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

# Solution: Freezing Point Depression Calculation **Given:** - $K_f$ for water = 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** $$\Delta T_f = T_f(\text{pure}) - T_f(\text{solution}) = 0°C - (-2.8°C) = 2.8\text{ K}$$ **Step 2: Use freezing point depression formula** $$\Delta T_f = K_f \cdot m$$ where $m$ is molality (mol solute/kg 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** $$\text{moles} = m \times \text{kg of water} = 1.505 \times 1.0 = 1.505\text{ mol}$$ **Step 5: Convert moles to grams** $$\text{mass} = 1.505 \text{ mol} \times 62\text{ g/mol} = \boxed{93.3\text{ g}} \approx \boxed{93\text{ g}}$$ **Why this works:** Ethylene glycol is a non-volatile, non-electrolyte solute that lowers the freezing point through colligative effects. The calculation uses the direct proportionality between freezing point depression and molality.