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Standard XII

Chemistry

Depression in Freezing Point

1 kg of an aq...

Question

$1$ kg of an aqueous solution of sucrose is cooled and maintained at $- 4^{o}$ C. How much ice will be separated out if the molality of the solution is $0.75$ ? $K_{f} (H_{2} O) = 1.86$ Kg $m o l^{- 1}$ K?

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Solution

Molar mass of sucrose= $342 g m o l^{- 1}$
So mass of sucrose per $k g$ of solvent= $0.75 \times 342 = 256.5 g$
Mass of sucrose in $1 k g$ of solution= $\frac{256.5}{1000 + 256.5} \times 1000 = 204.14 g$
Mass of solvent= $1000 - 204.14 = 795.86 g$
$Δ T_{f} = i K_{f} \times m = 1 \times 1.86 \times \frac{204.14}{342} \times \frac{1000}{W}$
$\Rightarrow 4 = \frac{1.86 \times 204.14 \times 1000}{342 \times W} \Rightarrow W = 277.55 g$
Amount of ice will be separated= $795.86 - 277.55 = 518.31 g$ .

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1 kg of an aqueous solution of sucrose is cooled and maintained at −4oC. How much ice will be separated out if the molality of the solution is 0.75? Kf(H2O)=1.86 Kg mol−1K?

$1$ kg of an aqueous solution of sucrose is cooled and maintained at $- 4^{o}$ C. How much ice will be separated out if the molality of the solution is $0.75$ ? $K_{f} (H_{2} O) = 1.86$ Kg $m o l^{- 1}$ K?