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A solution is made by mixing 50.0 g C3H6O and 50.0 g of CH3OH. What is the vapor pressure of this solution at 25 degrees C? What is the composition of the vapor expressed as a mole fraction? Assume ideal solution and gas behavior. (At 25 degrees C the vapor pressures of pure acetone and methanol are 271 torr and 143 torr, respectively.)

I calculated the vapor pressure [work shown at end] but I'm confused about the second part of the question. What does it mean to find the composition of the vapor expressed as a mole fraction? It can't be just the mole fraction of 50.0 g C3H6O to 50.0 g CH3OH can it?

P1= X1 * P*1
50.0 g C3H6O x 1 m/58.09g= .8607 m
50.0 g CH3OH x 1 m/32.05g= 1.560 m
.8607 m C3H6O/2.420 m total= .3556
P1= .3556*143 torr= 50.85 torr

I believe you have used Raoult's Law for a non-volatile solute dissolved in a volatile solvent. In this problem, however, BOTH liquids are volatile and what you have used is not appropriate (although some of the calculation is).
Calculate the mole fraction of acetone (which you have done) and the mole fraction of methanol (which you have done).
Then Pacetone=XacetonePoacetone


The total pressure, then, is
Ptotal=Pacetone + Pmethanol

Let me look to see that all of this copied correctly and I will post the second part of the problem separately.
Check my thinking.

To do the second part, one must realize that the VAPOR will be richer in the more volatile component. The problem is asking you to calculate the mole fractions of the components in the vapor. mole fraction of a component = ratio of partial pressure of component/total pressure. You have just calcualted the total P and you have the partial pressures of each component.



Check my thinking.

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