# 13.3 Vapor Pressure Depression and Raoult's Law

### Chad's General Chemistry Videos

- 19.1 Oxidation Reduction Reactions and Oxidation States
- 19.2 Balancing Oxidation Reduction Reactions
- 19.3 Galvanic Cells
- 19.4 Standard Cell Potentials aka emf or Voltage
- 19.5 Nonstandard Cell Potentials the Nernst Equation
- 19.6 Reduction Potentials and the Relationship between Cell Potential, Delta G, and the Equilibrium Constant
- 19.7 Electrolytic Cells
- 19.8 Electrolysis Calculations

### Chad's Ultimate General Chemistry Prep

Mastery Through Practice- 90 Quizzes Arranged by Topic
- 17 Chapter Tests
- 3 Practice Final Exams
- Over 1200 Questions
- Detailed Solutions
- Gen Chem Study Guides

Free Trial includes:

All videos, study guides, and quizzes for chapters 1 and 2

Try before you buy!

##### Video Transcript

Okay. So here we've gone through the freezing point depression and the boiling point elevation. Now we've got to talk about vapor pressure depression. And before I give you the equation that defines this, it's called Raoult's Law, we're just going to talk about it. Because the more you think about something conceptually and the less it seems like an equation, the easier it is to remember. So let’s say I've got a glass here, or let's say beaker of pure water. Pure water, on the left. And then over here we've got a beaker that's got nine moles of water and one mole of methanol mixed together. And they're totally soluble, they mix together. Like an ideal solute here. Okay. Both of these will have a vapor pressure of water above the solution. There'll be some water vapor up here and there will be water vapor up here. Above the solution. Like again we talked about earlier you know, above a lake there's certain amount of water vapor in the air. Seems more humid when you're around a lake than when you're away from a lake. So to speak. So any of that water vapor above either one of these beakers as well; my question for you is above which of these beakers, and technically these beakers I really should probably make these sealed containers to really make this make complete sense, but above which of these solutions will there be a greater water pressure, if you will of water vapor? Well, whichever solution has a greater concentration of water, will have more water vapor above it. So in which of these cases is there a greater concentration of water? The pure water. And so the pressure here is definitely to be higher. And in fact for pure water, at a given temperature, I can just look it up in a book what that pressures going to be. Depending on whatever temperature it's at. And it turns out at higher temperatures it usually is higher, and stuff like this. That's why I like say, Thomas I said "If I put on a plate in front of you a frozen turd or a warm turd, which would you rather I put in front of you on a plate?" If you had to have one no matter what. Take the frozen one why. Yeah. It's not going to smell as bad because things usually get more vapor pressure at higher temperatures. And notice if there's higher vapor pressure, then there's more likely vapors will reach your nose with the warm turd. Notice it has to reach your nose if you're going to smell it. Kind of gross, think you actually get turd vapor molecules in your nose to smell it. Think about that next time you smell it. Alright so here's the deal. So at a certain temperature we can look up what pure water is. In this case, let's say we look up pure water and let's just say it happened to come out to exactly 20 Torr. 20 Torr. And so here what can you tell me about the vapor pressure at this point? It's going to be lower than 20. Awesome. And you can do a fair amount better than this. What percentage of the moles in the solution here are water? 90%. Then the vapor pressure here is only going to be 90% of what pure water would have had. What's 90 percent of 20? 18. Subtract off 10%. 10% of 20 is 2. So in this case it's 18 Torr. You guys just did that without Raoult's Law in front of you, a plug and chug equation. I wanted you to see it conceptually before I actually gave you the equation. The way this works, the partial pressure of a particular liquid will equal remember that guy? Yeah, mole fraction, ok. So a mole fraction of water times the partial pressure of water. And then we draw an asterisk, and that asterisk means pure. And so here the partial pressure of an impure water solution is equal to the mole fraction of water in that solution, times whatever the partial pressure of water would be above a pure sample of water. Cool. My question for you is, there's methanol in here as well. And methanol is actually volatile as well. It'll have vapor up in here as well. What would be the partial pressure of methanol up here is my question for you. 10 percent of what? It's not 10 percent of 20. The correct answer is, Chad I don't know. That was what I was looking for. It is 10 percent of something. Because there's only a, you know, 1/10 of the solution is water- I'm sorry 1/10 of the solution is methanol. And so it's only 1/10 of whatever pure methanol would be. Did I give you what pure methanol would be? No. So then you can't really tell me what 1/10 of whatever that number is, because you don't what that number is. And so the answer I really just wanted you to see is that we don't know. I didn't give you enough information to actually figure it out in this case. It would be up there I just don't know what it is. Okay. So back to this calculation, what's the mole fraction of water in this solution? 9 out of 10. So 9/10 times 20, gets us the same 18 we just kind of did before on a more conceptual level. Cool.

## Raoult's Law

Raoult's Law for vapor pressure depression describes how the vapor pressure of an ideal solvent decreases as the amount of dissolved solute increases. Generally, Raoult's Law is most accurate for dilute solutions that have a large mole fraction of solvent.

P_{A} = X_{A}P_{A}* X_{A} = mole fraction A P_{A}* = vapor pressure of pure A

Ultimately, this means that if a solvent has a mole fraction of 0.95, then 95% of the moles in that solution are the solvent and it will have a vapor pressure that is 95% of what the pure solvent would have. Similarly, if a solvent has a mole fraction of 0.90, then 90% of the moles in that solution are the solvent and it will have a vapor pressure that is 90% of what the pure solvent would have. The next examples show the calculations for just such aqueous solutions.

In this example we can see that as the amount of solute increases (left to right) the vapor pressure of water decreases.