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

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