Question

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"`?

Answer 1

I got `0.294`.

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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)`