The mass of required to depress the freezing point of 500 g water by 2 K is ( = 1.86, K = 39, Cl = 3 — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Molar mass of $KCl$ = 74.5 g/mol, i = 2. ΔT_f = i * $K_f$ * m => 2 = 2 * 1.86 * (w_KCl / 74.5) * (1000 / 500) => w_KCl = 74.5 / 3.72 ≈ 20.03 g.