We know, $$\Delta$$T_f = k_f $$\times$$ molality of solution (m)

Given, k_f for water = 1.86 K kg mol^$$-$$1

$$\therefore$$ 0.30 = 1.86 $$\times$$ molality of solution (m)

$$\therefore$$ m = $${{0.30} \over {1.86}}$$ or, m = 0.161 mol kg^$$-$$1

According to Raoult's law, $${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$$

Amount of water in solution n₁ $$\times$$ 18 g

Molality of the solution $${{{n_2}} \over {{n_1} \times 18}} \times 1000 = {{55.55 \times {n_2}} \over {{n_1}}}$$ mol kg^$$-$$1

But, molality calculated from f. p. depression = 0.161 mol kg^$$-$$1

$$\therefore$$ $${{55.55 \times {n_2}} \over {{n_1}}} = 0.0161$$ or, $${{{n_2}} \over {{n_1}}} = 1.09 \times {10^{ - 3}}$$

Given : p⁰ = 23.51 mm Hg

$$\therefore$$ $${{23.51 - p} \over {23.51}} = {{{n_2}} \over {{n_1}}} = 1.09 \times {10^{ - 3}}$$

$$\therefore$$ p = 23.48 mm Hg