Use freezing point depression formula

For a non-electrolyte like ethylene glycol, $\Delta T_f = K_f m$ where:

• $\Delta T_f = 24\,\mathrm{K}$ (from $0^\circ\mathrm{C}$ to $-24^\circ\mathrm{C}$)
• $K_f = 1.86\,\mathrm{K\,kg\,mol^{-1}}$
• $m$ = molality

Calculate molality required

$m = \frac{\Delta T_f}{K_f} = \frac{24}{1.86}$

$m \approx 12.903\,\mathrm{mol\,kg^{-1}}$

Find moles of ethylene glycol needed

Mass of water = $18.6\,\mathrm{kg}$

Since $m = \frac{\text{moles of solute}}{\text{kg of solvent}}$ therefore, $\text{moles of ethylene glycol} = m \times 18.6$

$= \frac{24}{1.86} \times 18.6$

$= 24 \times 10 = 240\,\mathrm{mol}$

Convert moles to mass

Molar mass of ethylene glycol = $62\,\mathrm{g\,mol^{-1}}$

$\text{mass} = 240 \times 62 = 14880\,\mathrm{g}$

$14880\,\mathrm{g} = 14.88\,\mathrm{kg}$

Final integer answer

$\boxed{15}$

(nearest integer value)

Comparison with stored answer

Stored correct answer = $15$

Our derived answer is also $15$, so they agree.