Will I get a round of applause at the end too?

Now? Just based on this. Okay, so I don't know if we'll get through the whole day because apparently Ed's game will have a lot of slides but we'll get through as much as we can. The test is next Tuesday and it covers chapters 5 and 11 as well as any review chapters like

1 through 3. Take a look at the syllabus if you're not sure. I would highly suggest to everyone that you read the textbook. The textbook has a whole lot of good information obviously. That's why we buy it. But today we're going to be covering a little bit further ahead in the course. So guest lecture 6, I'm that guest. Hello. Basic properties of solutions.

So when we think of chemistry, like you just picture what happens when we perform chemistry. You usually think of someone are you raising your hand there? Oh yeah sure. Is that better?

So typically if I picture someone trying to perform chemistry, I don't usually picture them taking two solid blocks and rubbing them together. I picture them mixing some solutions. Solutions are very convenient. And much of chemistry occurs in a solution.

So when we want a reaction to happen we get in general the two things that collide together. Yes, this isn't always the case, but we need two things to hit. And for that to happen it's a lot easier in solution where everything's rolling on top of each other. Yes, in a solid state they're fixed. In a fixed spot it's a lot harder to get them together to react. And in the gas phase

yes you have lots of collisions, but gases can be dangerous in general. We do lots of chemistry with gas phases. I'm just saying it's a lot easier to do it in solution. And so understanding the characteristics of solutions is very important. So for instance it says here that there's little chance of an uncontrolled escape of reactive chemicals. It's very true.

In the gas phase however we sometimes have some accidents. So for instance, and I don't know if you guys saw this in the news recently. You don't have to look in the science magazines to hear about this. This was a giant accident back in 1984 where methyl excuse me, methyl isocyanate was spilled as a gas

down in, where was it, Bhopal, India. Thousands and thousands of people were dead. It's because of things like this that we're always questioning. How can I do this chemistry in a safer way? And often times we want to do it in a liquid form. So for instance we know that the North America has a lot of natural gas yes. We could use it for fuel. But it's very dangerous

to transport natural gas. And so we always look for ways on how to convert that to a liquid fuel like methanol. Convert methane to methanol. So we can avoid accidents. So solutions can of course be dangerous too. I know I've spoke things on myself before.

But while there are solutions of gases uniform compositions throughout and solids sort of a frozen snapshot of a liquid solution most solutions are liquids. With a liquid solvent and gaseous liquid or solid solute. So if you think about that for a second, that last line liquid solvent and gaseous liquid or solid solute

hydrochloric acid. You guys usually use it as a liquid right? HCl is actually a gas from temperature. And we dissolve it into a liquid usually water. And that gives us the HCl acid that we want with in the lab. So that's actually a solution right? It's a gas mixed in. Same with a liquid-liquid solution. I could take

something any liquid or whatever Windex and mix it with more water. That's a liquid-liquid solution. And a solid with liquid that would be something like sodium chloride which is a solid from temperature in a liquid. So but a solution is a single phase. If I take sodium chloride and mix it in with some water I don't see two different phases present.

That's because the sodium chloride is fully dissolved. So I have one phase just the solution. So dissolving sodium chloride in water it gives us a single phase like I was saying up to about twenty-five percent sodium chloride. Now that's by weight. So in your first chemistry class you may remember and I'm not sure if I do it here at UCI

but they may do a solubility test where you take some water and you put in sodium chloride until all of a sudden it will no longer dissolve the sodium chloride. And it starts piling on the bottom. That's when it's reached its saturation point. And for water room temperature that's around twenty-five percent. Once

no more salt can dissolve it's saturated. Adding further salt will give two phases. The solution which is full sodium chloride and the solid sodium chloride on the bottom. The saturation point depends on temperature. If we think about that just in general we know that hot solvents tend to dissolve things faster and easier. They

dissolve more material. This makes sense to us. A surprising note here. Before this lecture I thought seawater was a higher weight percent and just an interesting bit. You know that if you drink seawater normally you will die of dehydration. And all you have is just that little bit of three point five weight percent salt. I thought that was interesting. So

now a phase diagram. If a phase diagram is not familiar to you then you should be reading the textbook more. This is a basic phase diagram but instead of pressure versus temperature like you may have seen we have temperature versus the weight

percent sodium chloride in water. I'm gonna go through this slowly. So what we're taking is water at zero percent here. No sodium chloride. And then we slowly change to one hundred percent. So let's start over here on the left where we have zero percent sodium chloride. Now that's just water right? So

at zero degrees Celsius just water well that's gonna be ice. It's gonna be frozen. And above that it'll be just liquid water until you get to one hundred degrees Celsius and then it'll boil. So you probably have seen or heard about when roads freeze over like in the mountain passes it's very dangerous to drive over the ice. And so what they'll have is they'll come and salt the

roads. You may have heard of that before. What they're doing is putting that salt into the water and there's a freezing point of depression. So you can see on this phase diagram. Let's start at zero percent and then I put in a little bit of salt. All of a sudden that freezing point starts to lower and lower and lower down here. And now I see that I can get all the way down to minus twenty degrees Celsius and the solution won't freeze. And this is

safe for the roads because when they put that in there the ice will melt in general depending on the temperature and it's a lot safer to drive on the road. It's a good application. So if we start up in here just as a demonstration let's say we're at room temperature around twenty, twenty-five degrees Celsius. We know that as we add salt then it's going to dissolve, yes? Until we reach

this around twenty-five percent, twenty-five percent by weight in chloride. One thing to note here is the bending of this line. So you can see that it's bending slightly to the right, yes? As we increase in temperature I can see that the weight percent, the total weight percent that I get for liquid solution

before I start getting liquid solution in sodium chloride it starts to increase. This is just a visual application to that thought where the hotter we hear the higher temperature our solvent is at the more we can solubilize that sodium chloride to get a higher percent. But depending on what temperature we're at we'll pass into this phase. The liquid solution

plus sodium chloride. Let me just mix this. So the notation A plus B is always in these two phases. So in this instance we would have liquid solution as one phase and the solid sodium chloride as another. If we took this liquid solution plus sodium chloride and start to cool it at one point

we're going to get liquid solution plus sodium chloride that's hydrated. It's simple to think of. You're cooling it down more and more material is going to crash out, you're solubilizing less. But as you get down to some really really really cold temperatures you start putting that water on sodium chloride. So if we think of this entropically

as you lower the temperature it's a lot easier to get things to hit together and stick so that sodium chloride tends to crash into one of those waters or rather it's water crashing into the same chloride and it'll stick so we get this hydrated form. And as we increase the percentage of sodium chloride more and more and more we see that we get the sodium

chloride 2H2O plus just sodium chloride solid. And that's because if you think of this as a phase diagram what we're saying is oh in this region I have almost no water left and it's almost all sodium chloride. So why do things dissolve? Same reason that gas

spreads out. It's to fill a container. It's statistics. It's hard to keep things in one spot. Things like to spread out. So dissolving increases the disorder of the solute. You can think of it just based on statistics. And you don't have to take a class to just think of it with common sense. So

let's look at the statistics with just six atoms. We're just gonna say krypton but it doesn't matter. This is just six particles. It could be anything. So we're taking six atoms of krypton gas in a container starting with all six on one side and the stop clock closed. There's a picture in just a moment. I believe it's next. Yes, it's next

here. So imagine we have this scenario where we have two containers and there's a stop clock here so it can effectively block it off. At one point we took, we put six atoms, which of course just mentally that's an obscurely small amount of atoms. This is just a demonstration. Six atoms of krypton on one side. And then we open the stop clock. We know just from common sense

that that's going to equilibrate until they're on both sides. So why does this happen? Why does the pressure equalize? It equalizes because this is the most likely outcome. If I go back for a moment and take the moment when there are

six atoms of krypton and then I open the stop clock, what's gonna happen? So I've got six atoms that are just crashing around all over the place and I have an opening. Naturally since there's nothing here, it's a vacuum those are going to statistically start bouncing over onto the other side. And then once I get onto the other side, they're also gonna have a chance of bouncing back

until of course I reach this final state where it's equal on both sides. Now this is an example with only six. What are the chances of it not being equal on both sides? So here we've taken these atoms and we've colored them. Again these are just krypton. They're just colors so you can see a difference in individual atoms.

So we take these six different colors of krypton and this is a certain configuration. It's one possibility. And when it talks about the weight this is a statistical weight. It's on mass. This is how much does this particular configuration contribute to the total possible outcomes? And we'll go over the weight in one moment. So we have this scenario

where I have six on one side. That's a six to zero. So there is only one way to have all six on one side right? That way right there. All six colors in one whole. Nothing anywhere else. So moving on. If I have, if I changed a little bit and bring over one

a more even distribution has a higher weight. So before let me demonstrate. We have a weight of one for this configuration. It has one possible way of happening. When I go to just five and one, I see that I have six possible ways to do this. The configuration of five to one. Taking any one of these colors

yellow, red, blue, green, sort of blue magenta, I don't know. Taking one on either side there are six different ways to do this. Just statistical randomness. Krypton on one side, krypton on the other, reality not colored.

Yeah, okay, I already said that. So by the time we get to four and two there are a whole lot more possibilities. We can take, obviously I'm not gonna go through each one of these, but you can take any combination of these colors with four on one side and two on the other and you find that there are fifteen different ways.

You can see how this grows exponentially and in fact it is exponential. Kind of configurations. So for the three to three configuration there are twenty different ways to color the atoms. The configurations two to four, well you know let me go back for just one second.

So you can imagine that with all these, all these possibilities with four and two, now we move one more over and with three and three there are twenty different ways to color those atoms. Twenty different possibilities. And then remember we had two containers, right? One on the left and one on the right. And we were starting with the one on the left being more filled. If you

go on the opposite side, one on the right being more filled. All the statistics are the same. You're just doubling that possibility. So the configurations two and four, five and one and zero and six have those same weights as four and two, five and one, six and zero. They contribute just as much. So the total number of configurations that we have is given by one plus six plus fifteen plus twenty plus fifteen

plus six plus one equals sixty-four. You can see what we were doing was starting with six and zero, one, right? Five and one and that contributes five and so forth until it's even. And then you can go to the other side. That's how many possible orientations of the system there were. And that's sixty-four total. Which as I mentioned earlier

no coincidence that's being exponential. It's an exponential statistical analysis. Two to the sixth. The number of containers two raised to the power of the number of atoms. You'll mention in a moment and I'll just say it now. You can imagine that with sixty-four and only six atoms, just think of how many

atoms are in a balloon, right? If those were just, if those were two metal containers and you had an atmosphere and one and it opened it up. If it's exponential like that, the number of possibilities that include atoms mostly on one side, they don't contribute anything at all. It's statistically insignificant. It doesn't matter. And so by the time

you get to any sort of significant amount of atoms, there's no there's no point in even considering a situation where the atoms are on mostly on one side. They're equal. They will be equilibrated. About fifty for sixty-fourth of the time, if the atoms are moving randomly, we'll look and see a four to two or a three

to three or a two to four. And that's just adding up those possibilities. Equals up to fifty to sixty-four. And about fourteen to sixty-four times, we'll see that six to zero, five to one or one to five or zero to six. Like I was saying, if we took that sixty-four and made it

ten billion, right, Avogadro's number, what is it? Ten point three. That's way beyond a billion. You're talking about huge, huge numbers now. So when we view it this way, the gas is spread out just because it's more likely. When the first scientists

started looking at these systems, they didn't have to have the chemicals. They could just envision what's mathematically correct to first approximations when you don't consider chemical interactions. Math describes almost everything. It goes into physics. It goes into chemistry. It describes the world. It's only when you start taking exact

scenarios where you have to start considering how atoms interact together etc. So it's more likely for them to be found equally distributed and jammed into a smaller volume. Likewise, and again it's just going back to statistics, if you roll a pair of dice, the most likely outcome is to get a seven. So when we have large

numbers of atoms, a nearly even distribution is overwhelmingly likely. Considering the solvent to be like a vacuum and the solute to be like a gas, so consider for a moment if I had a beaker full of water with solid sodium chloride on the bottom, right, before it's dissolved. It's just the instant before

you stir it and it dissolves. That's like a concentrated one balloon that's concentrated with gas atoms, right? It doesn't like to just sit there when there's room in the solution for the sodium chloride to dissolve. So considering the solvent to be like a vacuum and the solute to be like a gas, it is overwhelmingly likely that a solute

will dissolve if there is no energy of interaction between the particles like the ideal gas case. So they're saying in the event that sodium chloride had no interaction with another sodium chloride, well it would easily just disperse just like a gas molecule. So, but we know that this isn't entirely true,

right? Just like we know that the ideal gas equation isn't totally true because atoms can interact together, we know that a solute is going to have interactions together. So in sodium chloride we know that sodium and chloride have that ionic force together, yes? Well in the crystal there are lots of ionic forces. One sodium cation is not interacting with just one chlorine anion,

right? It's interacting with lots of chlorine anions. So there are lots of forces pulling them together. It takes energy to separate one sodium chloride from another sodium chloride. We know that there are forces between them. So there are several energies that we have to consider. If I take water in sodium chloride, how do I get that sodium chloride

to dissolve? Well there are solute-solute forces so the solute being sodium chloride and sodium chloride interacting with water. Solute-solvent forces, so sodium chloride interacting with water and there's solvent-solvent, water interacting with water. For me to get a sodium chloride from another sodium chloride and a water from another

water together, I have to break a water-water interaction a sodium chloride-sodium chloride interaction and then put them together. So you can imagine, let's say perhaps for a moment that sodium chloride and water weren't miscible. They didn't like each other. They do, I'm saying this and they didn't.

It would be hard to break sodium chloride which has a strong bond and water which has a strong bond together and put them together if they didn't have a good interaction. These are the sort of things we need to consider. So if all of the forces were the same if water had an equal energy interaction

with another water and sodium chloride has an equal energy of interaction with another sodium chloride and they had an equal energy of interaction with each other, then they just dissolve because entropy is increasing. Back to statistics, we have that increase in randomness. So the enthalpy, delta H dissolve measures the sum of the three interactions. We must first

separate the solute molecules, so that's sodium chloride-sodium chloride and the solvent molecules, that water and water, and then mix the two together. So water has a network of hydrogen bonds we should know this by now, that keep forming and breaking randomly in the liquid.

So that's something I have to consider when I break that water-water interaction it's not just dipole-dipole, there's also the hydrogen bonds. We already know that water has a strong intermolecular force, right? So I have to break that first and that costs energy. Other molecules that can form hydrogen bonds are then quite soluble in water. So, yes?

Oh, sure, am I going too fast you guys? You want me to jack up? Two slides? Just tell me when. When? When I heard it? Okay, alright this is my first time. Be gentle. So, let's see, we're gonna enter some interaction. Let me go back a couple slides and go through it again, okay? Alright, what's funny is I heard

Mahatma quickly and I was like, not me, not me. Are you gonna happen? Alright. Do you want me to just, what time is it? I got plenty of time, right? Eleven twenty-four. Wow, twenty-four minutes have passed

and I know that's most of the slides. Well, keep it down. So Mahatma went through most of my slides. That's my excuse. But we do have some sample problems at the end, some review questions for the exam and we're gonna go over those as well and that will take up some time. Or we can just have light conversation. I'm down for anything.

So I haven't been keeping track of the website. Has he gotten the other ones up yet? So the slides up to lecture four are on and we're at lecture

like six, right? That's true. So he hasn't been here all week. So I anticipate he'll have the slides up. I'll make sure to ask him about it too. Yeah, good. I'm glad we had this talk.

Okay, so we have something eleven twenty-four. Alright, let's go through this again. Okay. Too far? You guys are never happy with anything.

Okay, okay, we're good? No, I'm getting mixed signals from everyone. Okay, we're gonna start here. Anyone has a problem, get out and leave. Okay, we're gonna start here. I'll go through it again. A little more slowly, alright? Okay, so county configurations. Remember we're talking about these two containers.

Hey, hey, hey, yeah, alright, thank you. This is all on video. You're making me look bad. Okay, so we have two containers, right? And what did we do? We took one container that had six atoms

in it and we opened it up to another container that had zero. Just by statistics, we know that this is going to become equal. As I mentioned before, when we open up this container at first and we had six on one side, they're crashing around randomly

and they're just going to, by statistics, eventually find this hole and make their way to the other side. This is gonna keep happening until we have an equal amount going from one side to the other and equal amount coming from that second side to the first. It's in equilibrium. We have three on each. One switches

to the other, another switches back, and so forth. But this is only with six atoms. So, this is only with six atoms. So, in this distribution, where we have five to one, five on one side and one on the other, this configuration label

is five to one. Five in the first container and one in the other. And it's statistical weight is six. There are six ways that I can put five on one side and one on the other. Right? And I want to do a one to five, so one on one side and five

on the other. That also is going to have a weight of six. But it'll be five on this side and one on the other. So, we have to consider what are all the possible ways for me to arrange these atoms. So, with four and two, continuing that line of thought, we have fifteen

different ways. Where four are on this side and two here. So, green, blue, blue, purple, right, yellow, red. You just switch one of them. Keep the red on here, but switch the yellow with that light blue one. That's a whole different configuration. A whole other possibility for these atoms

when we consider their randomness. Continued on, there are fifteen different ways to arrange four, or excuse me, four and two. Those have four, two. That one has one, two, three, four, five, six, seven. Oh, there. I'm glad we redid this.

I'm sure this is why you were all confused. So, in any case, ways to put four and two. Fifteen different ways. And then for the three and three, there are twelve. You can do this yourself. When the numbers are small, you can just write out all the different combinations and add them up if you like math.

So, for that three for three, there are twenty different ways to clone the atoms. But then those configurations of two to four, five and one, and zero to six, as I mentioned, they have the same weight as four to five to one. There's another typo. I didn't write these. So, two to

four and five to one. First container has two. Second container has four. That has the same as the first container having four and the second container having two. This one, five and one, has the same as one and five. And then zero and six. Six and zero. What?

Why wouldn't each one have the same weight? Let's go back. So, two and four, five and one, zero and six. Why don't each of two and four, five and one, zero and six have the same weight? So, let's go back

to the images. So, compare six to zero. If the atoms are moving around totally randomly, there's only one possible way for me to have six and zero. With six in the first and zero in the second. If I go just to five and one, and we'll go back and forth

between this, there are six different ways to do this. So, if I just take a snapshot of this gas, right, totally random. If I took a million snapshots, I'm gonna get a statistical distribution between whether my snapshot sees zero and six or five and one. It is totally

statistical based on what the chances of me seeing six and zero or five and one. I know that six times more often I'm gonna see five and one just because that is more likely. There are more configurations for this than for six and zero.

Yeah, I must have said, I must have said quickly earlier, this weight is not a mass. Yeah, so I was blazing through before. This weight is not a mass at all. This is a statistical weight.

It's saying contributed to all the possible configurations. This is just a number that represents what chance do I have of seeing this situation. So, going back by one, this has a weight of one. There's only one chance of having six and zero.

One possibility. Going on to the next, I have five possibilities of having five in the first and one in the second. And so this has a statistical weight of six. It's six times more likely to see this than six and zero. Does that answer your question? Thank you for bringing that up.

I didn't realize that people had missed that. So, four and two. So you can see, as I was talking about, if you have five and one, there's five possible configurations. Take that a step further. Go to four and two.

Here are all the possible ways for me to write four and one and two in the second. There are fifteen different combinations. So if the atoms are totally at random, going to have some configuration, I have fifteen different ways that show me this exact arrangement. Remember, these atoms,

all the different colors, they're still just one atom. They're all just krypton. So if I took a picture of this right now, I wouldn't be able to distinguish the top left from the bottom from any points on the right. It would all look exactly the same to me. And so I would count that as another one of my

four and two as a second count. And if I keep doing that, snapshot, snapshot, snapshot, over and over again, I'm going to have fifteen different possible ways to see four and two, but only one possible way to see six and zero. And that's where this comes from. That's where the statistical reasons for equilibrium come from. So

for that three and three, which is taking the same thing to the next configuration, there are twenty different ways to color those atoms. And then you were asking why they don't have the same weight. So the configurations, well, I already answered that question before, I suppose. So the configurations

two and one, four and the other, five and one, one and the other, and zero and one, and six and the other, those have the same weights as flipping your diagram. So six and zero has one possible configuration. And zero and six has one

possible configuration. Together I have two possibilities of ever taking that image and seeing six in either with zero and the other. So the total possible the total configurations that are possible are sixty-four. So let's think about that six

and zero. I take a snapshot. There are two and sixty- four chances that I will take a snapshot with six and one and zero and the other. It's very small. And if I have

and fifty out of sixty-four of those times if the atoms are moving randomly, we'll look and see that four and one and two and the other. Or three and one and three and the other. Because what I'm doing is adding the twenty. So remember twenty corresponded to three and three, right? So three and three plus the chances to see four and two. So four and two, that was fifteen, right?

So four and two, fifteen. Three and three, that's twenty. And two and four, which is another fifteen. That's how many, right? That's the sum. Four and two, fifteen. Fifteen, thirty, fifty. Fifty out of sixty-four times, I'm gonna see one of these.

And fourteen out of sixty-four times we'll see six, zero, five, one five or zero, six. So the six and the zero, right? That contributes one. The zero and the six, that contributes another one. The five and the one, that contributes six. The one and the five that contributes six. Total. Fourteen out of sixty-four.

So the chance of me seeing very few on one side and more on the other is very small. Only fourteen and sixty-four when I only have six atoms. Since the number of possibilities increases exponentially, the chance of me having

most of a gas on one side and little gas on the other that becomes completely insignificant and we don't even consider the possibility. So they spread out because it's more likely for them to be found equally. If I take a million atoms the chances of me seeing a half a million on one and a half a million on the

other, that's incredibly high. Of course, you know, give or take a couple atoms. Who's counting right now? Likewise if you're a gambling person, we roll a pair of dice. Most likely outcome is to get a seven. Quick tangent. Went to Vegas for my

first time a few weeks ago from anniversary. Bet a hundred dollars. Lost it all. Never do that again. Thank you. I know. It was roulette. I'm stupid. I know. Who plays roulette table? You're gonna lose. I was up sixty and then ten minutes later I'm down fifty. I don't know how it works.

When we have large numbers of atoms, a nearly even distribution is overwhelmingly likely. So, considering that solvent to be like a vacuum, as you were mentioning before, and the solute to be like a gas, it's overwhelmingly likely that a solute will dissolve if there

is no energy of interaction between the particles, like the ideal gas case. So, let's picture this again. We have that beaker of water with a bunch of sodium chloride just sitting in the mass. It's not likely for that sodium chloride to just sit in the mass, as long as those interactions are the same, like the ideal gas case. No energy

of interaction for sodium chloride. However, we know that there are forces between atoms in condensed phases. As I was mentioning earlier, we know that sodium chloride has an interaction with energy. And we know that in a sodium chloride crystal, that sodium chloride is interacting with several other sodium chlorides. And it's hard to break them apart

and strong ionic forces that require energy to break one sodium chloride from another sodium chloride. So, there are three paralyzed energies to consider. Solute, that's the water, oh excuse me, sodium chloride in this case, so breaking

apart sodium chloride from another sodium chloride, that takes energy. The solute and solvents. When I take that sodium chloride and water approaches, then we have interacting forces. And in the case of water and sodium chloride, that's a positive energy. It's favorable. So, those attract. And I have solvent-solvents, so water and water.

Like I mentioned before, water, we know, has very strong ionic forces. So, I have water, it takes energy to break that apart. So, I have two unfavorable energies, breaking water away from water, breaking sodium chloride away from sodium chloride, and I have a favorable energy, taking sodium chloride and interacting with water. And if that interacting energy is strong enough, it can overcome

the energies that I'm having to break. So, if all of the energies are exactly the same, then breaking the water apart from the water and replacing it with sodium chloride, well that's no different energy wise. Right? If the interaction of water with water, I break it

apart, but then that sodium chloride comes in, and I get that energy back, well then it doesn't matter. And it's just going to be statistics. And it's going to dissolve, because we're increasing that entropy. Delta S is greater than zero. If this were the first time going through it, would that be a good place? Better? Okay.

Alright. Keep that in mind. So, the enthalpy delta H of dissolving measures the sum of those three interactions. We must first separate the solute molecules and the solvent molecules, and then mix the two together. And if that is a good energy

it will happen naturally. And if it takes more energy to do that than to not do, it's not going to happen. So, water has that network of hydrogen bonds that keep forming and breaking randomly in the liquid. Other molecules can also form those hydrogen bonds

and then are quite soluble in water. So, let's go back to what I was saying a moment ago. Right? I have to break water from water, and I have to break something else from something else. Let's say ammonia. Right? NH3. Well, I know NH3 has hydrogen bonds, right? So, when I take that ammonia and bring it closer to water, I know they're going to form strong hydrogen

bonds. This increases that interaction between water and that solute. So, I think it would be more likely to dissolve. In this case, it would, as long as it is soluble in water. So, ethanol methanol, ethylene glycol, they're miscible with water, right?

What do they have? Ethanol has a hydrogen bond. Yeah, it has an aliphatic region, but it also has that OH group, right? And that forms hydrogen bonds with water. And so it can dissolve. Same thing with ethylene glycol and methanol, forms hydrogen bonds. On the other hand, a long chain alkane, like hexane,

with H atoms only on carbon, it's not going to dissolve very much. We get two layers, right? Like salad dressing. Or if you're just in your lab and you have oil and water. Right? So, let's talk about this for a second then. If we have oil, we'll just

say it's hexane. Okay? Hexane and water. They don't mix. We all know that. Why don't they mix? Hexane, what kind of forces does it have? Interlocking forces, I mean, right? Just dispersion mostly, right? It doesn't have any H bonds. It doesn't have a dipole. It doesn't have any ionic bonds. It's dispersion. Hexane doesn't hold together very well. That's why it has such a low boiling point.

Water, however, holds together very powerfully. It has strong hydrogen bonds. It has dipole-dipole forces. It holds together. So, when I try to put the two together, I have to break a hexane-hexane interaction. That's easy. Because its dispersion forces aren't very strong. But I have to break a strong

water-water interaction. Now, if the hexane-water interaction were powerful, that would be okay. But it's not. Hexane doesn't have any hydrogen bonds if you get the water. It's not a dipole. So, it can't interact with water's dipole very well. It's just sort of like dispersion forces or ion induced dipole. That sort of thing.

It's a weak interaction. And so you're not gonna break apart strong water with weak hexane. Because their interaction is also weak. So, they separate. So, we're on to some practice problems. Exam time is next week. Remember what Dr. Shaka said.

Oh, it's not time to leave. We have some problems. I mean, you can leave if you want. I'm not keeping track. But let's see here. What time? We got eleven forty-three. So, we got plenty of time. There are about five questions here. And the exam is Tuesday. It's gonna cover chapters five and eleven plus the review

one through three. But he'll probably focus on five through eleven, I'm sure. But just in case you need to know one through three. So, we'll just go through some exam questions right now. Well, not exam questions, some review questions. And we'll see how it goes. What factors influence the boiling point and the melting point of a pure

substance? What influence does molecular shape have on the liquid range? So, I think what I'll do for these problems, especially if we have plenty of time, is I'll just give maybe thirty seconds for you guys to think about a problem and then we'll go over it together. So, go ahead and just think about this for about thirty seconds. I'll try to keep quiet. It's very hard for me. I like to talk, but I'll try.

It's very hard. Alright, so let's talk about this.

What factors influence the boiling point and the melting point of pure substance? And how do you affect that liquid range? So, we know the boiling point is gonna be influenced by molecular forces. Boiling point is

how much energy do I have to put in to get those to break apart from each other? Right, to put them in the gas phase. So, very strong interactions are going to increase my boiling point. Water has very strong interactions. It's got a high boiling point. Hexane does not have very strong interactions. It's gonna have a low boiling point, right? So, we consider

ion-ion, ion-dipole, dipole-dipole, London dispersion forces and hydrogen bonding. If you ever forget what those mean, go look in the textbook. I really encourage people to read the textbook. I read the textbook. Look where I'm at. Up here teaching you. It works. So, the melting point is influenced by similar factors, but

packing is very important. I don't recall if Dr. Stock ever mentioned packing in his lectures. But how well layers and layers of your club hound fit together influence how closely they can get together and how strong that interaction is, right? So, we know that if I just took an ion-ion,

like a positive and a minus charge, the closer I bring them together the harder it is for me to pull them apart. Same thing with solids. If I have things that pack really well together, then when it comes time to try and break them apart to go from the melting point to get into the liquid phase, well it's going to be harder to pull them apart

into the liquid phase if they're packed really tightly together. If something doesn't fit at all, well it's not as able to interact as well and you can break it apart a little bit easier. And that's going to lower its melting point. So, a molecule that is round or symmetrical may have a high melting point because it fits together well. So,

if a symmetrical molecule has only dispersion forces, it may also have a low boiling point and still have a small range of our witches in the liquid. Some molecules like CO2 have no liquid range of 1 atm pressure. They sublime. Again, a phase diagram. You should be getting very familiar with these. Pressure versus temperature. We know that with

dry ice, with CO2, that's minus 78.5 degrees Celsius. That's at about 195 right in that range. So, it's unfortunate this graph doesn't show that place, but here we're in the solid form, dry ice, right? So, at 1 bar is around 1 atm.

At a temperature of around 195 Kelvin, we're in this region right here, where we would be a solid. And as it warms up, we enter right into the gas phase instead of the liquid phase. So, we need almost 60 bar pressure to have CO2, be it liquid or room temperature. So, if I increase this pressure,

right, let's say I'm at 100 bar, and then I increase the temperature. Then I'll enter into the liquid phase, and eventually it's super-cooled. Yes? What was that?

So, if something has a melting point 0 degrees Celsius, like water, and a boiling point of 100 degrees Celsius, that's a 100 degree Celsius range. They're saying that when things have very, very, very

small amount of interruptive forces, sometimes it can't even stay in the liquid very well. Once you melt it, it goes into the gas phase pretty quickly. So, you have a very small range, let's say 10 degrees. I don't know what, well, for CO2 there's no range at all, right? It doesn't stay in the liquid. But they're saying there will be a much

smaller range at 10 degrees Celsius, something like that, over which it's in the liquid. I can't think of any examples that have a really small range, but, except for this one maybe. So, are there any more questions on that problem? All right, I don't see any hands.

Okay. So, a gas mixture has 90% argon by volume, and 10% helium by volume. You may treat these as ideal gases. If low pressure effusion through a pinhole is used to try to separate the two gases, what would the percentage of helium by volume be after the process? Now, we're not gonna sit here while people work out

the whole problem. Take a minute to think about it, write out whatever you want, think about how you address this problem, and then we'll go over it together, okay?

All right, let's talk about it. So, we're using Graham's Law, page 208. It tells us the relative rates

of effusion through a small orifice go like the inverse square root of the masses. So, we'll look at this for one moment. That is R1 over R2. Those are rates, not radii. The rate of the first thing going through a small hole divided by the rate of the second thing going through a small hole equals

the mass of the second thing over the mass of the first take the root. So, let's talk about it for one second. Does this make sense? I have argon and helium. Argon is very heavy. Helium, not so much. I put them into a container with a small hole, right?

If they're at the same temperature, which they should be, which one's moving faster? Helium, right? And we have our own equations that tell us that. Helium's moving very quickly. It's bouncing around very, very fast. Statistically, it has a better chance of getting through that hole

because it's gonna hit that hole more often. The size of the hole doesn't have anything to do with the difference in size between argon and helium. Argon is still infinitesimally small compared to the size of the hole. So, it's only based on the mass. Helium is bouncing around. It's gonna hit that hole more often

and get through. So, if we take this and we plug in those masses and take the rates, we can find out the ratio of the rates. We can't solve R1 and R2. We're taking a ratio. Helium is four grams per mole and argon is forty. And so what this means is that the ratio, that relative rate,

and remember that's R1 over R2, the ratio is three point one six two. That's where we're saying helium will go through that hole three point one six two times as fast as whatever argon does. We haven't

insinuated any kind of rate or a time. We're not saying how long this happens for. We're saying that for one moment, the rate is going to be three point one six two times the rate of argon. And so if we set that rate of argon diffusion to one in any units, we know that the rate of diffusion

of helium is gonna be three point one six two times whatever that rate is. And hence after one stage of a fusion so we could have this in seconds or hours, whatever, it depends on how accurate you want to be. After one stage of a fusion through an orthos we'll have relative numbers of atoms. So

argon at some rate going through, I'm going to have point nine of whatever my units are going through. Helium there's only point one worth of helium versus argon. Remember it was ninety percent argon and ten percent helium. And so the ten percent of helium at three times faster

rate I'm going to have point three one six two amount of something coming through. And so what happens is total I have the point nine plus the point three one six two that's how much got through and this is how much helium there is that's now twenty six percent. So we increase from ten

percent helium to twenty six percent helium. I believe that Dr. Shaka used uranium as an example for this in the first lecture on ways to separate things. It can be quite useful. Although they don't use uranium isotope separation by the method they're not anymore. They use centrifuges.

So, oh, any questions? I didn't realize that was the end of the answer. Say again? Yeah, sure. We're all just writing down or do we have questions? What? Oh, sure. The one with the actual question on it?

Yes. In the second, so we have one chamber and then another. In that second chamber, right, we let stuff transfer over. That's now twenty percent, or excuse me, twenty six percent helium. Now something to consider is that rate doesn't stay the same, right? In that first

stage we had nine percent argon and ten percent helium. In the second, after we've let that go through, we know that my second container has twenty six percent helium. What? Why is it important? That means that in the first container I now have even less helium by percentage than I did before. I had ten percent before. Now whatever the amount is is going to be less than ten percent. So if I

keep letting that going, the difference in rate, which was three point one six two, that's going to start decreasing. Because as I get closer and closer to being almost no helium, there's going to be more argon going through than helium. So. Your question asks what would be the percentage of helium

by volume be after the process. What were you looking at? The last one? Oh, it was saying the percentage of volume is going to be twenty six percent. That's the new container. Whatever went through is now twenty six percent

helium. Yes, so this makes sense, right? Let's say one is helium. So the rate for helium is going to

increase faster and faster relative to argon as the mass of argon, or whatever my second one is, it's heavier and heavier versus helium. So let's pretend it's not argon for a second. When this is argon, we have three point one six four, right? So m two forty over m one four, take the root, three point one six four. If I made m

wet, right? Something really heavy. Let's just say something two hundred grams per mole, okay? That now is two hundred over one and take the root. That's going to mean r one over two is now really heavier. Because helium is so much faster than

whatever that really heavy gas is that I put in there. Does that make sense? Okay, I can hear that whole thing. Oh, it just makes the math easier if we had a

real situation. We would want to know exactly how much argon we have. This is a demonstration of using ratios to compare relative rates. We just said that we say that r one over r two

is, there you are, three point one six two. I'm pretending that whatever the rate of argon is, I just call that one. I'm saying relative to that, so r two is one. If the ratio r one over r two is three point one six two, that lets me say that the rate of helium transfer is three point one six two

times whatever argon is. I think I'll have to limit that by what? Well, we've had plenty of time, right? Wish there was a clock this high, this guy. Eleven fifty eight, his class gets out at twelve twenty. We've got a few more questions. If you have more questions on it, come down and ask me afterwards.

Did anyone want just a moment to write something down real quick? You got it? Alright, so another practice problem. Recall that a Van der Waals gas follows the equation of state as follows. I'm not going to read through that. Remember that the ideal

gas equation is just p b equals n r t, and here we've said p b equals n r t is great, but it's not fantastic. So we've added in these correction factors, a n squared over v squared and minus n v. So where the gas specific parameters a and b depend on a gas

if for xenon we have a equals four point one nine two liters squared atm per mole and b equals point zero five one five six liters per mole. Then what is the volume of one mole of xenon at five degrees Celsius and two atm pressure to three significant digits. Now I'd like to take a moment here

to point out something about units. So if you were just looking at this you'd think, well what are the units of a? I know what the units of n are, right? N squared, it's moles, that's going to be moles squared. And the v, that's liters, that's liters squared. For me to add two things together they have to be the same units, right?

So automatically I know the units of a because for a n squared over v squared to equal atm, a has to be liters squared to cancel this out. Atm so that you have atm per mole squared to cancel out the mole squared there. That's just a quick way to address where the units in a situation.

Similar to nb, for volume equals liters, nb also will have to be liters. And so b is going to be liters per mole because n is mole. So mole times liters per mole gives you liters. So go ahead and take a moment

to try and find the, where was it, and what is the volume. Try and find the volume of one mole of xenon at 5 degrees celsius with 218 pressure to three significant digits. Tip,

don't try and solve for v in that equation. You're going to make a quick guess using the equation that you have.

That's not quite as accurate as that one, but it'll let you almost get the answer. So Dr. Shaka earlier in the quarter, remember he has the get really close and then punch in the calculator a couple guesses until you're spot on, right? So don't try and solve for v. Alright, so let's go ahead and

flip on over. So first thing, as Dr. Shaka is fond of telling you, remember to convert celsius to kelvin.

Keep everything in the same units. His advice is to use pb equals nrt to first guess almost the right answer. And then you're going to nudge your way to the correct answer. So figure out that nrt figure out nrt and do not round r cause we need the accuracy. So something he keeps on harping on in the class

is don't round too much, right? If he asks the accuracy, don't think at the end, just three digits, just make sure at the end I have three digits. If you cut out certain digits earlier in the problem, it's gonna ruin your accuracy down in the end. So you have to keep that many digits all the time.

So in this case we're using as many digits of r as is practical. Going out to four is plenty good. That's gonna give us all the accuracy we need and more to get our three digits in the end. So figure out nrt don't round.

R, and that equals 22.8243 liter atm. From here, you, so we started with the ideal gas estimate of the volume when we begin guessing. So, or using that it works for you, you can graph it, or whatever, but make sure you get three

digits correct, verify your answer by substituting it back into here. Okay, all right. So, he's saying, well we know NRT. If we go back to the, to the full equation, NRT doesn't have A and B in it. We're not making any assumption by just solving for NRT. So, we can do that right off the bat. It's this side with the more complex equation that gives, that causes us grief. So, we can start with

just PV as a quick guess, which doesn't have the A and B in it, and then we can, in his words, scoosh our way over to the correct answer. So, if we let our function of volume equal 2 atm plus 4.192 liter atm per volume squared

times volume minus 0.05156L. So, let me go back for a second. What we're doing is we're taking, oh, we have this equation, right? We're trying to figure out the volume. So, we can easily make this a function that depends on volume, and

that's just pressure plus an squared over volume times volume minus nb. So, let's go back up to that slide. So, our function pressure plus the 4.192 liters squared atm over v squared. So, that's a, that's an squared divided by

v squared times v minus the nb. So, he just figured out these numbers and put them in, and this is our function for v. So, if we plug in values of v in liters and see what we get, our starting guess is, using the ideal gas equation, just v equals nRT over P. We should all know that, definitely. And

that guess in it is 11.4121 liters. So, we set up a table, okay, where this is the volume in liters, this is the function of volume in liters atm, our target value, which should be 11.4121, and then I'll

comment on that value. So, when we do the ideal gas law, we have 11.4121, or excuse me, the target value is 22.2. When we do volume based on the ideal gas law, that's 11.4121, and the function, when we plug the volume into our function, is

23.0867 liters atm. Our target value, which is nRT, right, we're trying to get PV equals nRT. Ideal gas law wasn't good enough. Van der Waals equation has a and b. We're taking that first guess, putting in the function, we know that

nRT is 22.8243, we're trying to see how accurate we are. So, volume 11.4121 gives us too big a value. So, we go a little bit smaller in volume, right, 11.2. Well, put it into our function, and that's a little too

small. So, 11.3, we put that in, and just, excuse me, his words, not mine, I think this is a weird way to describe it. I'd rather use tiny. So, 11.25, that's a bit small. So, depending on what accuracy we're looking for, we know that the

answer is in between 11.25 and 11.3. So, rounding to three digits, like it asked, that's 11.3.

So, I have all of that, don't I? Okay, so, Van der Waals equation, accurate, ideal gas equation, not so much, right? We can replace all of this with PV, and that's

the ideal gas equation. So, going forward, your question was, why does, let me bring it up, why does volume equal NRT over P? That's just using the ideal gas law, right? Did that answer your question? I feel like I missed something in your question. Is there something that I needed to do? Okay, I did? Alright, good.

Any questions with that? No? I think there's one more problem. Let me just go ahead. Yeah, so we have an x-ray diffraction question, and we have 12.08.

Alright, we got 12 minutes. I think this is the last problem. An x-ray diffraction experiment using copper K alpha x-rays, so that's just describing what level of x-ray it's at, of wavelength 1.54 angstrom on a pure

metal crystal gave a first-order Bragg peak at an angle theta equals 14.17 degrees. If you're not familiar with this, go read the book. They have a quick and simple explanation on diffraction and how this is working. So we have a metallic crystal, it's just layer upon layer, a simple cube probably, of matter.

So the incident angle of radiation is 14.17 degrees, and they tell us the wavelength is 1.54 angstroms. Assuming the crystal has a cubic lattice, and that the peak arises from the atoms on the corners of the cube, what is the dimension of the unit cell? If the structure is simple cubic, what

is the radius of the atom? What further information would you need to identify the actual element? So I'll go ahead and give you one minute to work on this a little bit. Excuse me. And then we'll work over it. Yeah. It's

funny, it's funny to be down here and actually see people like sleeping with their heads and stuff like that. Like, I mean, I smell like laser pointer in the eye, but obviously I can't because their eyes are closed. All right, real

quick before we go over it, I want to ask you guys up there in the balcony, is my camcorder still going? It is? Okay. This is my first time teaching a lecture with this many people. I've never done it before. If you all just give a shout to my mom and dad, I'm going to send it to them. So I'm free. We're all going to say, hi mom, hi dad. All right, just for me. Good for me, good for me, good for me. Okay? One,

two, three, hi mom, hi mom, hi dad. Love you. Okay. All right. Oh, my wife. Hi Daniela. Thank you.

Okay, let's talk about this. All right, so we have a first order diffraction and that's code for n equals 1. If we're thinking about this equation, okay, 1 normally this would be n, n equals 1 in the Bragg equation. Hence we have 1

times the wavelength, which we already have, so they gave us that, equals 2 times d sine theta. They gave us theta while we're solving towards d. So let's go back one second here because there was an assumption that was made and we didn't talk about it. When it says that assuming the crystal has a cubic lattice and that

peak arises from the atoms in the corners of the cube, what that's saying is d was a measure of the distance in between my slit where the x-rays are being diffracted, right? It's not always the case that one set of layers makes that diffraction. We're making that assumption. So if I just solve for d, well

maybe that comes down as 10 angstroms and that could correspond to five or six layers. We made the assumption that it's from the first layer and so d is just the distance in between the two corners on that point, okay? So point three,

so if you solve for d, that's 0.3145 nanometers and for a simple cubic unit cell, this is 2r. Remember we were trying to solve the radius of the atom. If it's a cubic lattice, I have atoms in all the cubes and they're touching each other, right? If they're touching each other and I know the distance in between the middle point of those two atoms because

they're right at the corners, then I know the radius of each atom because I know the distance and the radius is just 0.157 nanometers. So we're trying, oh sorry, we're just writing down? Do we have questions? Questions so far? Just

moving on? I'm hearing a lot of murmurs which I don't know what that means, yes or no. So we were trying to find the, let me read the end of the question real quick. We're trying to find the mass, right? Yeah, what further information do you need to identify the actual element? So what do we need to identify the element?

Well, we already have the volume, right? We have a radius. What's the last thing that we would need? For a simple cubic cell, there is one atom per unit cell, right? That's because at each corner we have one eighth of that atom contributing to the unit cell. Eight corners, that's one atom. So we can figure out how many moles there are in a cubic

centimeter. So if we then knew the density of the material in grams per centimeter squared, then we would know how many grams there are per mole. And that lets us, of course, identify the atom. If we know the grams per mole, we just look over at the periodic table and that's all we need. So, and this is very,

very true. Real X-ray experiments are, of course, a lot more complex than this. Nobody takes crystal structures of things that are just one element anymore. They're incredibly, incredibly complex. Any questions?

Well, why don't you go ahead and start going. You can come down here. Okay, so notes for the exam. They can only be notes that you've taken in class.

So you can't go outside of class, write a book, come bring it in. You wouldn't want to do that anyway. How are you going to have time to look at it, right?