Question

Vapour pressure of chloroform $(CHCl _{3})$ and dichloromethane $\left( CH _{2} Cl _{2}\right)$ at $25^{\circ} C$ are $200 mmHg$ and $41.5 mmHg$ respectively. Vapour pressure of the solution obtained by mixing $25.5 g$ of $CHCl _{3}$ and $40 g$ of $CH _{2} Cl _{2}$ at the same temperature will be (Molecular mass of $CHCl _{3}=119.5 u$ and molecular mass of $\left. CH _{2} Cl _{2}=85 u \right):$

• A. 173.9 mmHg
• B. 615.0 mmHg
• C. 347.9 mmHg
• D. 90.63 mmHg

Answer 1

Answer: (D)

Number of moles of $CHCl _{3}$,

$n _{ A }=\frac{ W }{ M }=\frac{25.5}{119.5}=0.213$

Number of moles of $CH _{2} Cl _{2}$

$n _{ B }=\frac{40}{85}=0.47$

Mole fraction of $CHCl _{3}$

$x _{ A }=\frac{ n _{ A }}{ n _{ A }+ n _{ B }}=\frac{0.213}{0.684}=0.31$

mole fraction of $CH _{2} Cl _{2}$

$X_{B}= 1-x_{A}$

$= 1-0.31=0.69$

$P_{\text {Total}}= P_{A} x_{A}+P_{B} x_{B}$

$=200 \times 0.31+41.5 \times 0.69$

$=62+28.63=90.63\, mm Hg$