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:** - $K_f$ for water = 1.86 K·kg·mol⁻¹ - Mass of water = 1.0 kg - Freezing point of solution = –2.8°C - Solute: ethylene glycol ($C_2H_6O_2$) **Step 1: Calculate freezing point depression** $$\Delta T_f = T_{f,pure} - T_{f,solution} = 0°C - (-2.8°C) = 2.8 \text{ K}$$ **Step 2: Apply freezing point depression equation** $$\Delta T_f = K_f \cdot m$$ where $m$ is molality (mol solute/kg solvent) $$2.8 = 1.86 \times m$$ $$m = \frac{2.8}{1.86} = 1.505 \text{ mol/kg}$$ **Step 3: Calculate moles of ethylene glycol needed** $$\text{moles} = m \times \text{kg of solvent} = 1.505 \text{ mol/kg} \times 1.0 \text{ kg} = 1.505 \text{ mol}$$ **Step 4: Convert moles to grams** Molar mass of $C_2H_6O_2$ = 2(12) + 6(1) + 2(16) = 62 g/mol $$\text{mass} = 1.505 \text{ mol} \times 62 \text{ g/mol} = 93.3 \text{ g} \approx 93 \text{ g}$$ **Answer: Approximately 93 grams of ethylene glycol must be added.**