1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $$-4°$$C before freezing. The amount of ice (in g) that will be separated out is _________. (Nearest integer) [Given : $$K_f(H_2O) = 1.86$$ K kg mol$$^{-1}$$]

We are given a 0.75 molal ($$m = 0.75\;{\rm mol\,kg^{-1}}$$) aqueous solution of sucrose whose total mass is 1 kg. Molality is defined as

$$m \;=\; \frac{n_{\text{sucrose}}}{\text{mass of solvent in kg}}$$

Let $$W$$ = mass of water (solvent) in g, $$n$$ = moles of sucrose present, and recall that the molar mass of sucrose is $$M = 342\;{\rm g\,mol^{-1}}.$$

From the definition of molality we have

$$n \;=\; 0.75 \times \frac{W}{1000} \quad\quad (1)$$

The total mass of the solution is the sum of the masses of water and sucrose, so

$$1000 \;=\; W \;+\; n\,M \quad\quad (2)$$

Substituting the value of $$n$$ from (1) into (2) gives

$$1000 \;=\; W \;+\; \left(0.75 \times \frac{W}{1000}\right)\! \times 342$$ $$\Rightarrow\; 1000 \;=\; W \;+\; W\left(\frac{0.75 \times 342}{1000}\right)$$ $$\Rightarrow\; 1000 \;=\; W\!\left[1 \;+\; \frac{256.5}{1000}\right]$$ $$\Rightarrow\; 1000 \;=\; W \times 1.2565$$ $$\Rightarrow\; W \;=\; \frac{1000}{1.2565} \;=\; 795.9\;{\rm g}$$

Thus, before cooling, the solution contains $$n = 0.75 \times \frac{795.9}{1000} = 0.5969\;{\rm mol}$$ of sucrose dissolved in $$795.9\;{\rm g}$$ of water.

We now cool the solution to $$-4^{\circ}{\rm C}$$. Some water freezes out as ice until the remaining liquid solution attains its equilibrium freezing point. For a non-electrolyte such as sucrose ($$i = 1$$), the depression of the freezing point is given by

$$\Delta T_f = K_f \, m'$$

where $$m'$$ is the final molality. Here $$\Delta T_f = 4\;{\rm K}, \quad K_f = 1.86\;{\rm K\,kg\,mol^{-1}}$$ so

$$m' = \frac{\Delta T_f}{K_f} = \frac{4}{1.86} = 2.1505\;{\rm mol\,kg^{-1}}$$

Let $$w$$ (g) be the mass of water that remains in the liquid phase after partial freezing. Applying the definition of molality once more,

$$m' = \frac{n}{w/1000} \;=\; \frac{0.5969}{w/1000} = \frac{596.9}{w}$$

Equating the two expressions for $$m'$$,

$$\frac{596.9}{w} = 2.1505$$ $$\Rightarrow\; w = \frac{596.9}{2.1505} = 277.7\;{\rm g}$$

Originally the solution contained $$795.9\;{\rm g}$$ of water. Therefore the mass of water that freezes out as ice is

$$\text{mass of ice} = 795.9 - 277.7 = 518.2\;{\rm g}$$

Taking the nearest integer, the amount of ice separated is $$518\;{\rm g}$$.

So, the answer is $$518\;{\rm g}$$.