be able to say something about to be able to quantify what these things are. So the solution, the solute, is the substance present in the smaller amount 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 an 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? You 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 much salt that just falls out. So this is saturated. Unsaturated is if there had been no salt on the bottom, like the ocean. Ocean's got salt water, but it's not saturated, okay? But we're saturated, we 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. Just like that, just forms up and makes a solid and 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, there's too much crap floating around the house. You know, you get a little bit of a piece of lint or something that falls in that little thing of salt. You get that, it's got to be very, very clean for this experiment to work. Alright, so you booked us a little bit differently. This is the Hess's Law thing, okay? If I have the solvent, which is just water, we know water has hydrogen bonding, we know water's happy to be just water, and then you have a piece of salt, let's say, okay? Or something, this could be salt, it could be ethanol. So you can have it as a two-step or three-step. This breaks into pieces, this breaks into pieces, and it forms this, or you can 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, a certain amount of energy here that has to go into it 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? So in this 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 have heard of before, but will particularly when you get to organic chemistry, at least when I went to organic chemistry, that 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... Decahn, okay? That is C10H22. You don't need to write that down. If I take some of that and then I have some dodecahn, which is C12H26, if I mix one with the other, they mix completely, okay? If instead I take decahn, which is C10H22, and I add water to it, they separate completely, okay? There's... because one is totally dispersion, which is the carbon-hydrogen-based one. The other has dispersion. It's got hydrogen bonding. It's got 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 have something, if I had a stain right here, if I happen 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, I think, you know? Or some oil-based something, so that all this dispersion and stuff can go together and they grab onto each other and then 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. Let me just put these things up here. So once again, it's just... you have things that are similar, or these have positive charges, negative charges. This has a positive and negative end, so it's going to dissolve in this. This is ethanol in 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 miscible, okay? Or they're going to be able to be mixed together. So you could be given a question on an exam that says, which of these would dissolve this? Or in which would this be a solution? And you'd say, okay, not water, not this, not this. It would have to be something that was non-polar, and like that. So... so, folks, does this for us, okay? So so pass that one end that is greasy, and one end that's like water, okay? So you end up with 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, and this is bigger than that came. 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 onto 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 that were in office hours, because I grew up on a farm, and we had to do all of our... we had to fix all the motors and the tractors and the lawnmowers and everything else. And we would end up, my dad and I would end up with just probably almost just black grease 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. I totally have fun. And so you've got to wonder, I mean, I was doing this like, once or twice a week it seemed like. So anyway, but you know what is 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. 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 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 one, the density of gasoline is about 0.8. So the 0.8 stuff would float. So I could shake it up and I could make it look a little bit 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 degree C, not K, okay? So 0 degrees C, it's not very solid. 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, you know, it's increased in solubility, it's like about 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 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 and 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 is 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 had 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 over the bigger pile of sugar in her coffee, okay? Because of this, you move back this way. If you're 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, it's about 150, it's twice as soluble, going from zero to 100, you get twice as much sodium nitrate in solution, in the water, than you would at zero degrees, okay? So what that means is if I can get everything dissolved in a cup of water, hot water, here, that somewhere between here and zero 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 can jelly, warm things up, soluble in pieces. There are some outliers where that is not in fact the case. So this is just a, they've taken two examples from the last one. This is sodium chloride, which doesn't seem to change much. Okay? I mean, the amount of sodium chloride that you can get in at, so we can get 34.2 grams per 100 grams of water at zero 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 nine. And this has to do with lattice energy, solvation energy, and so on as to why that actually happens. Let me say that here. Okay, well that's all stuff that you can get off of the, off of the, this itself. Unfortunately, it's just the opposite for gases. So you know, all your life you've been able to do products minus reactants, and then that enthalpies thing, all of a sudden it was reactants minus products kind of screwed you up. This is just the opposite. Look at this. This is a solubility of oxygen at zero degrees, so it drops by more than a factor of two, going from zero 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 have to pop it first and let, because when they ship it to you, we're going to have a problem on this. When they ship it to you, there's a lot of carbon dioxide at the top. Okay, but let's say we pop it like this, and let's say that we're sitting at, you know, one degree Celsius. Okay, so right above freezing. Pop it like this, it's going to make a sound, and I do this. Let's say I take another one and I pop it. Okay, so now they both have been vented. And now I put one of them back in the refrigerator, and I take another one right here. Okay, so now one of them still sits right about here, the other goes to here. Look at the difference. This is oxygen, 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 of CO2, but let's say it is CO2. From here to there, you know, that's like about a factor of like maybe 40% or so. 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, nothing. That was the cold one. Here's the warm one. God damn. Okay. It wasn't controlled enough, okay. 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. Okay. So 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. It looked like this. It maximized at four degrees. Anywhere on either side of four degrees, if you went toward four, you became more dense. If you moved away from four, you became less dense. Okay. Why does this scare it on? Yes. Okay. Well, so he's, okay. So in this case, that's a very good thing. I mean, well, it's a bad thing, but it's a good observation. 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, one degree here. Then we kick out more CO2. CO2 is 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'll 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 the gas in liquid is proportional to the pressure of the gas over the solution. So folks, what's that 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 remake 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 log 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 log constant is X. Okay. So lots of Henry's log constants have been calculated or determined. Probably not the Henry's log constant for diapepsis, 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 log. So once again, C or S is the concentration in moles per liter of dissolved gas. Moles per liter of this molarity. It's perfect. Okay. P is the pressure of the gas over the solution. And K is the constant for each gas. Okay. That depends only on the temperature. So look at this picture right here. You 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 have a count of 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 going to actually be able to cram into here. And that is totally dependent on the Henry's log constant. Okay. If the Henry's log constant were twice as big, then if we had eight here, we'd have eight here. Okay. But as it is right now for this particular one, the Henry's log constant, because look at the concentration. Okay. The concentration is directly proportional to the Henry's log constant. So if I make the Henry's log constant twice as big, the concentration gets bigger. If I don't change the Henry's log 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'd better write that down. Sometimes it's just c equals kp. I put the h in there just so that, because we have a lot of k's in science, and so this is the Henry's log. 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 one liter bottle of carbonated water if the manufacturer uses a pressure of 2.4 atmospheres in the bottling process at 25 degrees C. And I'm going to say that the Henry's log 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. There we are. So the equation is equal to kp. That's going buddy. 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're all this straightforward and simple. Okay? So this kind of a problem on an exam is probably like four points. Okay? 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 units. If you don't know what to do, look at the units. Okay? The reason this is in moles per liter atmospheres 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 1000 tor. Now I'll do this problem. 1000 tor is p. 1000 tor. How many need a little help? So everybody knows that there's 760 tor in an atmosphere. So 1000 tor is about one third atmospheres. So I'm predicting that the number here is going to be about like 0.0547. Okay. So folks, you must convert if you're given two different kinds of units. You can either change this to moles per liter tor, or you can turn my value that I gave you here. Instead of p is equal to 2.4 atmospheres, you could, or what I said, 1.000 tor. You can convert tor to atmospheres either way. Okay? Alright, 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 one liter. We had one liter of soda. Okay? So the question was how many grams of CO2? Do it. CO2. Remember carbons 12, oxygen 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, oh, this was in atmospheres. This is in liter atmospheres. I can make this into molarity. And then you have one liter. So one liter, or I said somewhere here, you had one liter. And then you just go ahead and do your simple little math here, and you come up with about 3.5 grams. Okay? Straight forward 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. Okay, now why do we say the partial pressure of oxygen is 0.21? The partial pressure means what is the percentage of that particular gas? Okay? Because we don't, 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 0.21 of it is oxygen. So that's the P that you're going to use, okay? 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 it's not, the fish die. Or at least are unhappy. Okay? 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. Header's lock 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 times 10 to the 4, negative 4. This is also, this is 1.3 times 10 to the 4. Millimoles means 10 to the minus 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, milliming in a thousand, so it's a thousandth of a mole. So it's 0.13 thousandths of a mole, or 0.13 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 got 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. Mole fraction, folks, is where it's at. I mean, this is simple, okay? But 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 0.33, or 33.3%. Okay? That's like sixth grade math, okay? But it doesn't tell us a whole lot. Because we were going to world where moles go together. You need one mole of this plus two moles of this to make one mole of that. So you thought that you need six pounds of this and 12 pounds of this, okay? How many of you 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 one pound of flour. Am I doing okay? One pound of flour, one cup of milk, and three eggs. I don't know what I would make with that, but I would make something, okay? So I hope that sounds like something you might read in a recipe book. What you don't get is a pound of flour, a pound of milk, and a pound of eggs. Eggs don't come in pounds, right? It's a molar thing. You want three eggs. Three 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 where 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. 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? Now 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 50s, you know, in class, there might have been two of us in each classroom of like 30 kids. Don and John and Bob and Bill. I'm with my sisters, Nancy, Judy, and Patty, you know? And now, I don't know. Okay, so what is the mole fraction of Don's in this room right now? Do it. One 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 Don's and Rafe's in the room? There's probably nobody else in this room called Rafe, although I don't know, besides you. Good looking. Good looking, yes, yes. Is anybody named Rafe in this room? Besides you, okay. So there's two of us. So the mole fraction is two over 400. There's probably only one Josette. Probably, okay. But there's probably more than one Brittany and one Jasmine and one, let me couple the guy's name. It's Andy. How many Andes? Yeah, look at this. One, two, three, four, five Andes. Okay? I knew Andy Effenberger. That was it. You know? He's the only Andy in my school, I think. The point is, is that sometimes you need mole fractions. 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 learn this, I hope, in high school. I hope 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 then 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 a thousand grams, and then you add stuff to it. Okay? So that 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 one liter of water... Guess what? If it fills right to the top and is one 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 0.98. 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 we're going to be going over, not just today, but next week. Okay. Don't forget there's an exam a week from today. And it's really long. It's getting shorter though, right? Because we're going to work on it. Okay. So this is what your book gives me, folks. You know, 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, mol 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. Small fraction. Easy to do. As long as we know how many moles of this we had and how many moles of this we had, 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 one liter, now we've got moles per liter. Not a liter here. In the end, the liter. Okay. So it's after you've done the adding, you have one liter, and that is the molarity. 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're given something that you've got this many moles per liter, then obviously you're talking about molarity. Okay. So what 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 at the molarity right there. Dropping with temperature. Molality. Right there. Because a kilogram warming up is still a kilogram. Okay? So this is the molality, this is the molarity. Ta-da. That is why I'm a distinguished professor. Okay, right there. 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've given 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 resolving 90.5 grams of fructose in 250.2 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 zeros don't count unless there's a decimal point. Okay? So that has three sig figs there, that has three sig figs there, so that means you don't have to 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 have sworn I did that. So figure out yourself. Carboned, so maybe we have to revert to two sig figs. 12, 1, 16. Figure the molecular weight out with that. Okay? Are you teaching discussion today? No, I'm not saying. It's called lunch. We do a group lunch. Definitely. You don't have to do this. I was a... Hey. It's a lunch. I thought you were going to... I thought it was lunch. It's a group. It's a grantee. It's a grantee. It's only if she wants to, though. It's okay. We're going to have to. No. Okay, well if you prefer not to, we won't. But... Plus the door and the door, and Jason and I are running today. We're running. Okay. We need to find a time when we can do it. Okay? Next week for this kid's birthday, and then also for a great glorious birthday, which was nice. We'll do a threesome. And my birthday. What do you think of that? I like it. I like it. I like it a lot. Okay, all right. Oh, next Friday in that... Oh, next Friday. Thursday. Yeah, we forget. Okay, sorry. I got carried away here. So, folks, once you figured out what that number is, which I think it's like around 180... One is? Okay. When you say that's about half of that, or twice that? So, you got about half a mole. And then you look at this, you go, hmm, I got a quarter of a kilogram. So, the answer's got to be close to two. 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 two if you didn't have all these extra sig-tigs 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 C686 dissolved in toluene C6H5C7H8, okay? In a solution for which the mole fraction of benzene is 0.15. So, folks, you got two things, okay? If you got two things, and they tell you what the mole fraction of one of them is, what's the other guy to mean? Whatever's left, okay? So, if the mole fraction of benzene is 0.15, then the mole fraction of toluene has to be 0.85. They have to add it 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. Oops. So, you can figure out the molecular weights, and then do this. I mean, this is where he 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? When we're talking mole fraction, you got two things. It's going to be, if you're given what x is, then the other is just one minus x, okay? So, for toluene, this is saying you've got 0.85 moles times, this is the molecular weight of toluene, times 101 kilograms per 1,000 kilograms, or per 1,000 grams you got grams here. So, here's your 0.85, here's your toluene, and there is the amount of weight of toluene. So, once again, very straight forward. 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 that. So, the molality then, we already know from what we were given that the number of moles of the mole fraction for benzene was 0.15. So, it should be 0.15. This whole thing again is what we got from the page before. So, the molality is equal to about two moles per kilogram, where the molality is about two. So, folks, this would be probably, I mean, it's probably an eight-point question on the exam. Once again, you do have to know, folks, you've got to know, particularly get to Raoult's law, you've got to understand the one minus 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 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 it 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 from what class. And they said that it would be useful, Professor Blake, if you posted the answers to the discussion, because some of the discussion, the TAs don't get through all the questions, okay? 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, it's 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 then is that 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 and you need to be doing a lot of problems. I'm just giving you problems that are ones that I think are important. It doesn't mean that there's others that aren't given to you in the back of the chapter that you shouldn't be doing, okay? Alright, so, let's do this problem. So, first step, and it's not a bad idea, folks. I mean, I'm a kind of a do things my own way, but I find that on these, 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 one, find the mass of exactly one liter, okay, of the solution, and you know what the density is, okay? So, look at this. This many grams per mill, so that if you want one liter, you want to buy that by a thousand. So, you've got exactly 1.14 times 10 to the third grams. That's how many, that is how many grams you've got in this mix, okay? So, that's step one. Figure out how many grams you've got. Step two, find the mass of solvent in exactly one liter of solution from solute minus solvent, blah, blah, blah, okay? So, we had suprose, 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 one liter here, so you have 1.06 moles of that. So, the number is going to be about 350, 360, okay? What, I've got 363, okay? So, once again, you've got, you're given the molarity, molarity is a really 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 a half a liter is equal to 1 mole. This is a simple thing. So, coming up then with how many grams of suprose, 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 one liter you've got 1.140 times 10 to the 3 grams. So, now we have to pull the thing apart. How much of it's 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 weighed this much per liter. We had that many in one liter. In one liter we have this many grams of the sucrose. So, how much water was there? Well, it's whatever's left. Whatever's left. So, it's 1140 minus 363. You've got 780 grams. So, you had 780 grams of water. So, now folks, you've got the number of moles of 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 0.78 kilograms. So, now you go 1.06 moles over 0.078 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 of things to go wrong on this, okay? So, you've got at least a 10-point question on an exam, okay? Which means, in a 65-minute exam, you should spend five or eight minutes on this problem, okay? 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. That kind of cut and paste stuff, all right? So, please go back and do this and do this and do this. But you get... I don't know how to not follow the steps and get it right. Okay? I still don't know how to not follow the steps and get it wrong. All right. So, we have something that we call colligative properties. 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's just how many you got in there, okay? So, look at vapor pressure lowering, okay? Same type of little three-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 tor. Okay? So, this value would be 20. This is saying that at whatever condition, like, pure water, at 20 degrees is going to have a vapor pressure of 20 tor. Okay? So, if I have pure water, that is 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 of 20 degrees of 20 tor, what is the vapor pressure of water in this situation? When you have a mole fraction of 1 for water. It's 20 tor. Okay? It's 20 tor. This is the... what it would be if it were pure, and it turns out that if that's a 1, it is pure. Okay? So, 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 tor times the mole fraction. If it turns out that it's 0.5, then it's going to be 0.5 times 20 tor, so it's going to be 10 tor. It's going to be the pressure. So, we can lower the vapor pressure of something by adding stuff to it. Okay? Well, 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 can 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. 100 grams of water. 50 grams of sodium chloride, and 50 grams of glucose. No, let... Wait, 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'll say glucose. Okay, something that's got... Okay, so I'm... So, 100 grams of water, we're okay on that. Okay, let's just say 100 grams of water, 100 grams of glucose at 180 grams per mole, and 180 grams of sucrose at 342. 100, 100. They're all 100, I've decided. They're 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 the molecular weight of another one is 342. But since they're all 100, I have 100 grams. What is the vapor pressure of water? At 180 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 nobody's like working. So, are we confused or are we just 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 got 100 grams of water, 100 grams of this, and 100 grams of this. I want to know what is P1? Sounds good. Okay, who's got an answer besides this guy? What did you get? 17. So look, it didn't change much. It went from 20 to 17. Why? Because we had 100 grams of water at 18 grams per mole. That's like 5 moles. Everything else was tiny. You had 100 grams of something that's 180. That's 0.9 moles. No, 0.5 moles. 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 of a mole plus 5 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 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 to the pure water. So what that means, folks, is probably the vapor pressure of water in my soda is probably lower than I've been thinking all these years. I've been thinking that the vapor pressure of water at 20 tor in the room, that my 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 percent here. So we've got some Pepsi and some goo in it and some coloring. That is how you lower it. You absolutely have to be comfortable, folks, calculating mole fraction, because it just gets tougher after this. So once again, simple stuff, Routes law, we just did it. This is just a cartoon that 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 and more stuff to it, we can bring it down to zero in theory. That's the mole fraction. You can have a lot of things. We had sucrose, we had glucose, we can add salt and whatever else we want in there. This is for a non-electrolyte. What I mean by that, folks, is that sugar just dissolves in water, but does not split apart. 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 things break into other things. We'll be talking about that later. But anyway, this is 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. If you have pure solvent, you have this many. I've got a pressure of three somethings up here. You add some stuff to it, some non-volatile solute, and you reduce the actual number of water molecules that are there. This is pure. This is what you've added something. And we can make this pressure drop, drop, drop, drop, drop, drop by adding more and more stuff into here. Okay. Now, we can do this with liquids, too. Now, just something that is a non-volatile solute, I can put liquid this together, and we can calculate then 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. For some reason, this didn't... Okay. Let's do this. We've already kind of done it. I'm so in tune with things that we've already done this. 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. Okay. So I'm going to put this in the vapor pressure of water at 20 degrees in a solution prepared by dissolving 10 grams of the non-electrolyte sucrose in 100 grams of water. Okay. And I've given you this up here. Look 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. It should be quick. Once again, folks, the more you do these things, as bored as you are, the better you're 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, geez, water has to be like 90, 95 percent. So if I've got 17.56 torr, I've got to go into this thinking that the pressure is not going to drop more than a torr or something like this. So once you want to get just a feeling of not what your answer is 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, then you go, okay, I'm probably right. If you get something where you've done something incorrectly, and you get a value where this vapor pressure drops to 6, then you've probably done something wrong. So go ahead and keep going. Okay. 17.47. Perfect. Okay. What'd you get? 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 that this is 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, they actually give you what the mole fraction wants. I don't think they do. Okay. So the mole fraction 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. Okay. 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.
 So let's start.  Okay, solutions.  Multiple choice, okay?  It's a homogeneous mixture of two or 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, gases.  What am I breathing?  What is the percentage of nitrogen in this room right now?  Who's got a...what's up?  78%, that's right.  So we have 78% of nitrogen.  We got oxygen about 20%, okay?  So that's a mixture, okay?  Gas of liquid.  Right here, a soda.  Okay, it's got CO2 in it and it's got 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?  People are nuts.  Solid liquid, okay?  Salt in water and then solid, solid brass, okay?  So we have several different types of way to mix things up  and we're going to learn about mixing them  and in science we have to be able to say something  about to be able to quantify what these things are.  So the solution, the solute, is the substance present  in the smaller amount 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 an 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?  You 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 much salt that just falls out.  So this is saturated.  Unsaturated is if there had been no salt on the bottom,  like the ocean.  Ocean's got salt water, but it's not saturated, okay?  But we're saturated, we 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.  Just like that, just forms up and makes a solid  and 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, there's too much crap floating around the house.  You know, you get a little bit of a piece of lint  or something that falls in that little thing of salt.  You get that, it's got to be very, very clean  for this experiment to work.  Alright, so you booked us a little bit differently.  This is the Hess's Law thing, okay?  If I have the solvent, which is just water,  we know water has hydrogen bonding,  we know water's happy to be just water,  and then you have a piece of salt, let's say, okay?  Or something, this could be salt, it could be ethanol.  So you can have it as a two-step or three-step.  This breaks into pieces, this breaks into pieces,  and it forms this, or you can 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,  a certain amount of energy here that has to go into it  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?  So in this 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 have heard of before,  but will particularly when you get to organic chemistry,  at least when I went to organic chemistry,  that 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...  Decahn, okay?  That is C10H22.  You don't need to write that down.  If I take some of that and then I have some dodecahn,  which is C12H26,  if I mix one with the other, they mix completely, okay?  If instead I take decahn, which is C10H22,  and I add water to it, they separate completely, okay?  There's... because one is totally dispersion,  which is the carbon-hydrogen-based one.  The other has dispersion.  It's got hydrogen bonding.  It's got 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 have something,  if I had a stain right here,  if I happen 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, I think, you know?  Or some oil-based something, so that all this dispersion and stuff  can go together and they grab onto each other  and then 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.  Let me just put these things up here.  So once again, it's just...  you have things that are similar,  or these have positive charges, negative charges.  This has a positive and negative end,  so it's going to dissolve in this.  This is ethanol in 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 miscible, okay?  Or they're going to be able to be mixed together.  So you could be given a question on an exam that says,  which of these would dissolve this?  Or in which would this be a solution?  And you'd say, okay, not water, not this, not this.  It would have to be something that was non-polar, and like that.  So...  so, folks, does this for us, okay?  So so pass that one end that is greasy,  and one end that's like water, okay?  So you end up with 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, and this is bigger than that came.  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 onto 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 that were in office hours,  because I grew up on a farm, and we had to do all of our...  we had to fix all the motors and the tractors and the lawnmowers and everything else.  And we would end up, my dad and I would end up with just probably  almost just black grease 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.  I totally have fun.  And so you've got to wonder, I mean, I was doing this like,  once or twice a week it seemed like.  So anyway, but you know what is 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.  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 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 one,  the density of gasoline is about 0.8.  So the 0.8 stuff would float.  So I could shake it up and I could make it look a little bit 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 degree C, not K, okay?  So 0 degrees C, it's not very solid.  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, you know, it's increased in solubility,  it's like about 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 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 and 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 is 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 had 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 over the bigger pile of sugar in her coffee, okay?  Because of this, you move back this way.  If you're 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, it's about 150,  it's twice as soluble, going from zero to 100,  you get twice as much sodium nitrate in solution,  in the water, than you would at zero degrees, okay?  So what that means is if I can get everything dissolved  in a cup of water, hot water, here,  that somewhere between here and zero 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 can jelly, warm things up,  soluble in pieces.  There are some outliers where that is not in fact the case.  So this is just a, they've taken two examples from the last one.  This is sodium chloride, which doesn't seem to change much.  Okay?  I mean, the amount of sodium chloride that you can get in at,  so we can get 34.2 grams per 100 grams of water at zero 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 nine.  And this has to do with lattice energy,  solvation energy, and so on as to why that actually happens.  Let me say that here.  Okay, well that's all stuff that you can get off of the,  off of the, this itself.  Unfortunately, it's just the opposite for gases.  So you know, all your life you've been able to do products minus reactants,  and then that enthalpies thing, all of a sudden it was reactants minus products  kind of screwed you up.  This is just the opposite.  Look at this.  This is a solubility of oxygen at zero degrees,  so it drops by more than a factor of two, going from zero 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 have to pop it first and let, because when they ship it to you,  we're going to have a problem on this.  When they ship it to you, there's a lot of carbon dioxide at the top.  Okay, but let's say we pop it like this,  and let's say that we're sitting at, you know, one degree Celsius.  Okay, so right above freezing.  Pop it like this, it's going to make a sound, and I do this.  Let's say I take another one and I pop it.  Okay, so now they both have been vented.  And now I put one of them back in the refrigerator,  and I take another one right here.  Okay, so now one of them still sits right about here,  the other goes to here.  Look at the difference.  This is oxygen, 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 of CO2, but let's say it is CO2.  From here to there, you know,  that's like about a factor of like maybe 40% or so.  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,  nothing.  That was the cold one.  Here's the warm one.  God damn.  Okay.  It wasn't controlled enough, okay.  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.  Okay.  So 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.  It looked like this.  It maximized at four degrees.  Anywhere on either side of four degrees,  if you went toward four, you became more dense.  If you moved away from four, you became less dense.  Okay.  Why does this scare it on?  Yes.  Okay.  Well, so he's, okay.  So in this case, that's a very good thing.  I mean, well, it's a bad thing, but it's a good observation.  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, one degree here.  Then we kick out more CO2.  CO2 is 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'll 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 the gas in liquid is proportional to the pressure of the gas over the solution.  So folks, what's that 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 remake 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 log 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 log constant is X.  Okay.  So lots of Henry's log constants have been calculated or determined.  Probably not the Henry's log constant for diapepsis, 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 log.  So once again, C or S is the concentration in moles per liter of dissolved gas.  Moles per liter of this molarity.  It's perfect.  Okay.  P is the pressure of the gas over the solution.  And K is the constant for each gas.  Okay.  That depends only on the temperature.  So look at this picture right here.  You 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 have a count of 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 going to actually be able to cram into here.  And that is totally dependent on the Henry's log constant.  Okay.  If the Henry's log constant were twice as big, then if we had eight here, we'd have eight here.  Okay.  But as it is right now for this particular one, the Henry's log constant, because look at the concentration.  Okay.  The concentration is directly proportional to the Henry's log constant.  So if I make the Henry's log constant twice as big, the concentration gets bigger.  If I don't change the Henry's log 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'd better write that down.  Sometimes it's just c equals kp.  I put the h in there just so that, because we have a lot of k's in science, and so this is the Henry's log.  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 one liter bottle of carbonated water  if the manufacturer uses a pressure of 2.4 atmospheres in the bottling process at 25 degrees C.  And I'm going to say that the Henry's log 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.  There we are.  So the equation is equal to kp.  That's going buddy.  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're all this straightforward and simple.  Okay?  So this kind of a problem on an exam is probably like four points.  Okay?  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 units.  If you don't know what to do, look at the units.  Okay?  The reason this is in moles per liter atmospheres 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 1000 tor.  Now I'll do this problem.  1000 tor is p.  1000 tor.  How many need a little help?  So everybody knows that there's 760 tor in an atmosphere.  So 1000 tor is about one third atmospheres.  So I'm predicting that the number here is going to be about like 0.0547.  Okay.  So folks, you must convert if you're given two different kinds of units.  You can either change this to moles per liter tor, or you can turn my value that I gave you here.  Instead of p is equal to 2.4 atmospheres, you could, or what I said, 1.000 tor.  You can convert tor to atmospheres either way.  Okay?  Alright, 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 one liter.  We had one liter of soda.  Okay?  So the question was how many grams of CO2?  Do it.  CO2.  Remember carbons 12, oxygen 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, oh, this was in atmospheres.  This is in liter atmospheres.  I can make this into molarity.  And then you have one liter.  So one liter, or I said somewhere here, you had one liter.  And then you just go ahead and do your simple little math here, and you come up with about 3.5 grams.  Okay?  Straight forward 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.  Okay, now why do we say the partial pressure of oxygen is 0.21?  The partial pressure means what is the percentage of that particular gas?  Okay?  Because we don't, 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 0.21 of it is oxygen.  So that's the P that you're going to use, okay?  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 it's not, the fish die.  Or at least are unhappy.  Okay?  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.  Header's lock 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 times 10 to the 4, negative 4.  This is also, this is 1.3 times 10 to the 4.  Millimoles means 10 to the minus 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, milliming in a thousand, so it's a thousandth of a mole.  So it's 0.13 thousandths of a mole, or 0.13 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 got 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.  Mole fraction, folks, is where it's at.  I mean, this is simple, okay?  But 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 0.33, or 33.3%.  Okay?  That's like sixth grade math, okay?  But it doesn't tell us a whole lot.  Because we were going to world where moles go together.  You need one mole of this plus two moles of this to make one mole of that.  So you thought that you need six pounds of this and 12 pounds of this, okay?  How many of you 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 one pound of flour.  Am I doing okay?  One pound of flour, one cup of milk, and three eggs.  I don't know what I would make with that, but I would make something, okay?  So I hope that sounds like something you might read in a recipe book.  What you don't get is a pound of flour, a pound of milk, and a pound of eggs.  Eggs don't come in pounds, right?  It's a molar thing.  You want three eggs.  Three 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 where 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.  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?  Now 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 50s, you know, in class,  there might have been two of us in each classroom of like 30 kids.  Don and John and Bob and Bill.  I'm with my sisters, Nancy, Judy, and Patty, you know?  And now, I don't know.  Okay, so what is the mole fraction of Don's in this room right now?  Do it.  One 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 Don's and Rafe's in the room?  There's probably nobody else in this room called Rafe, although I don't know, besides you.  Good looking.  Good looking, yes, yes.  Is anybody named Rafe in this room?  Besides you, okay.  So there's two of us.  So the mole fraction is two over 400.  There's probably only one Josette.  Probably, okay.  But there's probably more than one Brittany and one Jasmine and one, let me couple the guy's name.  It's Andy.  How many Andes?  Yeah, look at this.  One, two, three, four, five Andes.  Okay?  I knew Andy Effenberger.  That was it.  You know?  He's the only Andy in my school, I think.  The point is, is that sometimes you need mole fractions.  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 learn this, I hope, in high school.  I hope 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 then 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 a thousand grams, and then you add stuff to it.  Okay?  So that 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 one liter of water...  Guess what?  If it fills right to the top and is one 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 0.98.  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 we're going to be going over, not just today, but next week.  Okay.  Don't forget there's an exam a week from today.  And it's really long.  It's getting shorter though, right? Because we're going to work on it.  Okay.  So this is what your book gives me, folks.  You know, 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, mol 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.  Small fraction.  Easy to do.  As long as we know how many moles of this we had and how many moles of this we had,  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 one liter,  now we've got moles per liter.  Not a liter here.  In the end, the liter.  Okay.  So it's after you've done the adding, you have one liter,  and that is the molarity.  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're given something that you've got this many moles per liter,  then obviously you're talking about molarity.  Okay.  So what 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 at the molarity right there.  Dropping with temperature.  Molality.  Right there.  Because a kilogram warming up is still a kilogram.  Okay?  So this is the molality, this is the molarity.  Ta-da.  That is why I'm a distinguished professor.  Okay, right there.  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've given 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 resolving 90.5 grams of fructose in 250.2 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 zeros don't count unless there's a decimal point.  Okay?  So that has three sig figs there,  that has three sig figs there,  so that means you don't have to 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 have sworn I did that.  So figure out yourself.  Carboned, so maybe we have to revert to two sig figs.  12, 1, 16.  Figure the molecular weight out with that.  Okay?  Are you teaching discussion today?  No, I'm not saying.  It's called lunch.  We do a group lunch.  Definitely.  You don't have to do this.  I was a...  Hey.  It's a lunch.  I thought you were going to...  I thought it was lunch.  It's a group.  It's a grantee.  It's a grantee.  It's only if she wants to, though.  It's okay.  We're going to have to.  No.  Okay, well if you prefer not to, we won't.  But...  Plus the door and the door, and Jason and I are running today.  We're running.  Okay.  We need to find a time when we can do it.  Okay?  Next week for this kid's birthday, and then also for a great glorious birthday, which was nice.  We'll do a threesome.  And my birthday.  What do you think of that?  I like it.  I like it.  I like it a lot.  Okay, all right.  Oh, next Friday in that...  Oh, next Friday.  Thursday.  Yeah, we forget.  Okay, sorry.  I got carried away here.  So, folks, once you figured out what that number is, which I think it's like around 180...  One is?  Okay.  When you say that's about half of that, or twice that?  So, you got about half a mole.  And then you look at this, you go, hmm, I got a quarter of a kilogram.  So, the answer's got to be close to two.  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 two if you didn't have all these extra sig-tigs 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 C686 dissolved in toluene C6H5C7H8, okay?  In a solution for which the mole fraction of benzene is 0.15.  So, folks, you got two things, okay?  If you got two things, and they tell you what the mole fraction of one of them is, what's the other guy to mean?  Whatever's left, okay?  So, if the mole fraction of benzene is 0.15, then the mole fraction of toluene has to be 0.85.  They have to add it 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.  Oops.  So, you can figure out the molecular weights, and then do this.  I mean, this is where he 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?  When we're talking mole fraction, you got two things.  It's going to be, if you're given what x is, then the other is just one minus x, okay?  So, for toluene, this is saying you've got 0.85 moles times, this is the molecular weight of toluene,  times 101 kilograms per 1,000 kilograms, or per 1,000 grams you got grams here.  So, here's your 0.85, here's your toluene, and there is the amount of weight of toluene.  So, once again, very straight forward.  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 that.  So, the molality then, we already know from what we were given that the number of moles of the mole fraction  for benzene was 0.15.  So, it should be 0.15.  This whole thing again is what we got from the page before.  So, the molality is equal to about two moles per kilogram,  where the molality is about two.  So, folks, this would be probably, I mean, it's probably an eight-point question on the exam.  Once again, you do have to know, folks, you've got to know, particularly get to Raoult's law,  you've got to understand the one minus 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 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 it 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 from what class.  And they said that it would be useful, Professor Blake, if you posted the answers to the discussion,  because some of the discussion, the TAs don't get through all the questions, okay?  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,  it's 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 then is that 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 and you need to be doing a lot of problems.  I'm just giving you problems that are ones that I think are important.  It doesn't mean that there's others that aren't given to you in the back of the chapter that you shouldn't be doing, okay?  Alright, so, let's do this problem.  So, first step, and it's not a bad idea, folks.  I mean, I'm a kind of a do things my own way, but I find that on these, 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 one, find the mass of exactly one liter, okay, of the solution, and you know what the density is, okay?  So, look at this. This many grams per mill, so that if you want one liter, you want to buy that by a thousand.  So, you've got exactly 1.14 times 10 to the third grams.  That's how many, that is how many grams you've got in this mix, okay?  So, that's step one.  Figure out how many grams you've got.  Step two, find the mass of solvent in exactly one liter of solution from solute minus solvent, blah, blah, blah, okay?  So, we had suprose, 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 one liter here, so you have 1.06 moles of that.  So, the number is going to be about 350, 360, okay?  What, I've got 363, okay?  So, once again, you've got, you're given the molarity, molarity is a really 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 a half a liter is equal to 1 mole.  This is a simple thing.  So, coming up then with how many grams of suprose, 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 one liter you've got 1.140 times 10 to the 3 grams.  So, now we have to pull the thing apart.  How much of it's 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 weighed this much per liter.  We had that many in one liter.  In one liter we have this many grams of the sucrose.  So, how much water was there?  Well, it's whatever's left.  Whatever's left.  So, it's 1140 minus 363.  You've got 780 grams.  So, you had 780 grams of water.  So, now folks, you've got the number of moles of 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 0.78 kilograms.  So, now you go 1.06 moles over 0.078 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 of things to go wrong on this, okay?  So, you've got at least a 10-point question on an exam, okay?  Which means, in a 65-minute exam, you should spend five or eight minutes on this problem, okay?  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.  That kind of cut and paste stuff, all right?  So, please go back and do this and do this and do this.  But you get...  I don't know how to not follow the steps and get it right.  Okay?  I still don't know how to not follow the steps and get it wrong.  All right.  So, we have something that we call colligative properties.  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's just how many you got in there, okay?  So, look at vapor pressure lowering, okay?  Same type of little three-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 tor.  Okay?  So, this value would be 20.  This is saying that at whatever condition, like, pure water, at 20 degrees is going to have a vapor pressure of 20 tor.  Okay?  So, if I have pure water, that is 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 of 20 degrees of 20 tor,  what is the vapor pressure of water in this situation?  When you have a mole fraction of 1 for water.  It's 20 tor.  Okay?  It's 20 tor.  This is the... what it would be if it were pure, and it turns out that if that's a 1, it is pure.  Okay?  So, 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 tor times the mole fraction.  If it turns out that it's 0.5, then it's going to be 0.5 times 20 tor, so it's going to be 10 tor.  It's going to be the pressure.  So, we can lower the vapor pressure of something by adding stuff to it.  Okay?  Well, 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 can 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.  100 grams of water.  50 grams of sodium chloride, and 50 grams of glucose.  No, let...  Wait, 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'll say glucose.  Okay, something that's got...  Okay, so I'm...  So, 100 grams of water, we're okay on that.  Okay, let's just say 100 grams of water, 100 grams of glucose at 180 grams per mole, and 180 grams of sucrose at 342.  100, 100.  They're all 100, I've decided.  They're 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 the molecular weight of another one is 342.  But since they're all 100, I have 100 grams.  What is the vapor pressure of water?  At 180 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 nobody's like working.  So, are we confused or are we just 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 got 100 grams of water, 100 grams of this, and 100 grams of this.  I want to know what is P1?  Sounds good.  Okay, who's got an answer besides this guy?  What did you get?  17.  So look, it didn't change much.  It went from 20 to 17.  Why?  Because we had 100 grams of water at 18 grams per mole.  That's like 5 moles.  Everything else was tiny.  You had 100 grams of something that's 180.  That's 0.9 moles.  No, 0.5 moles.  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 of a mole plus 5 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 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 to the pure water.  So what that means, folks, is probably the vapor pressure  of water in my soda is probably lower  than I've been thinking all these years.  I've been thinking that the vapor pressure of water at 20 tor in the room,  that my 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 percent here.  So we've got some Pepsi and some goo in it and some coloring.  That is how you lower it.  You absolutely have to be comfortable, folks, calculating mole fraction,  because it just gets tougher after this.  So once again, simple stuff, Routes law, we just did it.  This is just a cartoon that 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 and more stuff to it,  we can bring it down to zero in theory.  That's the mole fraction. You can have a lot of things.  We had sucrose, we had glucose, we can add salt and whatever else we want in there.  This is for a non-electrolyte.  What I mean by that, folks, is that sugar just dissolves in water,  but does not split apart.  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 things break into other things.  We'll be talking about that later.  But anyway, this is 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.  If you have pure solvent, you have this many.  I've got a pressure of three somethings up here.  You add some stuff to it, some non-volatile solute,  and you reduce the actual number of water molecules that are there.  This is pure.  This is what you've added something.  And we can make this pressure drop, drop, drop, drop, drop, drop  by adding more and more stuff into here.  Okay.  Now, we can do this with liquids, too.  Now, just something that is a non-volatile solute,  I can put liquid this together,  and we can calculate then 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.  For some reason, this didn't...  Okay.  Let's do this.  We've already kind of done it.  I'm so in tune with things that we've already done this.  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.  Okay.  So I'm going to put this in the vapor pressure of water at 20 degrees  in a solution prepared by dissolving 10 grams of the non-electrolyte  sucrose in 100 grams of water.  Okay.  And I've given you this up here.  Look 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.  It should be quick.  Once again, folks, the more you do these things,  as bored as you are, the better you're 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, geez,  water has to be like 90, 95 percent.  So if I've got 17.56 torr, I've got to go into this thinking  that the pressure is not going to drop more than a torr  or something like this.  So once you want to get just a feeling of not what your answer is 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, then you go, okay, I'm probably right.  If you get something where you've done something incorrectly,  and you get a value where this vapor pressure drops to 6,  then you've probably done something wrong.  So go ahead and keep going.  Okay.  17.47.  Perfect.  Okay.  What'd you get?  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 that this is 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, they actually give you what the mole fraction wants.  I don't think they do.  Okay.  So the mole fraction 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.  Okay.  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.
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