The mass of KCl required to depress the freezing point of 500 g water by 2 K is (, ) — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

$\Delta T_f = i \times K_f \times m$. For KCl, $i = 2$. Molar mass of KCl = $39 + 35.5 = 74.5 \text{ g/mol}$. $2 = 2 \times 1.86 \times (w / (74.5 \times 0.5))$. $1 = 1.86 \times (w / 37.25) \Rightarrow w = 37.25 / 1.86 = 20.026 \text{ g} \approx 20.03 \text{ g}$. Therefore, correct answer is D.