Vapour pressure of chloroform ($CHCl_3$) and dichloromethane ($CH_2Cl_2$) at 25°C are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of the solution was obtained by mixing 25.5 g of $CHCl_3$ and 40 g of $CH_2Cl_2$ at the same temperature will be: (Molecular mass of $CHCl_3$ = 119.5 u and molecular mass of $CH_2Cl_2$ = 85 u)

According to Raoult's law for a solution of volatile liquids, the total vapour pressure is the sum of the partial vapour pressures of each component ($p_{total} = p_1^0 x_1 + p_2^0 x_2$) .

Step 1: Calculate the number of moles. Moles of $CH_2Cl_2$ ($n_1$) = $\frac{40 \text{ g}}{85 \text{ g mol}^{-1}} = 0.4706 \text{ mol}$ Moles of $CHCl_3$ ($n_2$) = $\frac{25.5 \text{ g}}{119.5 \text{ g mol}^{-1}} = 0.2134 \text{ mol}$

Step 2: Calculate mole fractions. Total moles = $0.4706 + 0.2134 = 0.6840 \text{ mol}$ Mole fraction of $CH_2Cl_2$ ($x_1$) = $\frac{0.4706}{0.6840} \approx 0.688$

• Mole fraction of $CHCl_3$ ($x_2$) = $1 - 0.688 = 0.312$

Step 3: Calculate Total Vapour Pressure. Given $p_1^0 (CH_2Cl_2) = 41.5 \text{ mm Hg}$ and $p_2^0 (CHCl_3) = 200 \text{ mm Hg}$. $p_{total} = 41.5(0.688) + 200(0.312)$

• $p_{total} = 28.55 + 62.4 = 90.95 \text{ mm Hg}$

The calculated value (90.95 mm Hg) is closest to option A (90.40 mm Hg). (Note: This question contains data values different from NCERT Example 1.5, which uses 415 mm Hg for $CH_2Cl_2$).