1. Initial composition of the solution

Given: 1 kg of 0.75 molal aqueous sucrose solution.

Molality is defined as: $m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

Let mass of water initially be $w$ kg. Then moles of sucrose are: $n = 0.75w$

Total mass of solution is 1 kg: $w + (0.75w)(342\,\text{g mol}^{-1})/1000 = 1$ $w + 0.2565w = 1$ $1.2565w = 1$ $w = 0.7959\,\text{kg}$

So,

• mass of water initially $= 0.7959$ kg
• moles of sucrose: $n = 0.75 \times 0.7959 = 0.5969\,\text{mol}$

2. Condition at $-4^\circ C$

At the temperature where freezing starts again, the remaining solution must have freezing point depression equal to $4^\circ C$.

Using: $\Delta T_f = K_f m$ $4 = 1.86\, m$ $m = \frac{4}{1.86} = 2.1505\,\text{mol kg}^{-1}$

3. Mass of water remaining in solution

Let mass of water left unfrozen be $W$ kg. Since sucrose does not freeze out, its moles remain $0.5969$ mol.

Thus, $2.1505 = \frac{0.5969}{W}$ $W = \frac{0.5969}{2.1505} = 0.2776\,\text{kg}$

4. Mass of ice separated

Initial water mass $= 0.7959$ kg

Water remaining in solution $= 0.2776$ kg

So ice separated: $0.7959 - 0.2776 = 0.5183\,\text{kg}$ $= 518.3\,\text{g}$

Nearest integer: $\boxed{518}$