A solution containing 62 g ethylene glycol in 250 g water is cooled to $$-10^{\circ}$$C. If $$K_f$$ for water is 1.86 K kg mol$$^{-1}$$, the amount of water (in g) separated as ice is:

We start by recalling the formula for depression of the freezing point of a dilute solution:

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

where $$\Delta T_f$$ is the lowering of the freezing point, $$K_f$$ is the cryoscopic constant of the solvent (for water, $$K_f = 1.86\ \text{K kg mol}^{-1}$$), and $$m$$ is the molality of the solution.

The solution contains ethylene glycol, $$C_2H_6O_2$$, which is a nonelectrolyte, so its van ’t Hoff factor $$i = 1$$. The given data are:

Mass of ethylene glycol added = $$62\ \text{g}$$
Molar mass of ethylene glycol = $$62\ \text{g mol}^{-1}$$
Initial mass of water = $$250\ \text{g} = 0.250\ \text{kg}$$
Observed final temperature = $$-10^{\circ}\text{C}$$
Pure water freezes at $$0^{\circ}\text{C}$$, so $$\Delta T_f = 0 - (-10) = 10\ \text{K}$$.

First we determine the moles of solute present:

$$n_{\text{solute}} = \frac{62\ \text{g}}{62\ \text{g mol}^{-1}} = 1\ \text{mol}$$

When the solution is cooled, some water separates out as ice. Let $$x\ \text{g}$$ be the mass of water that freezes. The mass of liquid water left behind (the solvent for the remaining solution) is therefore $$250 - x\ \text{g}$$, or $$\dfrac{250 - x}{1000}\ \text{kg}$$.

Hence the molality of the solution after separation of ice is

$$m = \frac{\text{moles of solute}}{\text{kilograms of liquid water}} = \frac{1}{\dfrac{250 - x}{1000}} = \frac{1000}{250 - x}\ \text{mol kg}^{-1}.$$

Substituting this molality and the given values of $$\Delta T_f$$ and $$K_f$$ in the freezing-point depression formula, we have

$$\Delta T_f = K_f\, m \quad\Longrightarrow\quad 10 = 1.86 \left(\frac{1000}{250 - x}\right).$$

Now we solve for $$x$$ step by step:

$$10(250 - x) = 1.86 \times 1000$$

$$2500 - 10x = 1860$$

$$2500 - 1860 = 10x$$

$$640 = 10x$$

$$x = \frac{640}{10} = 64\ \text{g}.$$

This value of $$x$$ represents the mass of water that has crystallised out as ice.

Hence, the correct answer is Option B.