Question

A solution is prepared by mixing $8.5$ g of $CH_{2}Cl_{2}$ and $11.95\text{g}$ of $CHCl_{3}$ . If vapour pressure of $CH_{2}Cl_{2}$ and $CHCl_{3}$ at $298$ K are $415$ and $200$ mm Hg respectively, the mole fraction of $CHCl_{3}$ in vapour form is: $\left(\text{M} \text{o} \text{l} \text{a} \text{r} \, \text{m} \text{a} \text{s} \text{s} \, \text{o} \text{f} \, Cl = 35.5 \, \text{g} \, \text{m} \text{o} \left(\text{l}\right)^{- 1}\right)$

• A. $0.162$
• B. $0.675$
• C. $0.325$
• D. $0.486$

Answer 1

Answer: (C)

$\text{mole of CH}_{2}\text{C}\text{l}_{2} \, \text{in liquid phase}=\frac{8.5}{85}=0.1$
$\text{mole of CHCl}_{\text{3}} \, \text{in liquid phase}=\frac{11.95}{119.5}=0.1$
$\text{mole fraction of CH}_{\text{2}}\text{Cl}_{\text{2}} \, \text{in liquid phase}=\frac{0.1}{0.2}=\frac{1}{2}$
$\text{mole fraction of CHCl}_{\text{3}} \, \text{in liquid phase}=\frac{0.1}{0.2}=\frac{1}{2}$
$\text{P}_{\text{T}}=\text{X}_{\text{C} \text{H}_{2} \text{C} \text{l}_{2}}\times \, \, \text{vapour pressure}_{\text{C} \text{H}_{2} \text{C} \text{l}_{2}}+ \, \text{X}_{\text{C} \text{H} \text{C} \text{l}_{3}}\times \, \text{vapour pressure}_{\text{C} \text{H} \text{C} \text{l}_{3}}$
$= 415 \times \frac{1}{2} + 200 \times \frac{1}{2} = 307.5$
$\left(\text{P}\right)_{\text{T}} \left(\right. \left(\text{X}\right)_{\text{A}} \left.\right)_{\text{V} \text{P}} = \left(\right. \left(\text{X}\right)_{\text{A}} \left.\right)_{\text{L} \text{P}} \times \, \left(\text{ Vapour pressure of CHCl}\right)_{3}$
$307.5 \times \, \left(\right. \left(\text{X}\right)_{\text{C} \text{H} \text{C} \left(\text{l}\right)_{3}} \left.\right)_{\text{V} \text{P}} = 200 \times \frac{1}{2}$
$\text{X}_{\text{C} \text{H} \text{C} \text{l}_{3}}=\frac{100}{307.5}=0.325$