QJEEM21Ch7b24: JEE Main 2021 Chemistry - Class 12 JEEM21Ch7b24 Solution: 1 kg of 0.75 molal aqueous solution of sucrose can be cooled up …

Let mass of water initially present = x gm

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

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

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

$\Rightarrow$ 256.5x = 10⁶ $-$ 1000x

$\Rightarrow$ x = 795.86 gm

$\Rightarrow$ moles of sucrose = 0.5969

New mass of H₂O = a kg

$\Rightarrow 4 = {{0.5969} \over a} \times 1.86 \Rightarrow$ a = 0.2775 kg

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