Vapour pressure of chloroform $(CHCl_3$) and dichloromethane ($CH_2Cl_2$) at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5g of $CHCl_3$ and 40 g of $CH_2Cl_2$ at 298 K .

Calculation of vapour pressure
Molar mass of $CH_2Cl_2 = 12 \times 1 + 1 \times 2 + 35.5 \times 2 = 85 g mol^{–1}$
Molar mass of $CHCl_3 = 12 \times 1 + 1 \times 1 + 35.5 \times 3 = 119.5 g mol^{–1}$
Moles of $CH_2Cl_2 = \large\frac{40}{85}$$=0.47mol Moles of CHCl_3= \large\frac{25.5}{119.5}$$=0.213mol$
Total number of moles = 0.47 + 0.213 = 0.683 mol
$X_{CH_2Cl_2}=\large\frac{0.47}{0.683}$$=0.688$
$X_{CHCl_3}=1.00 – 0.688 = 0.312$
$P_{Total}=P_1^0+(P_2^0-P_1^0)X_2=200+(415-200)\times 0.688$
$\Rightarrow$200 + 147.9 = 347.9 mm Hg
Hence (A) is the correct answer.