Vapour pressure of chloroform $\left(\mathrm{CHCl}_3\right)$ and dichloromethane $\left(\mathrm{CH}_2…
Vapour pressure of chloroform $\left(\mathrm{CHCl}_3\right)$ and dichloromethane $\left(\mathrm{CH}_2 \mathrm{Cl}_2\right)$ at $25^{\circ} \mathrm{C}$ are $200 \mathrm{~mm} \mathrm{Hg}$ and $415 \mathrm{~mm} \mathrm{Hg}$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \mathrm{~g}$ of $\mathrm{CHCl}_3$ and $4 \mathrm{O} \mathrm{g}$ of $\mathrm{CH}_2 \mathrm{Cl}_2$ at the same temperature will be:
[Molecular mass of $\mathrm{CHCl}_3=119.5 \mathrm{~g} / \mathrm{mol}$ and molecular mass of $\mathrm{CH}_2 \mathrm{Cl}_2=85 \mathrm{~g} / \mathrm{mol}$ ]
$173.9 \mathrm{mmHg}$
$615.0 \mathrm{mmHg}$
$347.9 \mathrm{mmHg}$
$28.5 \mathrm{mmHg}$
Solution
Molar mass of $\mathrm{CH}_2 \mathrm{Cl}_2=12 \times 1+1 \times 2+35.5 \times 2=85 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of $\mathrm{CHCl}_3=12 \times 1+1 \times 1+35.5 \times 3=119.5 \mathrm{~g} \mathrm{~mol}^{-1}$
Moles of $\mathrm{CH}_2 \mathrm{Cl}_2=40 \mathrm{~g} / 85 \mathrm{~g} \mathrm{~mol}^{-1}=0.47 \mathrm{~mol}$
Moles of $\mathrm{CHCl}_3=25.5 \mathrm{~g} / 119.5 \mathrm{~g} \mathrm{~mol}^{-1}=0.213 \mathrm{~mol}$
Total number of moles $=0.47+0.213=0.683 \mathrm{~mol}$
Mole fraction of component 2
$=0.47 \mathrm{~mol} / 0.683 \mathrm{~mol}=0.688$
Mole fraction of component 1
$=1.00-0.688=0.312$
We know that:
$\begin{aligned}
& \mathrm{P}_{\mathrm{T}}=\mathrm{p}_1^0+\left(\mathrm{p}_2^0-\mathrm{p}_1^0\right) \mathrm{x}_2 \\
& =200+(415-200) \times 0.688 \\
& =200+147.9 \\
& =347.9 \mathrm{~mm} \mathrm{Hg}
\end{aligned}$