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 ($m$) = 0.05 mol/kg - Non-electrolyte (van't Hoff factor $i = 1$) - Solvent = water **Step 1: Apply freezing point depression formula** $$\Delta T_f = i \cdot K_f \cdot m$$ where $K_f$ (cryoscopic constant for water) = 1.86 K·kg/mol **Step 2: Calculate depression** $$\Delta T_f = 1 \times 1.86 \times 0.05 = 0.093°C \approx 0.09°C$$ **Step 3: Find freezing point** $$T_f = 0°C - 0.093°C = -0.093°C \approx -0.09°C$$ **Why C is correct:** Option C likely states **−0.093°C** (or approximately **−0.09°C**), which matches our calculation. The freezing point of pure water (0°C) is depressed by the solute, resulting in a negative temperature. **Note:** If other options included: - Positive values → Wrong (depression means lower temperature) - Values like −0.46°C → Wrong (would require molality of 0.25 or electrolyte behavior)