1. **State the problem:** We need to find the mass of ethylene glycol (C2H6O2) to add to 15.0 L of water to make a solution that freezes at -23.3 \degree C. 2. **Relevant formula:** The freezing point depression formula is $$\Delta T_f = i K_f m$$ where $\Delta T_f$ is the freezing point depression, $i$ is the van't Hoff factor, $K_f$ is the freezing point depression constant for water, and $m$ is the molality of the solution. 3. **Known values:** - $\Delta T_f = 0 - (-23.3) = 23.3$ \degree C - For ethylene glycol, $i = 1$ (it does not ionize) - $K_f$ for water = 1.86 \degree C kg/mol - Volume of water = 15.0 L = 15,000 mL - Density of water = 1 g/mL, so mass of water = 15,000 g = 15.0 kg 4. **Calculate molality $m$:** $$m = \frac{\Delta T_f}{i K_f} = \frac{23.3}{1 \times 1.86} = 12.52 \text{ mol/kg}$$ 5. **Define molality:** $$m = \frac{\text{moles of solute}}{\text{kg of solvent}}$$ Rearranged: $$\text{moles of solute} = m \times \text{kg of solvent} = 12.52 \times 15.0 = 187.8 \text{ mol}$$ 6. **Calculate mass of ethylene glycol:** Molar mass = 62.1 g/mol $$\text{mass} = \text{moles} \times \text{molar mass} = 187.8 \times 62.1 = 11666.4 \text{ g}$$ 7. **Convert to kilograms:** $$11666.4 \text{ g} = 11.67 \text{ kg}$$ **Final answer:** The mass of ethylene glycol needed is approximately **11.7 kg**.