The vapour-pressure lowering upon forming solution can be used for determining molecular weight of nonvolatile solutes. Consider a solvent with known P1* and molar mass M1. Upon adding m2 kg of a solute to m1 kg of the solvent, the solution vapour pressure has been measured as P.

(A) Obtain M2 in terms of the specified values (the answer should contain only P1*, P, M1, m1,2), assuming an ideal solution.


I found M2 ={(P1* /(P1*- P))-1}*M1*m2 /m1

How would I approach Part B below? Any help would be appreciated


Correct the result for the case of a volatile solute with known P2*.

I traced your answer back to its starting point, which must have been P = x1·P1*, which is correct for dilute or ideal solutions. Working forward from this by substituting x1 = n1/(n1+n2), then n1=m1/M1 and n2=m2/M2 gives your answer to the first question.

It seems to me that you can proceed in the same way for the second question, but you should start with P = x1·P1* + x2·P2*. Since the problem states that the solution is ideal, this is correct even when the solution is not dilute. Since at least one of x1 or x2 is greater than 0.5, the solution is obviously not dilute from the point of view of that chemical species, so you do need the assumption that the solution is ideal for this to work.

Substituting x1 = n1/(n1+n2), x1 = n2/(n1+n2) as in the first problem gives (n1+n2)P = n1·P1* + n2·P2*, from which you can solve to get n2 = n1(P1*-P)/(P-P2*) and then substituting n1=m1/M1 and n2=m2/M2 as before will let you solve for M2. You get M2 = M1·(m2/m1)·(P-P2*)/(P1*-P). This should be the answer you want.

Two things to notice: First, if you set P2* to 0, that makes material #2 nonvolatile, so this formula should turn into the formula for the first question in that case. Check it out and see. (It does, I checked). Second, notice how the pressures are arranged. In the new situation, the total pressure will not necessarily be lower than P1* as it was in the previous case, because P2* might actually be more than P1*. The actual pressure will be somewhere between P1* and P2*; in fact it will be the weighted average of those two pressures. So if P1* > P2*, then you get P-P2* > 0 and P1*-P > 0, so the formula gives a POSITIVE value for M1 (which it had better do, since negative molar masses make no sense at all!); on the other hand, if P1* < P2*, then you get P-P2* < 0 and P1*-P < 0, so again M1 comes out positive.

Hope that helps!

--Stuart Anderson

Thank you very much Mr. Anderson