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
Answer & explanation
Correct answer: option 2
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:
- $200 \times 0.2857 \approx 57.14$
- $415 \times 0.7143 \approx 296.43$
$P_\text{total} \approx 57.14 + 296.43 = 353.57\ \text{mmHg}$