100 g of sucrose solution in water is cooled to $- {0.5}^{o} C$ . What is the mass(g) of ice in decagram (nearest integer) that would be separated out at this temperature, if solution to freeze at $- {0.38}^{o} C$ ? ( $K_{f}$ for $H_{2} 6 = 1.86 K m o l^{- 1} k g)$ .

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Solution

Let 100 g solution contains 'w' g solute in 'W' g solvent
$∴ w + W = 1000$ .....(i)
Now, $Δ T_{f} = \frac{1000 \times K_{f} \times w}{m \times W}$
$0.38 = \frac{1000 \times 1.86 \times 2}{W \times 342}$ .....(ii)
$∴ \frac{w}{W} = 0.07$
Solving eqs. (i) and (ii), we get
$w = 6.6 g$
$W = 93.40 g$
Now, at $- 50^{o} C$ , some water separates out as ice and 6.6 g solute exists in solution.
$∴ 0.5 = \frac{1000 \times 1.86 \times 6.6}{W \times 342}$
$∴ W = 71.78 g$
$∴$ Mass of ice separated out is
$(93.40 - 71.78) g = 21.62 g \approx 2 d a g$

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100 g of sucrose solution in water is cooled to −0.5oC. What is the mass(g) of ice in decagram (nearest integer) that would be separated out at this temperature, if solution to freeze at −0.38oC? (Kf for H26=1.86Kmol−1kg).

100 g of sucrose solution in water is cooled to $- {0.5}^{o} C$ . What is the mass(g) of ice in decagram (nearest integer) that would be separated out at this temperature, if solution to freeze at $- {0.38}^{o} C$ ? ( $K_{f}$ for $H_{2} 6 = 1.86 K m o l^{- 1} k g)$ .