A quantity of benzene, $\ce{C6H6}$, and toluene, $\ce{C6H5CH3}$, is placed in a $\pu{1 L}$ evacuated vessel at $\pu{25 ^\circ C}$. At equilibrium, a small volume of liquid is visible at the bottom of the container. A sample of the vapour phase is analysed and found to contain $\pu{53 mol-\%}$ benzene. What is the mole fraction of benzene in the liquid phase?
Vapour pressures at $\pu{25 ^\circ C}$: $P_\mathrm{vap}(\text{benzene})= \pu{0.125 atm}$, $P_\mathrm{vap}(\text{toluene}) = \pu{0.037 atm}$
First of all, I am not even certain if this a Raoult's law problem. I just assumed so, since Henry's constant was not given in the problem.
First I'm getting the total liquid pressure using Raoult's law. \begin{align} P(\text{benzene}) &= X(\text{benzene}) \cdot P_\mathrm{vap}(\text{benzene}) & &= 0.53 \times 0.125 &&= 0.06625~\mathrm{atm}\\ P(\text{toluene}) &= X(\text{toluene}) \cdot P_\mathrm{vap}(\text{toluene}) & &= 0.47 \times 0.037 &&= 0.01739~\mathrm{atm}\\ P_\mathrm{total} &&&= 0.06625 + 0.01739 &&= 0.08364~\mathrm{atm}\\ \end{align}
Then using Dalton's law, I am getting the mole fraction of Benzene in the liquid phase.
\begin{align}
P(\text{benzene}) &= Y(\text{benzene}) \cdot P_\mathrm{total}\\
Y(\text{benzene}) &= \frac{P(\text{benzene})}{P_\mathrm{total}}\\
&= \frac{0.06625}{0.08364} = 0.792.
\end{align}
However, my answer is incorrect according to the answer key that has $0.25$ as the answer. I have a feeling my whole approach may be wrong since my answer is off by a lot. What am I doing wrong?