# What is the mass fraction of C2HCl3 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}) &= \pu{73.0 mmHg}\\[0.5em] P_\text{vap}(\ce{CHCl3}) &= \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: $$w(\ce{C2HCl3})=\frac{\pu{35.47 gmol-1}}{\pu{122.61 gmol-1}}=0.29$$ $$w(\ce{CHCl3})=\frac{\pu{87.14 gmol-1}}{\pu{122.61 gmol-1}}=0.71$$