Kf for water is 1.86 K kg mol–1. If your automobile radiator holds 1.0 kg of water, how many grams o — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

# Freezing Point Depression Solution **Given:** - $K_f = 1.86$ K·kg·mol⁻¹ - Mass of water = 1.0 kg - Freezing point of solution = –2.8°C - Solute: ethylene glycol ($C_2H_6O_2$), $M = 62$ g/mol **Step 1: Calculate freezing point depression** $$\Delta T_f = T_f(\text{pure}) - T_f(\text{solution}) = 0°C - (-2.8°C) = 2.8°C = 2.8 \text{ K}$$ **Step 2: Apply freezing point depression formula** $$\Delta T_f = K_f \times m$$ where $m$ is molality (moles of solute per kg of solvent). **Step 3: Solve for molality** $$m = \frac{\Delta T_f}{K_f} = \frac{2.8}{1.86} = 1.505 \text{ mol/kg}$$ **Step 4: Calculate moles of ethylene glycol needed** $$\text{moles} = m \times \text{mass of solvent (kg)} = 1.505 \times 1.0 = 1.505 \text{ mol}$$ **Step 5: Convert to grams** $$\text{mass} = 1.505 \text{ mol} \times 62 \text{ g/mol} = 93.3 \text{ g} \approx \boxed{93 \text{ g or } 94 \text{ g}}$$ **Why this is correct:** Ethylene glycol is a non-electrolyte (doesn't dissociate), so it exhibits ideal colligative behavior. The calculation directly uses the freezing point depression equation with the given constant.