First slide

General discussion of solutions

Question

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)

Moderate

A

95.5 g
B

4.5 g
C

45.5 g
D

47.8 g

Solution

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

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