Vapor pressures of solutions

Non-volatile solutes; that is, solutes that do not evaporate.

We have discussed vapor pressure, and we know that the vapor pressure of a substance arises from its ability to evaporate or sublime. (Remember, all this means is that some of the molecules have enough energy to overcome the IMF that hold the molecule together.) What happens to the vapor pressure of a substance if there is something disolved in that substance.

 

Flask A has a certain vapor pressure because at any given time there are some molecules at the surface which have enough energy to become vapor. Eventually the rate at which the vapor condenses equals the rate at which the molecules are evaporating. In flask B because there are some salt paricles there are fewer water molecules at the surface. Since there are fewer water molecules at the surface the rate of evaporation decreases. Since the rate of condensation stays the same the amount of vapor present decreases. So, the solution in flask B has a lower vapor pressure than the water in flask A.

Flask A contains water Flask B contains salt water.

So, if only half of the molecules at the surface are water molecules then we would expect that the number of vapor phase molecules would be cut in half also; in fact, this is what happens.

mathematically...

That is, the vapor pressure of the solution, P_soln, equals the mole fraction of the solvent, , times the normal vapor pressure of the pure solvent, P°_solvent.

This is Raoult's Law (like most chemical laws it works most of the time, but there are exceptions...keep going we are getting there).

This analysis also implies that the nature of the solute is unimportant, just how much solute is present is important; this is also true...most of the time.

But what if the solute is volatile?