Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at $$-24°C$$ is ______ kg (Molar mass in g mol$$^{-1}$$ for ethylene glycol 62, $$K_f$$ of water = $$1.86 \text{ K kg mol}^{-1}$$)

Find the mass of ethylene glycol needed to protect 18.6 kg water from freezing at $$-24°C$$.

$$ \Delta T_f = K_f \times m $$

where $$\Delta T_f$$ is the depression in freezing point, $$K_f$$ is the molal freezing point depression constant, and $$m$$ is molality.

$$\Delta T_f = 0 - (-24) = 24$$ K.

$$ m = \frac{\Delta T_f}{K_f} = \frac{24}{1.86} = 12.903 \text{ mol/kg} $$

$$ \text{Moles} = m \times \text{mass of solvent (kg)} = 12.903 \times 18.6 = 240 \text{ mol} $$

Molar mass of ethylene glycol = 62 g/mol.

$$ \text{Mass} = 240 \times 62 = 14880 \text{ g} = 14.88 \text{ kg} \approx 15 \text{ kg} $$

The answer is 15 kg.