Solutions

Given,

Vapour pressure of pure $\mathrm{CHCl}_3\left[p_1\right]$ and $\mathrm{CH}_2 \mathrm{Cl}_2\left[p_2\right]=200 \mathrm{~atm}$ and $41.5 \mathrm{~atm}$ respectively. Weight $\left(w_1\right)$ of $\mathrm{CHCl}_3$ and $\mathrm{CH}_2 \mathrm{Cl}_2\left(w_2\right)=11.9 \mathrm{~g}$ and $17 \mathrm{~g}$ respectively. Molar mass of $\mathrm{CHCl}_3\left(M_1\right)$ and $\mathrm{CH}_2 \mathrm{Cl}_2\left(M_2\right)=119 \mathrm{~g} \mathrm{~mol}^{-1}$ and $85 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively.

Thus, number of moles $(n)$ of $\mathrm{CHCl}_3=\frac{11.9}{119}=0.1 \mathrm{~mol}$

Number of moles of $\mathrm{CH}_2 \mathrm{Cl}_2=\frac{17}{85}=0.2 \mathrm{~mol}$

$\therefore p_{\text {total }}=p_1 \chi_1+p_2 \chi_2$

$\begin{aligned} \chi_1 & =\frac{n_1}{n_T}=\frac{0.1}{0.3}=0.33 \\ \chi_2 & =\frac{n_2}{n_T}=\frac{0.2}{0.3}=0.66 \\ p_{\text {Total }} & =200 \times 0.33+41.5 \times 0.66=94.33 \end{aligned}$

Hence, option (c) is the correct.

Key idea For the solutes, which can undergo association or dissociation in a given solvent. Elevation in boiling point $\left(\Delta T_b\right)=i K_b \times m$, where, $K_b=$ molar elevation constant $m=$ molality, $i=$ van't Hoff factor

Given,

$\begin{aligned} & \Delta T_b=1^{\circ} \mathrm{C} \text { and } M(\mathrm{NaCl})=58.5 \\ & K_b\left(\mathrm{H}_2 \mathrm{O}\right)=0.52 \mathrm{~kg} \mathrm{~mol} \\ & \text { for } \mathrm{NaCl}=n=(2) \text { and } w_{\text {solvent }}=500 \mathrm{~g} \end{aligned}$

$\begin{aligned} \text{Thus,} \quad \Delta T_b & =i \times K_b \times \frac{w(\text { solute })}{M(\text { solute })} \times \frac{1000}{w(\text { solvent })} \\ 1^{\circ} & =2 \times 0.52 \times \frac{w}{58.5} \times \frac{1000}{500} \\ w & =\frac{58.5 \times 500}{2 \times 0.52 \times 1000}=28.125 \mathrm{~g} \end{aligned}$

Hence, option (b) is the correct.

$\begin{aligned} & \text { Total }=1-x+\frac{x}{2}=1-\frac{x}{2} \\\\ & \therefore \quad i=\frac{1-x / 2}{1} \\\\ \text{or,} & i=1-\frac{x}{2} \\\\ \end{aligned}$