Question

Vapour pressure of chloroform ${(CHCl_3)}$ and dichloromethane ${(CH2Cl2)}$ at $25^{\circ}C$ are 200 mmHg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of ${CHCl_3}$ and 40 g of ${CH2Cl2}$ at the same temperature will be: (Molecular mass of ${CHCl_3}$ = 119.5 u and molecular mass of ${CH2Cl2 = 85 \, u}$)

• A. 615.0 mm Hg
• B. 347.9 mm Hg
• C. 285.5 mm Hg
• D. 90.38 mm Hg

Answer 1

The Correct Option is D

Number of moles of $CHC{{l}_{3}},$
${{n}_{A}}=\frac{w}{M}$
$=\frac{25.5}{119.5}$
$=0.213$
Number of moles of $C{{H}_{2}}C{{l}_{2}},$
${{n}_{B}}=\frac{40}{85}=0.47$
Mole fraction of $CHC{{l}_{3}}$
${{\chi }_{A}}=\frac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}}$
$=\frac{0.213}{0.683}=0.31$
Mole fraction of $C{{H}_{2}}C{{l}_{2}},$
${{\chi }_{B}}=1-{{\chi }_{A}}$
$=1-0.31=0.69$
$\because \; {{p}_{total}}={{p}_{A}}{{\chi }_{A}}+{{p}_{B}}{{\chi }_{B}}$
$=200\times 0.31+41.5\times 0.69$
$=62+28.63$
$=90.63mm\,Hg$