If 1 mole of a non-volatile, non-electrolyte solute in 1000 g of water depresses the freezing point — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

In 1000 g water, $m = 1 \text{ m}$, so $\Delta T_f = 1.86^\circ\text{C}$. In 500 g water, $m = 1 \text{ mol} / 0.5 \text{ kg} = 2 \text{ m}$. Thus, $\Delta T_f = 2 \times 1.86^\circ\text{C} = 3.72^\circ\text{C}$. The freezing point of the solution is $0 - 3.72 = -3.72^\circ\text{C}$. Therefore, correct answer is D.