Question

One mole of ethylene glycol was dissolved in 1 kg water and the resulting solution was cooled to -6°C. How many grams of ice is separated in the process? [$K_f$ of $H_2O = 1.86K kg mol^{-1}$]

Answer 1

690

Answer 2

The solution freezes at -6°C. The freezing point depression is 6°C. Using $\Delta T_f = K_f \cdot m$, the molality of the solution at -6°C is $m = \frac{6}{1.86}$. Since the moles of solute (1 mole) remain constant, the mass of water remaining ($W_{remaining}$ in kg) can be found using $m = \frac{1}{W_{remaining}}$. Solving for $W_{remaining}$, we get $W_{remaining} = \frac{1.86}{6} = 0.31$ kg = 310 g. The initial mass of water was 1000 g. The mass of ice separated is $1000 \text{ g} - 310 \text{ g} = 690 \text{ g}$.