# 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"?

Jan 10, 2017

I got $0.294$.

Raoult's law states:

${P}_{j} = {\chi}_{j}^{l} {P}_{j}^{\text{*}}$,

where:

• ${P}_{j}$ is the partial vapor pressure above the solution coming from the component $j$ in solution.
• ${P}_{j}^{\text{*}}$ 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 $\text{1 atm}$ $=$ $\text{760 torr}$ 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:

$\Delta P = {P}_{i} - {P}_{i}^{\text{*}}$,

where ${P}_{i}^{\text{*}}$ 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:

$\Delta P = {\chi}_{i}^{l} {P}_{i}^{\text{*" - P_i^"*}}$

$= {P}_{i}^{\text{*}} \left({\chi}_{i}^{l} - 1\right)$

$= - {\chi}_{j}^{l} {P}_{i}^{\text{*}}$

The change in pressure was:

$\text{760 torr}$ $-$ $\text{1077 torr" = -"317 torr}$

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

$\textcolor{b l u e}{{\chi}_{j}^{l}} = - \frac{\Delta P}{{P}_{i}^{\text{*}}}$

$= - \left(- \text{317 torr")/("1077 torr}\right)$

$= \textcolor{b l u e}{0.294}$