For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

## Discussion

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books & Services

### Vapor Pressure of Solutions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Vapor Pressure of Solutions 2:07
- Vapor Pressure & Raoult's Law
- Example 1
- When Ionic Compounds Dissolve
- Example 2
- Non-Ideal Solutions
- Negative Deviation
- Positive Deviation
- Example 3

### AP Chemistry Online Prep Course

### Transcription: Vapor Pressure of Solutions

*Hello, and welcome back to Educator.com; welcome back to AP Chemistry.*0000

*Today, we are going to be talking about the vapor pressure of solutions.*0003

*Now, we have talked about vapor pressure before, and that is basically: at any given temperature, there is always going to be some of the molecules on the surface of a liquid that have escaped into the gas phase.*0008

*So, in a closed container, there is always going to be some gas above the particular liquid--like, if I have water in a closed container, yes, you see the liquid water there; but on top of the liquid water, there is a little bit of water vapor, at every temperature--not just at 100 degrees Celsius.*0018

*100 degrees Celsius is the boiling temperature of water; that is when every single molecule of liquid water has enough energy to go and stay in the vapor state.*0036

*But, in every liquid, there is always some amount of gas above it.*0045

*Well, again, pressure is just a measure of the number of particles in a particular closed container; so we call that thing the vapor pressure of the solvent.*0051

*Well, today we are going to be talking about the vapor pressure of solutions; if you take a non-volatile solute (like if you take salt and dissolve it in some water), that actually ends up changing the vapor pressure of the solvent.*0059

*Now, you can call it different things, and in the process of discussing it, sometimes I'm going to say "vapor pressure of solute"; sometimes I'm going to say "vapor pressure of solution"; we'll know what we are talking about if we are talking about pure solvent or some solution.*0071

*But, in essence, when I put a solute (like salt) in water and dissolve it, now the water molecules are actually competing with other particles (namely, the sodium ion and the chloride ion) to make their way up to the surface, so they can actually escape into the gas phase.*0086

*Well, because these ions are actually running interference, not as many water molecules make it to the surface; therefore, not as many jump up into the vapor phase; therefore, the vapor pressure (which is a measure of how much is on top of the liquid at any given time) is going to be lower, in general.*0104

*Let's go ahead and get started; that is just a quick description of what it is that we are going to do.*0122

*OK, so let's write down: the vapor pressure of a...no, not "the vapor pressure of a"...what am I saying here?--I can't write that.*0127

*The presence (I'm getting ahead of myself) of a non-volatile solute (and non-volatile means, generally, it is not a liquid; it's a solid of some sort--a salt or sugar--some covalent compound that forms a crystal) lowers the vapor pressure of the solvent.*0148

*And again, if you need to do a quick review of vapor pressure, just go back some lessons, where we actually talk about what vapor pressure is.*0184

*We have actually also talked about it twice, if you remember--when we did gases, when we were collecting a gas above a liquid and using that technique of displacement to measure the volume of a gas, we talked about vapor pressure then.*0191

*So, we have actually discussed it a couple of times.*0203

*All right, vapor pressure of a solvent...now, this is our important equation: it's something called Raoult's Law, and it goes something like this.*0207

*You know what, I'm always worried when I use different variables; I actually like to write my variables out; I don't really care for letters, myself, personally--because there are so many letters in science that you collect, after a certain number of classes, that they tend to get confusing.*0218

*Well, you know what, let me just go ahead and leave it as I have.*0236

*The vapor pressure of the solution is equal to the mole fraction of the solvent (it's very, very important to identify, because now we have solvent and solute; we have to specify what we are talking about), times the initial vapor pressure of the solvent.*0240

*So, Raoult's Law says that, if I have a particular solution, and there is some solute that has been put into solvent to create the solution, the vapor pressure that I end up measuring under ideal conditions--it will be: I take the normal vapor pressure of the pure solvent, and I multiply it by the mole fraction of the solvent.*0258

*So now, the mole fraction is no longer 1; there is actually solvent and solute in the solution; therefore, the mole fraction is going to be less than 1.*0283

*Remember what mole fraction is: the moles of the A over the total moles.*0295

*So, in this case, it's going to be the moles of solvent over moles of solute plus moles of solvent.*0299

*It is going to be less than 1, so your pressure is actually going to be less than the normal pressure, which makes sense, because again, molecules--the solvent molecules--are interfering with the solvent molecules' ability to get to the surface so they can jump up into the vapor phase.*0305

*Let's do an example.*0321

*We'll do Example 1: OK, now, what is the expected vapor pressure (I'll just call it V/P) at 25 degrees Celsius, for a sucrose solution of 120.0 grams of sucrose, dissolved in (oops, it's not 572; I want...) 527.0 milliliters of water?*0326

*OK, now we're going to have some information that we need here: At 25 degrees Celsius, the density of water (that's a little Greek delta) equals 0.9971 grams per milliliter.*0374

*The vapor pressure, the normal vapor pressure, of H _{2}O (the solvent) equals 23.76 torr.*0393

*That means that, at 25 degrees Celsius (just really briefly), if I have a glass of water that is covered, at 25 degrees Celsius, the amount of vapor--water vapor--above the liquid in the glass actually accounts for 23.76 torricelli.*0407

*Not a lot, but it is some pressure; it is some gas particles that are bouncing around, hitting the walls of the container, causing a pressure; that is what vapor pressure is.*0425

*It is the measure of how much vapor is actually above a liquid at a given temperature.*0434

*Sorry to keep repeating myself, but I find that repetition sort of enforces comfort with these things.*0439

*Now, sucrose (which is just normal table sugar--so we're just making a sugar solution) is 342.3 grams per mole; so these are the things that we are going to need.*0447

*OK, so Raoult's Law: this is the law that we are going to use; let me go ahead and circle this in red.*0460

*That is what we want to use; so we need to find--we have this one already; that is the 23.76 torr; we need to find the mole fraction of water, the solvent.*0465

*We go...so we are going to need to find, first of all, the moles of sucrose; well, the moles of sucrose--we said we had 120 grams, times 1 mole (nice stoichiometric conversion), over 342.3 grams; that gives us 0.3506 mol.*0478

*You know, let me write my numbers a little bit bigger; I think they might be a little tiny here: 0.3506 moles of sucrose.*0500

*OK, now we need the moles of H _{2}O.*0509

*Well, H _{2}O...we have 527 milliliters, so we need to convert that to grams first, so that is why we need the density.*0513

*We have 0.9971 grams per milliliter, times (grams goes down here; mol goes up there) 18 grams in 1 mole; so it's going to be 527, times .9971, divided by 18; and we get 29.193 moles.*0521

*29.193 mol: therefore, the mole fraction of H _{2}O is equal to the number of moles of water, divided by the total moles (which is 29.193+0.3506).*0543

*When I do that, I get 0.9881; so it's going to be less than 1; it's a fraction--it's a part over the whole.*0565

*If we wanted the mole fraction of sucrose, well, we would say .3506 in the numerator; the denominator is the total moles; I hope that is OK.*0574

*OK, so now, when we plug this into the equation, the vapor pressure of the solution is going to be 0.9881 (which is the mole fraction of water), times the normal vapor pressure at that temperature (23.76), and we end up with 23.47 torricelli (or torr, or millimeters of mercury...if you want to convert that to atmospheres, psi, kilopascal, whatever you need; it's just a unit of pressure).*0583

*Notice, 23.47 is not a lot less than 23.76, but it is less.*0615

*That is what is important: when you add a solute to a solvent (in other words, when you create a solution--a non-volatile solute), your vapor pressure will lower.*0621

*What that means is that fewer molecules of liquid will hang around in the vapor state on top of the liquid, because fewer molecules will be able to get to the surface, because the molecules of solute are getting in their way, so they can't jump up as much.*0633

*OK, now, this was for a covalent compound: sugar is a covalent compound--it dissolves, but when it dissolves, it doesn't actually break up into its individual atoms.*0650

*It just dissolves: the crystal comes apart--the molecules separate.*0661

*Ionic compounds are different; so, when ionic compounds dissolve, they dissociate; in other words, they separate into their ions; so we must account for the total number of dissolved particles--that is what is important: the total number of free particles.*0665

*For example, if I had some magnesium chloride that I dropped into water, well, I have: 1 mole of magnesium chloride produces 1 mole of magnesium ions, plus 2 moles of chloride ions.*0713

*So, I have to account for...whatever the number of moles is that I get of that, it is actually 3 times that many particles.*0727

*1 mole of magnesium chloride produces 3 moles of free-flowing particles--of free particles--so it's actually 3 that is what I use when I calculate the mole fraction, OK?*0737

*That is it; that is the only thing that is different--you have to differentiate between ionic compounds and covalent compounds by accounting for the number of free particles that are floating around.*0751

*Let's do an example.*0761

*This is going to be Example #2.*0765

*OK, we have: 40.0 grams of solid potassium sulfate is dissolved (let's actually write it so we can read it; that might be nice) in 180.0 grams of water.*0770

*So, this time, they went ahead and they gave you the mass of the water directly, instead of volume--not a big deal.*0796

*OK, what is the vapor pressure of solution?*0802

*OK, well, again, we are going to use Raoult's Law; so, since I don't have it on this page, let me write it one more time.*0812

*The vapor pressure of solution is equal to the mole fraction of solvent, times the initial vapor pressure of solvent.*0819

*OK, well, we have the vapor pressure of solvent; that is going to end up being...it looks like this is going to be...I'm sorry, I didn't specify: this is 25 degrees Celsius, so we are going to use the 23.46; that is water.*0830

*The initial of the solvent is going to be 23.76 torr; so that is that one.*0842

*We need to find the mole fraction of the solvent; OK.*0856

*Well, let's find the number of moles of water; so moles of H _{2}O: we have 180.0 grams; 1 mole is 18 grams, 18.0 grams; so we are looking at 10 moles of water.*0859

*OK, now let's find the number of moles of K _{2}SO_{4}, potassium sulfate.*0876

*Well, we have 40 grams of the potassium sulfate; and we want to do 1 mole of it; when we check the molar mass using a periodic table, we get 174.2 grams (I hope I did the arithmetic correctly--if not, not a problem--it's just a number), equals 0.2296 moles.*0882

*OK, now here is where we have to be careful: K _{2}SO_{4}...when K_{2}SO_{4} dissociates, it produces 2 moles of potassium ions, plus 1 mole of sulfate ion.*0905

*1 mole of potassium sulfate produces 3 moles of particles--3 moles of free particles.*0919

*That is what matters: the identity of the species don't matter; these are colligative properties--these are...what matters is the number of particles that are floating around in solution.*0930

*The more number of particles you have floating around in solution, the more they get in the way of the solvent doing what it needs to do, which is escape into the vapor phase.*0940

*3 moles of free particles are formed.*0949

*So, when we do our mole fraction, our mole fraction is going to look like this.*0954

*The mole fraction of H _{2}O is equal to...well, 10 moles of H_{2} over 10 plus three times the .2296 (right? .2296 is the number of moles of potassium sulfate, times 3 because, for every mole of potassium sulfate, three moles of free particles are going to be floating around in that solution).*0962

*So, you get: 0.9356; therefore, the vapor pressure of solution is equal to 0.9356 (let me make the numbers a little bit bigger; I don't know why--I usually have a habit of doing big numbers, and lately I know I have been doing small numbers), times 23.76 torricelli, which gives us 22.23 torr.*0986

*There you go: the only difference--handle it exactly the same way, but when you are dealing with an ionic compound, make sure you account for the dissociation of the ion and the number of free particles that are floating around--very important.*1023

*OK, so the identity of the dissolved species doesn't matter, just how many; it doesn't matter if it's sucrose or potassium sulfate or whatever.*1035

*It is how many particles, not the identity--just like gases, remember?--the identity of the gas doesn't matter--1 mole of gas at standard temperature and pressure always occupies 22.4 liters.*1043

*All gases--it doesn't matter what the identity of the gas is; what matters is the number of particles.*1055

*OK, so now, let's discuss non-ideal solutions.*1063

*What we have been discussing are ideal solutions; now, we will discuss non-ideal solutions.*1066

*This is analogous to discussing the non-ideal gas law versus the ideal gas law: the ideal gas law can be...actually, for most purposes, it's just fine; and Raoult's Law, for most purposes, is actually just fine.*1075

*But this is an experimental science--it is chemistry--so when we do the experiment, it doesn't always fit with the theory, with the numbers, and we need to actually explain why that is the case.*1088

*So, let's explain why.*1099

*OK, so: so far (well, you know what, I'm not going to write this part out), what we have assumed is that the solute is non-volatile--that is some solid crystal that you drop in the water; you mix it up; you dissolve it; it's non-volatile.*1102

*In other words, the solute itself that you put in the solvent doesn't have a vapor pressure of its own.*1115

*But, if you mix a liquid and a liquid (like methanol in water, or hexane in benzene, or sodium acetate in pentane--something like that), it is a liquid-liquid.*1122

*Well, all liquids have a certain vapor pressure; so now, when you mix a liquid and a liquid, it is still just a solution; you have some liquid that you are putting into another liquid, some solute that you are putting into a solvent.*1142

*Now, both of those liquids--the molecules can escape into the vapor phase.*1155

*The vapor pressure of solution is now a sum of the mole fractions of each, times the vapor pressures of each.*1162

*Let's see--so non-ideal solutions: for liquid-liquid solutions, where both are volatile, Raoult's Law becomes: the vapor pressure of solution is equal to the mole fraction of A, times the vapor pressure of A (pure A), plus the mole fraction of B, times the vapor pressure of pure B.*1171

*That is it: we are just sort of combining the Raoult's Law for the individual liquids, but now we are mixing both liquids, so we have to account for both of their partial pressures.*1222

*The total pressure of the solution--the total number of molecules of vapor--some of the vapor molecules are going to be of A; some of the molecules are going to be of B; each one of them contributes to the total pressure (the total number of particles above the liquid at a given temperature).*1233

*OK, so let me draw this out so you see what is going on.*1248

*What I am going to draw is an ideal situation: so under ideal conditions, this is what we would expect; this is the kind of behavior we would expect.*1255

*Let me go back to blue.*1263

*Now, I'm going to go from here; I'm going to go up to here; then, I'm going to go here; I'm going to go down this way, and I'll...the mole fraction of A increases as we move in this direction; the mole fraction of B increases as we move in this direction.*1265

*Therefore, at this point, this is pure B; and at this point, we have pure A, right?*1285

*So, if we start with absolutely no A and pure B, well, that is what this represents.*1295

*As we add more A, less B, more A, less B...this here represents the mixture, if you will: a certain amount of B, a certain amount of A; a certain amount of B, a certain amount of A; a certain amount of B, a certain amount of A; that is all this is.*1301

*OK, and they are represented by mole fraction: as mole fraction goes from 0 to 1...well, for pure A, when it's 1, it is pure A; when the mole fraction of B is 1, it's pure B.*1320

*There is any other combination; so if you want to draw vertical lines, that just represents how much B we have and how much A we have in a given solution.*1331

*Well, the vapor pressure is going to look like this.*1339

*It is going to be this number, plus that number.*1349

*Again, you are just adding graphs; that is all that is happening here.*1352

*It is this number plus that number, and you are going to get this number; so this actually, down here, doesn't really matter all that much; what we want you to realize is that the vapor pressure of the solution is the sum of the mole fractions of each, times the standard vapor pressure of that liquid alone.*1356

*That is what this is saying: you are just adding two functions to get a final function.*1376

*You have been doing this for years in your precalculus and your algebra classes, so it shouldn't be a problem.*1380

*Now, this is ideal behavior right here.*1385

*In other words, all we have done is take 2 things that are volatile and added them together.*1394

*This is what should happen--what we expect to happen, theoretically.*1399

*OK, when the two liquids are very much alike, this happens--when the two liquids are very much alike.*1403

*What we mean by "very much alike": we mean they have similar properties and similar structures--"very much alike."*1415

*So, for example, if you had something like water and ethanol: ethanol is just regular alcohol--the alcohol that you drink; the structures of those two are actually very, very similar, and because they are very, very similar, they are actually going to behave in an ideal fashion.*1422

*Now, we will talk about how, actually, it deviates from ideal behavior; but any time you put two liquids together--when you mix two liquids to make a solution--if the two liquids have a lot of things in common (a lot of properties, and are very much alike structurally), what you are going to get is ideal behavior.*1441

*In other words, your vapor pressure of solution is going to equal this right here.*1459

*Now, let's talk about deviation from ideal behavior.*1464

*There are two ways to deviate from ideal behavior: there is something called negative deviation, and what that means is: when the two liquids have a special affinity for each other, such as hydrogen bonding (and more often than not, it is going to be hydrogen bonding that sort of creates the negative deviation), the P _{solution} measured is lower than expected by Raoult's Law.*1468

*So, the equation that I just gave--that will give you an expected vapor pressure, but because the two liquids have a special affinity for each other (for example, water and ethanol--there is hydrogen bonding that takes place between those two)--because of that, the vapor pressure that we measure is going to be less than what we calculate by this law.*1560

*That is negative deviation, and what this looks like is the following.*1582

*I'm going to draw the same graph that I had...well, not graph--the same diagram that I had on the previous page.*1588

*And so, let me go this way, except--notice that I have reversed the colors this time.*1595

*I have one going that way; so this is...that would be ideal behavior; negative deviation looks like this.*1600

*This is lower; this is lower than expected; therefore, this is going to be lower than expected.*1615

*Let me go over that one in black.*1625

*Don't worry so much about these; this is the one that I am concerned about.*1633

*Ideal behavior would be right there: this is what the equation that I just gave you would predict.*1636

*However, because these things actually have a greater affinity for each other (in this case, hydrogen bonding--for example, hydrogen bonding), the vapor pressure that we measure is less than what we expect or calculate from this equation.*1641

*This is called negative deviation, and again, it makes perfect sense.*1655

*If you have two liquids that you would expect to just sort of treat...one liquid mixes with another liquid, so you just add up their mole fractions times their standard vapor pressures, and you should get a number; that number should be somewhere on that line, depending on the relative amounts of each liquid that you use.*1659

*However, because of the hydrogen bonding, this ethanol and this water actually are attracted to each other more; therefore, they will not let go of each other all that much so that they can go off into the vapor phase.*1678

*You actually need more energy to make that happen; but at a given temperature, the energy is fixed; therefore, the vapor pressure will actually be lower than you expect.*1689

*That is all that it is saying; it is saying that, when you measure a vapor pressure that is lower than you expect from calculation, that means it's a negative deviation--that means the two liquids have an extra-special affinity for each other.*1700

*That means they want to be together--they like being together.*1712

*The opposite of this is positive deviation.*1716

*Oh, let me give you an example of this one; so, for example, if I mix acetone (which has a molecular structure like this--wow, you know what, I like black--I think I'll stick with that) and water--if you mix acetone and water, well, there is some serious hydrogen bonding (let me draw it another way, so you can actually see the hydrogen bonding): this is going to be H; this is going to be O; this is going to be H.*1718

*There is hydrogen bonding; because of that hydrogen bonding, the acetone and the water are actually more attracted to each other; therefore, the vapor pressure we measure of their mixture is less than what we would expect: negative deviation.*1746

*OK, now let's do positive deviation.*1759

*Positive deviation is the opposite: it is when the two liquids' interactions are weaker than the solvents alone--weaker than the solvent and solute alone.*1765

*So, for example, if I had hexane--pure hexane--the hexane molecules interact with each other.*1802

*If I had water, the water molecules interact with each other.*1808

*If I put hexane and water together, hexane and water are not very much alike; in fact, they do not like to be together (you know this: oil and water--they don't mix).*1811

*Hexane is just a smaller version of oil, if you will--it's a hydrocarbon, not polar...and we will talk about what polar means; but water and hexane are not very much alike; they don't like to be together.*1819

*Therefore, they actually, literally, want to be apart; because they want to be apart, it is more likely that they will push away from each other and actually go into the vapor phase sooner, because they don't like being together.*1831

*Whereas, up here, these molecules that have a special affinity do like being together; so a positive deviation (if we stick to this graph, I'm going to do the positive deviation in blue)--what the Raoult's Law equation says for a given concentration--I should be here; but when I measure the vapor pressure, there is actually more vapor above the liquid.*1845

*What that means is that the molecules, mixed together--they don't want to be together, so they actually will go out of their way to escape into the vapor phase to avoid being so close together in the liquid form.*1876

*The vapor pressure is actually up here: the measured is higher than what I expect.*1888

*This is called positive deviation.*1893

*I'll let you think about that a little bit, and I will finish off with an example here.*1896

*Let's do this in black; so we have Example 3.*1901

*5.8 grams of acetone is mixed with 12 grams of CHCl _{3}; OK, now, at 35 degrees Celsius, the measured vapor pressure of solution is equal to 265 torr.*1909

*OK, is this an ideal solution?*1949

*In other words, does it actually obey Raoult's Law?*1958

*Well, let's use Raoult's Law, calculate what we expect, compare it to what we measured (which is 265 torr), and see if it is lower or higher than that.*1961

*If it's lower than that, it's negative deviation; if it's higher than that, it's positive deviation; in either case, it's a non-ideal solution.*1969

*If it is close to 265 (you know, in the 260 to 270 range), it's an ideal solution.*1976

*OK, so our ideal is: solution equals the mole fraction of acetone (I'll just call it A...yes, that is fine), times the normal vapor pressure of acetone, plus the mole fraction of (I'll call this...I'll just do CHCl _{3}...here we go again: chemistry with all of its symbolism, all over the place).*1982

*OK, so let's calculate the number of moles of CHCl _{3}.*2014

*We have 12 grams, and 1 mole of that is actually equal to 119.4, equals 0.1005 mol.*2021

*And then, moles of C _{3}H_{6}O (that is the acetone): we have 5.8 grams, times 1 mole, which is 58.1 grams; OK, and we get 0.0998 mol.*2037

*Therefore, the mole fraction of acetone is equal to 0.0998, divided by 0.1005 plus 0.0998; you end up with 0.5.*2056

*And the mole fraction of the CHCl _{3} is equal to 0.1005, divided by 0.1005, plus 0.0998; and you actually end up with, again, .5.*2076

*So you see that the mole fraction of each is .5.*2093

*Now, our P solution is equal to the mole fraction of A (0.5), times its vapor pressure of the pure solvent at that temperature.*2097

*In the case of acetone at 35 degrees Celsius, it is 293 torr.*2115

*That is information that is actually given to you, or you can look it up.*2120

*...plus 0.5, which is the mole fraction of the CHCl _{3}, and its vapor pressure (35 degrees Celsius, pure solvent) is 345 torr--that is going to equal 319 torr.*2124

*So, what I would expect under normal circumstances, without accounting for affinity or disaffinity, is 319 torr.*2140

*Well, what I measured was 293 torr; clearly, 293 torr is less than 319 torr, so this is negative deviation.*2148

*Negative deviation means that the acetone and the CHCl _{3} have some sort of an extra affinity to each other.*2158

*Well, let's see what that affinity is.*2165

*As it turns out, it is going to be exactly what you think: hydrogen bonding.*2167

*This is acetone; this is the H; this extra bonding--the hydrogen bonding--this intermolecular force between the acetone and the CHCl _{3}--causes them to actually stick together more than they normally would.*2173

*Because they stick together more than they normally would, there are fewer molecules jumping out of solution into the vapor phase.*2195

*Because there are fewer molecules in the vapor phase, the vapor pressure that I measured is lower than what I expected.*2204

*That is what is going on; so that is it--standard Raoult's Law; I think the equation is pretty simple, and hopefully you understand what it is that is going on.*2211

*It is just: when you mix things, because now you don't have a pure solvent--you have other particles floating around in solution--they are actually getting in the way of the solvent particles from reaching the surface in order to break out into the vapor phase.*2221

*That is all that is going on: if you understand that physically, you will understand everything else.*2234

*Thank you for joining us here at Educator.com.*2239

*We'll see you next time; goodbye.*2241

3 answers

Last reply by: Professor Hovasapian

Fri Nov 18, 2016 8:43 PM

Post by Kaye Lim on October 24, 2016

I have 2 questions:

1/ The vapor pressure of water at 25*C (P* water solvent)= 23.76 torr. Is this fixed value the measurement for 1 mol or 18g of water at 25*C? If not, then what is the amount of water for this P*solvent value?

2/ Why the density of solid water is 1, but liquid water is less than 1? Lets say for 1 mol of water (18g), if this amount of water is ice, doesn't it occupy more volume than the same 18g of water in liquid based on the fact that H2O expands going from liquid to solid? Then why would density of ice bigger if the Volume on the denominator is bigger for ice?

Thank you!

1 answer

Last reply by: Professor Hovasapian

Thu Dec 17, 2015 1:46 AM

Post by Gaurav Kumar on December 14, 2015

Hi Professor,

I have one more question about the last example. Did you use the measured value of 265 torr in your work? Wouldn't you have to compare 265 torr with 319 torr to see whether or not it is an ideal solution or not?

Thank you!

1 answer

Last reply by: Professor Hovasapian

Thu Dec 17, 2015 1:38 AM

Post by Gaurav Kumar on December 14, 2015

Hi Professor,

i am a little confused on what you said about how we need to take in account for the total number of free particles when ionic compounds dissolve. Can you please clarify what a free particle is and what it means to take them into account.

Thank you so much!

1 answer

Last reply by: Professor Hovasapian

Thu Dec 17, 2015 1:32 AM

Post by Tammy T on December 13, 2015

Hello Prof. H!

For the Raoult's law equation, I was given a somewhat different equation with extra variables in it.

The equation given in my class is Pi = ai.P*i = yi xi P*i

i stands for component i

ai = activity of component i or also "effective concentration"

yi = Raoult's Law activity coefficient for component i

xi = mol fraction of component i

I understand the equation given in your lecture, but how come the equation given in my class is so different. I was wonder if you could please run through the equation above to explain the meaning for those extra variables and why the need for those when everything was fine using just your given Raoult's law.

Your lectures help me understand chemistry so much. Thank you!