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

💡 Solution & Explanation

I need to calculate the freezing point depression for a 0.05 molal non-electrolyte solution in water. **Step 1: Identify the freezing point depression formula** $$\Delta T_f = K_f \times m \times i$$ where: - $K_f$ = cryoscopic constant of water = 1.86 K·kg/mol - $m$ = molality = 0.05 mol/kg - $i$ = van't Hoff factor = 1 (non-electrolyte doesn't dissociate) **Step 2: Calculate freezing point depression** $$\Delta T_f = 1.86 \times 0.05 \times 1 = 0.093 \text{ K} \approx 0.09°C$$ **Step 3: Determine new freezing point** The normal freezing point of pure water is 0°C (273.15 K). Since solute addition **lowers** the freezing point: $$T_f = 0 - 0.09 = -0.09°C$$ or approximately **-0.1°C** (depending on significant figures) **Why this is correct:** Non-electrolytes don't dissociate in solution, so they produce minimal particle interactions. The freezing point depression depends directly on molality and $K_f$. The depression is relatively small (~0.09°C) because the molality is low (0.05 m). Option C likely shows **-0.09°C** or **-0.1°C**, which matches this calculation.