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? (Kf = 1.86 kg mol-1)

Molality of solution = P°-PP×1000M ⇒40.18-4040×100018⇒0.251000 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 100090×4.31 = 47.88 gH2O Mass of ice formed = 95 .69 - 47 .88 = 47 .8 g