A solution of '' mole of sucrose in 100 g of water freezes at . As ice separates out, the freezing p — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Let $K_f$ be the constant for water. At $-0.2^\circ\text{C}$, $\Delta T_f = 0.2 = K_f \times (x / 0.1) \Rightarrow x K_f = 0.02$. At $-0.25^\circ\text{C}$, the remaining mass of liquid water is $W$ kg, so $\Delta T_f = 0.25 = K_f \times (x / W) \Rightarrow x K_f = 0.25W$. Equating the two expressions for $x K_f$ gives $0.02 = 0.25W \Rightarrow W = 0.02 / 0.25 = 0.08 \text{ kg} = 80 \text{ g}$. Since the initial mass of water was 100 g, the mass of ice separated is $100 - 80 = 20 \text{ g}$. Therefore, correct answer is B.