Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene ($\ce{C2HCl3}$) and trichloromethane ($\ce{CHCl3}$) is established at $25\,°C$. 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.
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 $25\,°C$$\pu{25 °C}$ are:
$$P_{vap}(\ce{C2HCl3}) = 73.0\,mmHg$$
$$P_{vap}(\ce{CHCl3}) = 199.1\,mm Hg$$$$\begin{align}P_\text{vap}(\ce{C2HCl3}) &= \pu{73.0 mmHg}\\[0.5em] P_\text{vap}(\ce{CHCl3}) &= \pu{199.1 mmHg}\end{align}$$
So, what I did was I found mole fraction of C2HCl3$\ce{C2HCl3}$ and then used the two mole fractions along with the vapour pressures to find the total pressure of the solution.
$$P_{vap}= \frac{o.73}{199.1}+\frac{0.27}{73} = 165.053mmHg$$$$P_\text{vap}= \frac{0.73}{199.1}+\frac{0.27}{73} = \pu{165.053mmHg}$$
Then, from Raoult's Law I know that the mole fraction in liquid phase is equal to Molemole 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? Please help!
