Previous Year NEET Question- Solution L-4

Question: The correct option for the value of vapour pressure of a solution at $45^{\circ} \mathrm{C}$ with benzene to octane in molar ratio $3: 2$ is#

[At $45^{\circ} \mathrm{C}$ vapour pressure of benzene is $280 \mathrm{~mm} \mathrm{Hg}$ and that of octane is $420 \mathrm{~mm} \mathrm{Hg}$. Assume ideal gas].

A) $160 \mathrm{~mm}$ of $\mathrm{Hg}$

B) $168 \mathrm{~mm}$ of $\mathrm{Hg}$

C) $336 \mathrm{~mm}$ of $\mathrm{Hg}$

D) $350 \mathrm{~mm}$ of $\mathrm{Hg}$

Answer: $336 \mathrm{~mm}$ of $\mathrm{Hg}$#

Solution:#

Molar ratio of benzene to octane, $\frac{n_B}{n_0}=\frac{3}{2}$ Let $n_{\mathrm{B}}=3 \times \mathrm{mol}, n_{\mathrm{O}}=2 \times \mathrm{mol}$ Total number of moles $$ =n_B+n_0=3 x+2 x=5 x \mathrm{~mol} $$

Mole fraction of benzene, $$ \chi_B=\frac{n_B}{n_B+n_0}=\frac{3 x}{5 x}=\frac{3}{5} . $$

Mole fraction of octane, $$ \chi_0=\frac{n_0}{n_B+n_0}=\frac{2 x}{5 x}=\frac{2}{5} $$

Vapour pressure of benzene, $$ p_{\mathrm{B}}^{\circ}=280 \mathrm{~mm} \mathrm{Hg} $$

Vapour pressure of octane, $$ p_0^{\circ}=420 \mathrm{~mm} \mathrm{Hg} $$

Total vapour pressure of solution, $$ p_S=\chi_B p_B^{\circ}+\chi_0 p_0^{\circ} $$ $\begin{aligned} & =\frac{3}{5} \times 280+\frac{2}{5} \times 420 \ & =3 \times 56+2 \times 84 \ & =168+168 \ & =336 \mathrm{~mm} \text { of } \mathrm{Hg}\end{aligned}$