Question

A solution is prepared by mixing $8.5$ g of $C{H}_{2}C{l}_2$ and $11.95\text{g}$ of $CHC{l}_3$ . If vapour pressure of $C{H}_{2}C{l}_2$ and $CHC{l}_3$ at $298$ K are $415$ and $200$ mm Hg respectively, the mole fraction of $CHC{l}_3$ 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)

${\text{mole of CH}}_2\text{C}{\text{l}}_2\text{in liquid phase}=\frac{8.5}{85}=0.1$
${\text{mole of CHCl}}_{\text{3}}\text{in liquid phase}=\frac{11.95}{119.5}=0.1$
${\text{mole fraction of CH}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{in liquid phase}=\frac{0.1}{0.2}=\frac{1}{2}$
${\text{mole fraction of CHCl}}_{\text{3}}\text{in liquid phase}=\frac{0.1}{0.2}=\frac{1}{2}$
${\text{P}}_{\text{T}}={\text{X}}_{\text{C}{\text{H}}_2\text{C}{\text{l}}_2}\times {\text{vapour pressure}}_{\text{C}{\text{H}}_2\text{C}{\text{l}}_2}+{\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_3}\times {\text{vapour pressure}}_{\text{C}\text{H}\text{C}{\text{l}}_3}$
$=415\times \frac{1}{2}+200\times \frac{1}{2}=307.5$
${\left(\text{P}\right)}_{\text{T}}({\left(\text{X}\right)}_{\text{A}}{)}_{\text{V}\text{P}}=({\left(\text{X}\right)}_{\text{A}}{)}_{\text{L}\text{P}}\times {\left(\text{ Vapour pressure of CHCl}\right)}_3$
$307.5\times ({\left(\text{X}\right)}_{\text{C}\text{H}\text{C}{\left(\text{l}\right)}_3}{)}_{\text{V}\text{P}}=200\times \frac{1}{2}$
${\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_3}=\frac{100}{307.5}=0.325$