Vapour pressure of chloroform $(CHCl_{3})$ and dichloromethane $(CH_{2}Cl_{2})$ at $298\,K$ are $200\,mm\,Hg$ and $415\,mm\,Hg$ respectively. $(i)$ Calculate the vapour pressure of the solution prepared by mixing $25.5\,g$ of $CHCl_{3}$ and $40\,g$ of $CH_{2}Cl_{2}$ at $298\,K$ and,$(ii)$ mole fractions of each component in vapour phase.

$(i)$ Molar mass of $CH_{2}Cl_{2} = 12 \times 1 + 1 \times 2 + 35.5 \times 2 = 85\,g\,mol^{-1}$.
Molar mass of $CHCl_{3} = 12 \times 1 + 1 \times 1 + 35.5 \times 3 = 119.5\,g\,mol^{-1}$.
Moles of $CH_{2}Cl_{2} = \frac{40\,g}{85\,g\,mol^{-1}} = 0.47\,mol$.
Moles of $CHCl_{3} = \frac{25.5\,g}{119.5\,g\,mol^{-1}} = 0.213\,mol$.
Total number of moles $= 0.47 + 0.213 = 0.683\,mol$.
$x_{CH_{2}Cl_{2}} = \frac{0.47\,mol}{0.683\,mol} = 0.688$.
$x_{CHCl_{3}} = 1.00 - 0.688 = 0.312$.
$P_{\text{total}} = P_{CH_{2}Cl_{2}}^{0} \times x_{CH_{2}Cl_{2}} + P_{CHCl_{3}}^{0} \times x_{CHCl_{3}} = 415 \times 0.688 + 200 \times 0.312 = 285.52 + 62.4 = 347.92\,mm\,Hg$.
$(ii)$ Mole fraction in vapour phase $(y_i = P_i / P_{\text{total}})$.
$y_{CH_{2}Cl_{2}} = \frac{285.52}{347.92} = 0.82$.
$y_{CHCl_{3}} = \frac{62.4}{347.92} = 0.18$.