What is the mass fraction of trichloroethene in the liquid phase using vapour pressure and mole fraction in gas phase?
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:
$$w(\ce{C2HCl3})=\frac{\pu{35.47 gmol-1}}{\pu{122.61 gmol-1}}=0.29$$
$$w(\ce{CHCl3})=\frac{\pu{87.14 gmol-1}}{\pu{122.61 gmol-1}}=0.71$$