Rob Lederer: Raoult's law to calculate the vapor pressure of a solution not a solid but a solution it's quite elegant you put together. Take a look. Something you are aware of, that if you have a solvent at a given temperature, well it come to certain vapor pressure above that solvent if in a closed container in an equilibrium condition. Okay, what if we dump in solute and let's make it 50% in terms of its number of particles. 50% of the solution is now made up of solute and 50% solvent. Okay, then how much blockage is going to occur from the solute? Well probably block the evaporation of solvent to above 50% of what its vapor pressure was before. sure it will, and so Raoult's law quite simply for calculating the vapor pressure of the solution is, pressure of the solution equals the original pressure of the solvent times the mole fraction because, we are dealing with particles so we will go to moles, the mole fraction of the solvent over solute plus solvent. Because look, see if this is a 50% so this mole fraction would be 0.5, 0.5 times the original vapor pressure, here would give us half the vapor pressure from the solvent going to the solution. Okay now, if this was actually 25% solute then the mole fraction here would be 75% or 0.75 is what this would equal solvent over solute plus solvent. 0.75 times the original vapor pressure gives us three quarters of what we had before because, one quarter solute here is going to do an amount of damage or blocking the solvent to about one quarter of its original vapor pressure. Isn't that cool? By the way, this is all if the solutions behave properly, there are no attractive forces or repelling forces between solute, solvent or the solute is not too volatile or evaporate. They have to be ideal solutions. Let's look at the couple of formulae manipulations here. Okay, we all agree that PV equals nRT. No discussion that's absolutely right. oh okay, divide each side by V and you get P equals nRT over V. sure okay, now you agree with me that moles is actually grams divided by grams per mole, which is molar mass. So, we can say that why don't we just substitute that then, grams over grams per mole in for moles. Watch what I am getting that? Then of course, because grams is divided by grams per mole, we can just put that in the denominator over here and get that as a formulae. Recognize this, grams per unit of volume, that's called the density. Now, in this case the density is in grams per litre because, the volume is going to be in liter. so anytime you do these types of problems and get the formula that were going to be in the -- here, the density must be in grams per liter. Okay, so this is just density. So, we put that in there. Pressure equals dRT over M; I am calling that molar mass. Big M can be concentration but it can be molar mass. and look, when we manipulate by multiplying both sides by n and dividing both sides by P, we get a formula, molar mass equals dRT over P. that's right, we can calculate the molar mass of a gas if we know its density with that formula, molar mass equals dRT over P. just an another nice little relationship that you should know about. Do you agree that kai which is mole fraction equals number of moles in the solute over the number of moles totaled, which is usually the solute plus the solvent? Sure, that's an another way writing it, right? n over n totaled. Okay, now ideal gas law, PV=nRT. so n equals PV over RT. So we can substitute in here PV over RT divided by PV over RT or let's take the pressure of the solute there, PV over RT. I will just call that P1, that's the solute's pressure. that will be divided whatever the total amount of pressure is, number of moles totaled times VRT because, this is PV over RT totaled, but here is the thing, if you keep the volume and temperature constant, volume and temperature constant throughout, then these two constants cancel and what are you are left with? Kai equals P or that P1 over the P totaled. So, listen h