Question

Vapour pressure of chloroform $\left(CH{Cl}_3\right)$ and dichloromethane $\left({CH}_2{Cl}_2\right)$ at ${25}^{o}C$ are ${25}^{o}C$ are $200mm$ $Hg$ and $41.5mm$ $Hg$ respectively. Vapour pressure of the solution obtained by mixing $25.5g$ of $CH{Cl}_3$ and $40g$ of ${CH}_2{Cl}_2$ at the same temperature will be:
[Molecular mass of $CH{Cl}_3=119.5u$ and molecular mass of ${CH}_2{Cl}_2=85u$]

• A. $615.0mm$ $Hg$
• B. $347.9mm$ $Hg$
• C. $285.5mm$ $Hg$
• D. $\mathrm{1.73.9}mm$ $Hg$
• E. $90.38mm$ $Hg$

Answer 1

Answer: (E)

$90.38mm$ $Hg$

$CH{Cl}_3$ ${CH}_2{Cl}_2$
(A) (B)

$n_A=\frac{25.5}{119.5}=0.213$

$n_B=\frac{40}{85}=0.47$

$X_A=\frac{n_A}{n_A+n_B}=\frac{0.213}{0.648}=0.31$

$X_B=1-0.31=0.69$

$P=P_A^o{X}_A+P_B^o{X}_B$

$=200\times \left(0.31\right)+41.5\left(0.69\right)$

$=62.28+28.63=90.38$

This question will be a bonus because V.P of Dichloromethane cannot be less than chloroform (it would be $415mm$ of $Hg$)