 So it says, this problem says, since we're using raw load slots, it says calculate the vapor pressure lowering and the final vapor pressure, the new vapor pressure, when 10 mls of glycerol, which is C3H8O3, is added to 500 mls of water at 50 degrees Celsius. At this temperature, the vapor pressure of pure water is 92.5 Tor, and its density is 0.988 grams per mil. The density of glycerol is 1.26 grams per mil. So let's go ahead and solve this problem. So if you recall, the vapor pressure lowering is the formula is delta P equals mole fraction of the solute times the original vapor pressure. So we're going to have to figure out what the mole fraction is of these particular substances. So I calculated the molar mass of glycerol earlier. So glycerol here, so 10 mls of it, and what is it? 1.26 grams per 1 mil. So that's going to cancel. And then what is it? 92.09 grams per 1 mole of glycerol. So this is the molar mass. It's canceled. The three safe things, 0.137 water. The density at that temperature, what it would say was 0.988 grams per 1 mil. 18.02 grams. Three safe things, 27.4 mole. So now we want to figure out, well, what's the mole fraction of the solute, okay? So how do we do that? Well, we're going to have to erase some of this stuff in a second. But the mole fraction is going to be the moles of what you're looking for over the total number of moles, okay? As you might imagine. So what's on the top? 0.137 moles. Like that. Everybody okay with that? So this 0.137 moles plus 27.4 moles. And that's going to give us the mole fraction. What's the units of the mole fraction? No units. Okay, so let's add these two together. Okay, so the mole fraction I got is going to be 0.00. So now we want to figure out, well, what's the vapor pressure lower, okay? So remember, vapor pressure lowering is going to be the mole fraction of the solute times the pressure at the temperature of that particular solvent, okay? So at this case, in this case it's water. And this solute, well, was water. So in this case, the vapor pressure is 92.54, okay? So we've got the mole fraction there. So this is going to be 0.00498 times 92.5 tor, like that. So let's just multiply this times 92.5. So that gives us 0.460 tor, okay? So that's the first answer, right? That's how much it got lowered. So what would be the new pressure, right? So the pressure new is going to be the original pressure minus the change in pressure. Does that make sense? Okay, so what was the original pressure? 92.5 tor minus 0.460 tor. So, well, when we do that, so 92.5 minus 0.460. And I get 92.0 tor. So when we added glycerol to water, did it lower the vapor pressure? Yeah, just like it was supposed to, okay? So this is Raoult's law. This is the first of the colligative properties that you get to attend. There's a couple of more that I want to go over with you. I'm going to have to wait until tomorrow to do that. So good job, guys. I'll see you tomorrow. Any questions about this one before we go? I know everybody wants to leave, so okay.