100 g of C₆H₁₂O₆(aq) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H₂O(1) is 40.18 torr at same temperature. If this solution is cooled to -0.93$°$C, what mass of ice will be separeted out? (K_{f =} 1.86 kg mol^-1)

Molality of solution = $$\begin{gathered} \frac{P°-P}{P}\times \frac{1000}{M} \\ \Rightarrow \frac{40.18-40}{40}\times \frac{1000}{18}\Rightarrow 0.25 \end{gathered}$$

1000 g water present with 45 g glucose or 100 g solution has 4.31 g glucose and 95.69 g H₂O. Final molality is 0.5, 1000 g solvent contain 90 g glucose or 4.31 g glucose present with $\frac{1000}{90}\times 4.31 = 47.88 g$

H₂O Mass of ice formed = 95 .69 - 47 .88 = 47 .8 g