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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?

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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 $\ce{CHCl3}$. $$M(\ce{C2HCl3}) = \pu{131.38 \frac{g}{mol}}$$ $$M(\ce{CHCl3}) = \pu{119.37 \frac{g}{mol}}$$ 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 $\pu{1mol}$ of molecules. Now if we calculate the absolute masses, we get: $$M(\ce{C2HCl3}) = \pu{131.38 \frac{g}{mol}} * \pu{0.27\frac{mol}{mol}} = \pu{35.47 \frac{g}{mol}}$$ $$M(\ce{CHCl3}) = \pu{119.37 \frac{g}{mol}} * \pu{0.73\frac{mol}{mol}} = \pu{87.14 \frac{g}{mol}}$$ And the total mass in the system results in: $$m_\text{total}=\pu{35.47 \frac{g}{mol}}+\pu{87.14 \frac{g}{mol}}=\pu{122.61 \frac{g}{mol}}$$ And with the total relative mass given, we can calculate the mass fraction of $\ce{C2HCl3}$, as follows: $$w(\ce{C2HCl3})=\frac{35.47\frac{g}{mol}}{122.61\frac{g}{mol}}=\pu{0.29\frac{g}{g}}$$ $$w(\ce{CHCl3})=\frac{87.14\frac{g}{mol}}{122.61\frac{g}{mol}}=\pu{0.71\frac{g}{g}}$$

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