How many grams of ethylene glycol should be mixed in 795 g water to form an antifreeze solution that — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

The freezing point depression is $\Delta T_f = 30^\circ\text{C}$. Molality $m = \Delta T_f / K_f = 30 / 1.86 = 16.129 \text{ mol/kg}$. Mass of water $W = 795 \text{ g} = 0.795 \text{ kg}$. Moles of ethylene glycol ($C_2H_6O_2$, M=62 g/mol) required = $m \times W = 16.129 \times 0.795 = 12.8225 \text{ mol}$. Mass of ethylene glycol = $12.8225 \text{ mol} \times 62 \text{ g/mol} = 794.995 \text{ g} \approx 795 \text{ g}$. Therefore, correct answer is 0795.