The freezing point of a 0.05 molal solution of a non- electrolyte in water is — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

# Freezing Point Depression Calculation **Given:** - Molality of non-electrolyte solution: $m = 0.05$ mol/kg - Solvent: water - Need to find: freezing point **Step 1: Apply freezing point depression formula** $$\Delta T_f = K_f \times m \times i$$ where: - $K_f$ (cryoscopic constant for water) = $1.86$ °C/mol - $m$ = molality = $0.05$ mol/kg - $i$ = van't Hoff factor = $1$ (non-electrolyte doesn't dissociate) **Step 2: Calculate depression** $$\Delta T_f = 1.86 \times 0.05 \times 1$$ $$\Delta T_f = 0.093 \text{ °C} \approx 0.09 \text{ °C}$$ **Step 3: Find freezing point** Normal freezing point of water = $0$ °C New freezing point: $$T_f = 0 - 0.093 = -0.093 \text{ °C} \approx -0.09 \text{ °C}$$ **Answer: The freezing point is approximately –0.09°C (or –0.093°C)** The solution freezes at a lower temperature than pure water due to colligative property of freezing point depression caused by the dissolved solute particles.