1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $-$4$^\circ$C before freezing. The amount of ice (in g) that will be separated out is __________. (Nearest integer)

[Given : K_f(H₂O) = 1.86 K kg mol^$-$1]

Let mass of water initially present = x gm<br><br>$\Rightarrow$ Mass of sucrose = (1000 $-$ x) gm<br><br>$\Rightarrow$ moles of sucrose = $\left( {{{1000 - x} \over {342}}} \right)$<br><br>$$ \Rightarrow 0.75 = {{\left( {{{1000 - x} \over {342}}} \right)} \over {\left( {{x \over {1000}}} \right)}} \Rightarrow {x \over {1000}} = {{1000 - x} \over {342 \times 0.75}}$$<br><br>$\Rightarrow$ 256.5x = 10⁶ $-$ 1000x<br><br>$\Rightarrow$ x = 795.86 gm<br><br>$\Rightarrow$ moles of sucrose = 0.5969<br><br>New mass of H₂O = a kg<br><br>$\Rightarrow 4 = {{0.5969} \over a} \times 1.86 \Rightarrow$ a = 0.2775 kg<br><br>$\Rightarrow$ ice separated = (795.86 $-$ 277.5) = 518.3 gm