Question

Vapour pressure of chloroform $\left(CHC{l}_3\right)$ and dichloromethane $\left(C{H}_{2}C{l}_2\right)$ at ${25}^{o}C$ are $200mmHg\text{and}41.5mmHg$ respectively. Vapour pressure of the solution obtained by mixing $25.5g$ of $CHC{l}_3$ and $40g$ of $C{H}_{2}C{l}_2$ at the same temperature will be:
(Molecular mass of $CHC{l}_3=119.5u$ and molecular mass of $C{H}_{2}C{l}_2=85u$)

• A. 615.0 mm Hg
• B. 173.9 mm Hg
• C. 285.5 mm Hg
• D. 347.9 mm Hg

Answer 1

The correct option is D 347.9 mm Hg

$\text{Molar mass of}C{H}_{2}C{l}_2=85gmo{l}^{–1}\text{Molar mass of}CHC{l}_3=119.5gmo{l}^{-1}\text{Moles of}C{H}_{2}C{l}_2=\frac{40g}{85gmo{l}^{-1}}=0.47mol\text{Moles of}CHC{l}_3=\frac{25.5g}{119.5gmo{l}^{-1}}=0.213mol\text{Total number of moles}=0.47+0.213=0.683mol\text{Mole fraction of}C{H}_{2}C{l}_2=\frac{0.47}{0.683}=0.688\text{Mole fraction of}CHC{l}_3=1.00–0.688=0.312$

We know that:

$P_T=p_1^o+(p_2^o-p_1^o){\chi }_2$
$p_1^o$ is vapour pressure of pure $CHC{l}_3$
$p_2^o$ is vapour pressure of pure $C{H}_{2}C{l}_2$
${\chi }_2$ is mole fraction of $C{H}_{2}C{l}_2$

$P_T=200+(415–200)\times 0.688$

$=200+147.9=347.9mmHg$

Hence, option (c) is correct.