What is the mass fraction of trichloroethene in the liquid phase using vapour pressure and mole fraction in gas phase? Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene ($\ce{C2HCl3}$) and trichloromethane ($\ce{CHCl3}$) is established at $\pu{25 °C}$. The mole fraction of $\ce{CHCl3}$ in the vapour phase is $0.73$. What is the mass fraction of $\ce{C2HCl3}$ in the liquid phase?  Round your answer to two significant figures.


The vapour pressures of trichloroethene and trichloromethane at $\pu{25 °C}$ are:

$$\begin{align}P_\text{vap}(\ce{C2HCl3}) &amp;= \pu{73.0 mmHg}\\[0.5em]
P_\text{vap}(\ce{CHCl3}) &amp;= \pu{199.1 mmHg}\end{align}$$

So, what I did was I found mole fraction of $\ce{C2HCl3}$ and then used the two mole fractions along with the vapour pressures to find the total pressure of the solution. 

$$P_\text{vap}= \frac{0.73}{199.1}+\frac{0.27}{73} = \pu{165.053mmHg}$$

Then, from Raoult's Law I know that the mole fraction in liquid phase is equal to mole fraction in vapour phase, multiplied by vapour pressure, divided by total pressure. From that, I found the mole fraction of both things in liquid phase. I use the mole fraction to find mass of both, and then did mass of $\ce{C2HCl3}$ divided by the total mass that I calculated. I got an answer of 

$\frac{15.68}{120}=0.13$ but it says that it's wrong. I'm not sure where I messed up?
As far as the question is concerned, you do not need the vapor pressure. In order to calculate the mass fraction of $\ce{C2HCl3}$, we first have to calculate the molar mass of $\ce{C2HCl3}$ and $\ce{CHCl3}$.
$$M(\ce{C2HCl3}) = \pu{131.38 gmol-1}$$
$$M(\ce{CHCl3}) = \pu{119.37 gmol-1}$$
With these two calculated we can calculate the absolute mass relative to $\pu{1 mol}$, due to the fact of $\frac{\pu{0.73 mol}}{\pu{1 mol}}$ being relative to a total of $\pu{1 mol}$ of molecules. Now if we calculate the absolute masses, we get:
$$M(\ce{C2HCl3}) = \pu{131.38 gmol-1} * \pu{0.27\frac{mol}{mol}} = \pu{35.47 gmol-1}$$
$$M(\ce{CHCl3}) = \pu{119.37 gmol-1} * \pu{0.73\frac{mol}{mol}} = \pu{87.14 gmol-1}$$
And the total mass in the system results in:
$$m_\text{total}=\pu{35.47 gmol-1}+\pu{87.14 gmol-1}=\pu{122.61 gmol-1}$$
And with the total relative mass given, we can calculate the mass fraction of $\ce{C2HCl3}$, as follows:
$$w(\ce{C2HCl3})=\frac{\pu{35.47 gmol-1}}{\pu{122.61 gmol-1}}=0.29$$
$$w(\ce{CHCl3})=\frac{\pu{87.14 gmol-1}}{\pu{122.61 gmol-1}}=0.71$$