Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2K. (K_f for water = 1.86 K kg mol^-1) 

40.05 g

Since one mol of KCl gives 2 mole particles,
the value of i = 2, ΔT_f = 2K, K_f = 1.86 kg mol^-1
Applying equation, ΔT_f = iK_fm

m = \(\frac{ΔT_f}{iK_f}\) = \(\frac{2}{2×1.86}\) = 0.54 mol kg^-1
∴ 0.54 mole of KCl should be added to 1 kg of water
Molar mass of KCl = 39 + 35.5 = 74.5 g
∴ Amount of KCl = 0.54 × 74.5 g = 40.05 g