How to determine the mole fraction of the liquid phase from a given mole fraction of the vapour phase for a benzene/toluene mixture? 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}) &amp;= X(\text{benzene}) \cdot P_\mathrm{vap}(\text{benzene}) &amp;
                  &amp;= 0.53 \times 0.125 &amp;&amp;= 0.06625~\mathrm{atm}\\
P(\text{toluene}) &amp;= X(\text{toluene}) \cdot P_\mathrm{vap}(\text{toluene}) &amp;
                  &amp;= 0.47 \times 0.037 &amp;&amp;= 0.01739~\mathrm{atm}\\
P_\mathrm{total}  &amp;&amp;&amp;= 0.06625 + 0.01739 &amp;&amp;= 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}) &amp;= Y(\text{benzene}) \cdot P_\mathrm{total}\\
Y(\text{benzene}) &amp;= \frac{P(\text{benzene})}{P_\mathrm{total}}\\
                  &amp;= \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?
The question gives you the mol percentage of benzene ($53 \%$) in the vapour phase, and not in the liquid phase.

It's always good to write down what you have first. (note: I use $y$ to denote mole fractions in the vapour phase and $x$ to denote mole fractions in the liquid state)

First of all, we have $y_\text{benzene} = 0.53$. Additionally, we have been given the vapour pressure of pure benzene and toluene (denoted with a superscript $*$):

$$p^*_\text{benzene} = 0.125\ \mathrm{atm}  \qquad p^*_\text{toluene} = 0.037\ \mathrm{atm} $$

and we require $x_\text{benzene}$

We also, know $p_\text{benzene} = x_\text{benzene}\ p^*_\text{benzene}$, and the total vapor pressure is given by $$P = p_\text{benzene} + p_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene} + x_\text{toluene}\ p^*_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}$$

Finally, 

$$y_\text{benzene} = \frac{p_\text{benzene}}{P} = \frac{x_\text{benzene}\ p^*_\text{benzene}}{x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}}$$

It is now, trivial to solve this equation. Since I am lazy, and was not in the mood to write down two lines of algebra and reach for a calculator, I chucked it into Mathematica, and the result is: 



Thus, $x_\text{benzene} = 0.25$, which is indeed the result given in your answer key.