Vapour pressures of chloroform ($CHCl_3$) and dichloromethane ($CH_2Cl_2$) at 298 K are 200 mmHg and 415 mmHg, respectively. Calculate the vapor pressure of the solution prepared by mixing 0.2 mol of $CHCl_3$ and 0.5 mol of $CH_2Cl_2$ at 298 K

353.57 mmHg

The correct answer is Option (2) → 353.57 mmHg

Step 1: Recall Raoult’s law for a binary solution

For an ideal solution:

$P_\text{total} = P_A^0 \, x_A + P_B^0 \, x_B$

Where:

• $P_A^0, P_B^0$​ = vapor pressures of pure components
• $x_A, x_B$​ = mole fractions of components

Step 2: Calculate mole fractions

Number of moles:

• CHCl₃ = 0.2 mol
• CH₂Cl₂ = 0.5 mol

Total moles:

$n_\text{total} = 0.2 + 0.5 = 0.7 \ \text{mol}$

Mole fractions:

$x_\text{CHCl_3} = \frac{0.2}{0.7} \approx 0.2857$

$x_\text{CH_2Cl_2} = \frac{0.5}{0.7} \approx 0.7143$

Step 3: Apply Raoult’s law

$P_\text{total} = P_{CHCl_3}^0 \, x_{CHCl_3} + P_{CH_2Cl_2}^0 \, x_{CH_2Cl_2}$

$P_\text{total} = (200)(0.2857) + (415)(0.7143)$

Step-by-step calculation:

1. $200 \times 0.2857 \approx 57.14$
2. $415 \times 0.7143 \approx 296.43$

$P_\text{total} \approx 57.14 + 296.43 = 353.57\ \text{mmHg}$