A one kg aqueous solution of sugar ( molality, $0.8$ was cooled and maintained at $-4°\mathrm{C}$. Calculate the amount of ice that would separate from it ${\mathrm{K}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-}$

Molar mass of sugar is:

 $\left.({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)=(12\times 12)+\left(22\times 1\right)+\left(11\times 16\right)=342 \mathrm{g}/\mathrm{mol}$ 

Molality is 0.8 m which means that 0.8 moles of solute present in one kg of solvent. Mass of sugar is:

$\text{Mass of sugar}=0.8 \mathrm{mol}\times 342 \mathrm{g}/\mathrm{mol}=273.6 \mathrm{g}$

$+(22\times 1)+(11\times 16)=342.\text{ Molality, }\mathrm{m}=0.8 \mathrm{m}=0.8\mathrm{mol}$  sugar per kg solvent

Mass of solution is equal to mass of solute and solvent.

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{solution}=\mathrm{Mass} \mathrm{of} \mathrm{solute}+\mathrm{mass} \mathrm{of} \mathrm{solvent} \\ \mathrm{Mass} \mathrm{of} \mathrm{solution}=273.6 \mathrm{g}+1000 \mathrm{g}=1273.6 \mathrm{g} \end{gathered}$$

Mass of sugar present in 1273.6 g of solution is 273.6 g. Then, the mass of solution present in one kg (1000 g) is:

$\mathrm{Mass} \mathrm{of} \mathrm{sugar}=\frac{273.6 \mathrm{g}}{1273.6 \mathrm{g}} \times 1000 \mathrm{g}=214.82 \mathrm{g}$

Mass of  solvent $\left({\mathrm{H}}_2\mathrm{O}\right)$ in $1 \mathrm{kg}$ solution

$\mathrm{Mass} \mathrm{of} \mathrm{solvent}=1000-214.82 \mathrm{g}=785.18 \mathrm{g}$

Using the formula:

${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{iK}}_{\mathrm{f}}\mathrm{m}={\mathrm{iK}}_{\mathrm{f}}\frac{\mathrm{w}}{\mathrm{m}}\times \frac{1000}{\mathrm{W}}$

Substitute the values: (For sugar, the value of van't Hoff factor (i) is 1.)

$4=\frac{1 \times 1.86 \times 214.82 \times 1000}{342 \times \mathrm{W}}$

$\mathrm{W}=\frac{1.86 \times 214.82 \times 1000}{4 \times 342}=292.1 \mathrm{g}$

291.2 g of solvent required to maintain this solution at $-4°\mathrm{C}$. The remaining weight of water will change into ice. So, weight of ice formed

$\mathrm{weight} \mathrm{of} \mathrm{ice}=785.18 - 292.1 = 493.08 \mathrm{g}$