# 1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to –4°C before freezing.

Question:

1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to –4°C before freezing. The amount of ice (in g) that will be separated out is ________ . (Nearest integer)

$\left[\right.$ Given : $\left.\mathrm{K}_{\mathrm{r}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$

Solution:

Let mass of water initially present = x gm

$\Rightarrow$ Mass of sucrose $=(1000-\mathrm{x}) \mathrm{gm}$

$\Rightarrow$ moles of sucrose $=\left(\frac{1000-x}{342}\right)$

$\Rightarrow 0.75=\frac{\left(\frac{1000-x}{342}\right)}{\left(\frac{x}{1000}\right)} \Rightarrow \frac{x}{1000}=\frac{1000-x}{342 \times 0.75}$

$\Rightarrow 256.5 x=10^{6}-1000 x$

$\Rightarrow x=795.86 \mathrm{gm}$

$\Rightarrow$ moles of sucrose $=0.5969$

New mass of $\mathrm{H}_{2} \mathrm{O}=\mathrm{a} \mathrm{kg}$

$\Rightarrow 4=\frac{0.5969}{\mathrm{a}} \times 1.86 \Rightarrow \mathrm{a}=0.2775 \mathrm{~kg}$

$\Rightarrow$ ice separated $=(795.86-277.5)=518.3 \mathrm{gm}$

Vk
May 31, 2024, 6:35 a.m.
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March 23, 2024, 12:15 p.m.
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Arpit
Jan. 17, 2024, 7:04 a.m.
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