Question

The vapor pressure of pure benzene is 750.0 torr and the vapor pressure of toluene is 300.0 torr at a certain temperature. You make a solution by pouring “some” benzene with “some” toluene. You then place this solution in a closed container and wait for the vapor to come into equilibrium with the solution. Next, you condense the vapor. You put this liquid (the condensed vapor) in a closed container and wait for the vapor to come into equilibrium with the solution. You then condense this vapor and find the mole fraction of benzene in this vapor to be 0.714. Determine the mole fraction of benzene in the original solution, assuming the solution behaves ideally.

Short answer

The mole fraction of Benzene in the original solution is 0.286.

Step-by-step solution

Step 1:

The condensed vapor from container 1 is transferred to container 2.

The mole fraction of vapor from container 1 is equal to the mole fraction in vapor from container 2.

$$\begin{gathered} {\chi }_{B1}={\chi }_{B2} \\ {\chi }_{T1}={\chi }_{T2} \end{gathered}$$$$\begin{gathered} {\chi }_{B2}=0.714 \\ {\chi }_{T2}=0.286 \end{gathered}$$

Step 2:

Calculate more fractions in the solutions for container 2.

$$\begin{gathered} {\chi }_{B2}=\frac{300.0torr\times 0.714}{750.0torr\times 0.286+300.0\times 0.714} \\ {\chi }_{B2}=0.500 \\ {\chi }_{T2}=1-{\chi }_{B2} \\ {\chi }_{T2}=0.500 \end{gathered}$$

Step 3:

Calculate more fractions in the solutions for container 1.

$$\begin{gathered} {\chi }_{B1}=\frac{300.0torr\times 0.500}{750.0torr\times 0.500+300.0\times 0.500} \\ {\chi }_{B1}=0.286 \\ {\chi }_{T1}=1-{\chi }_{B1} \\ {\chi }_{T1}=0.71 \end{gathered}$$