A Solution of a non–volatile solute in water freezes at –0.30ºC. The vapour–pressure of pure water at 298K is 23.51 mm Hg and $k_f$ for water is 1.86 K. kg mol–1 Calculate the vapour–pressure of this solution at 298K.

Solution of non-volatile solute in water

Change in freezing point ($∆T_f$) = -0.30°C

V.P of pure water (P°) = 23.51 mmHg

Temperature is same as 298K 

Molal freezing point depression constant ($K_f$) = 1.86 Kg$mo{l}^{-1}$

V.P. of solution (P) = ? 

We know that, 

   $∆T_f=K_f.m_2$

   $m_{2 }= \frac{0.3}{1.86}=0.1613$

So, 

    $\frac{n_2}{n_1}=\frac{0.1613\times 18}{1000}=2.9\times {10}^{-3}$

    $\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}+1 = \frac{{\mathrm{n}}_2+{\mathrm{n}}_1}{{\mathrm{n}}_1} = 2.9\times {10}^{-3}+ 1$

    $\frac{n_1}{n_1+n_2}$ = $X_1$ =$\frac{1}{1+ 2.9\times {10}^{-3}}$

    = 0.997

So,  

P = P° $X_1$ = 23.51× 0.997 = 23.44 mm

Hence, the correct option is option (A) 23.4 mm.