Question

You have a solution of two volatile liquids, A and B (assume ideal behavior). Pure liquid A has a vapor pressure of 350.0 torr and pure liquid B has a vapor pressure of 100.0 torr at the temperature of the solution. The vapor at equilibrium above the solution has double the mole fraction of substance A as the solution does. What is the mole fraction of liquid A in the solution?

Short answer

The vapor pressure of the pure liquid is${\chi }_A=0.3$ .

Step-by-step solution

Step 1:

Given ${\chi }_A=2{\chi }_A$

$$\begin{gathered} {\chi }_A=\frac{P_A}{P_{TOT}} \\ 2{\chi }_A=\frac{P_A}{P_{TOT}} \\ {P}_{TOT}=\frac{P_A}{2{\chi }_A}=\frac{{\chi }_A{P°}_A}{2{\chi }_A} \\ {P}_{TOT}=\frac{{P°}_A}{2} \end{gathered}$$$2{\chi }_A$

Step 2:

${\chi }_A=\frac{{P°}_A-2{P°}_B}{2\left({P°}_A-{P°}_B\right)}$

The vapor pressure of the pure liquid.

$$\begin{gathered} {\chi }_A=\frac{350torr-2\times 100torr}{2\left(350torr-100torr\right)} \\ {\chi }_A=0.3 \end{gathered}$$