IIT JEE Advanced 2012 Paper 2

When a non-volatile solute is added to pure solvent, there is a decrease in vapour pressure. So, according to relative lowering in vapour pressure, $\frac{p^{\circ} - P_s}{p^{\circ}}$ = $x_2$ where $x_2$ is the number of moles of solute, $p_o$ is the vapour pressure of the pure solvent and $p_s$ is the vapour pressure of the solution. Now, $x_2$ can also represented as $\frac{n_2}{n_1 + n_2}$. Thus, we have $\frac{p^{\circ} - p_s}{p^{\circ}}$ = $\frac{nn_2}{n_1 + n_2}$ Neglecting $n_2$ (as $n_2$ << n1) from the denominator, and substituting $p^{\circ}$ = 1 atm = 760 mm Hg, we get $\frac{760 - ps}{760}$ = $\frac{\frac{2.5}{M_B}}{\frac{100}{18}} \times \frac{1000}{1000}$ = $\frac{18}{1000} \times m$ (1) According to elevation in boiling point: $\Delta T_b$ = $K_b \times m$ ⇒ 2 = 0.76 × m ⇒ m = $\frac{2}{0.76}$ Putting this value in Eq. (1), we get $\frac{760 - p_s}{760}$ = $\frac{8}{1000} \times \frac{2}{0.76}$ ⇒ $p_s$ = $760 - \frac{18 \times 2 \times 760}{1000 \times 0.76}$ = 760 - 36 = 724 mm Hg