What is the mol fraction of ethylene glycol in the solution phase for an aqueous solution with a vapor pressure of "760 torr" if the pure vapor pressure was "1077 torr"?

1 Answer
Jan 10, 2017

I got 0.294.


Raoult's law states:

P_j = chi_j^lP_j^"*",

where:

  • P_j is the partial vapor pressure above the solution coming from the component j in solution.
  • P_j^"*" is the vapor pressure above a sample of pure liquid j at the same T as the solution.
  • chi_j^l is the mol fraction of j in the liquid phase.

Since the total vapor pressure of the solution was given, which is lower than the pure vapor pressure of water, there was a decrease in vapor pressure.

I assume however, that the "1 atm" = "760 torr" is for the WATER in the solution, not for the ENTIRE solution (though the question was written in such a way that it seemed to be for the overall solution...).

To track that change:

DeltaP = P_i - P_i^"*",

where P_i^"*" was the vapor pressure of water before adding ethylene glycol, and P_i is the vapor pressure of water after adding ethylene glycol.

Then, plugging in Raoult's law:

DeltaP = chi_i^lP_i^"*" - P_i^"*"

= P_i^"*"(chi_i^l - 1)

= -chi_j^lP_i^"*"

The change in pressure was:

"760 torr" - "1077 torr" = -"317 torr"

So, the mol fraction of ethylene glycol in the solution (not in the vapor phase) would be:

color(blue)(chi_j^l) = -(DeltaP)/(P_i^"*")

= -(-"317 torr")/("1077 torr")

= color(blue)(0.294)