Vapour pressure of ${\mathrm{CCl}}_4$ at ${25}^{o}C$ is 143 mmHg ,0.05 g of a non-volatile solute (mol. wt. = 65) is dissolved in 100 ml ${\mathrm{CCl}}_4$. Find the vapour pressure of the solution (Density of ${\mathrm{CCl}}_4=1.58\mathrm{g}/{\mathrm{cm}}^3$)

$\begin{array}{l}\frac{{\mathrm{P}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{s}}}{{\mathrm{P}}^{\mathrm{o}}}=\frac{\mathrm{w}\times \mathrm{M}}{\mathrm{mW}}\\ =143-1.03=141.97\mathrm{mm}\end{array}$

$\left[OR\right]$

$$\begin{gathered} \frac{{\mathrm{P}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{S}}}{{\mathrm{P}}^{\mathrm{o}}}= \frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}} \\ {\mathrm{n}}_{\mathrm{B}}=\frac{ 0.05}{65}=0.00076 \end{gathered}$$

$$\begin{gathered} {\mathrm{w}}_{{\mathrm{CCl}}_4} = \mathrm{V}\times \mathrm{d} = 100\times 1.58=158 \\ {\mathrm{n}}_{\mathrm{A} }= \frac{158}{154} \end{gathered}$$

$\frac{{\mathrm{P}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{S}}}{{\mathrm{P}}^{\mathrm{o}}}= \frac{0.05}{65}\times \frac{154}{158}$

$\frac{{\mathrm{P}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{S}}}{{\mathrm{P}}^{\mathrm{o}}}=0.007$

$$\begin{gathered} 143-{\mathrm{P}}_{\mathrm{S}}=0.007\times 143 \\ 143 -{\mathrm{P}}_{\mathrm{S} }=1.001 \\ {\mathrm{P}}_{\mathrm{S}}=143-1.001=141.99\mathrm{mm} \mathrm{of} \mathrm{Hg} \end{gathered}$$