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 (depression = 2.8 K) - Molar mass of $C_2H_6O_2$ = 62 g/mol **Step 1: Use the freezing point depression equation** $$\Delta T_f = K_f \times m$$ where $m$ is molality (moles of solute per kg of solvent). $$2.8 = 1.86 \times m$$ $$m = \frac{2.8}{1.86} = 1.505 \text{ mol/kg}$$ **Step 2: Calculate moles of ethylene glycol needed** $$\text{moles} = m \times \text{mass of solvent (kg)}$$ $$\text{moles} = 1.505 \times 1.0 = 1.505 \text{ mol}$$ **Step 3: Convert moles to grams** $$\text{mass} = \text{moles} \times \text{molar mass}$$ $$\text{mass} = 1.505 \times 62 = 93.3 \text{ g}$$ **Answer: ~93 g of ethylene glycol** (or approximately 94 g depending on rounding) This is option **B**. The colligative property of freezing point depression depends only on the number of solute particles, making ethylene glycol an ideal antifreeze since it's non-volatile and non-electrolytic.