Question

How many grams of KCl should be added to 1kg of water to lower its freezing point to ‐8.0 °C (kf = 1.86 K kg /mol)

Answer 1

Since KCl dissociate in water completely L = 2

ΔTf = ikf x m;

m = ΔTf/ikf

m = 8 / 2 x 1.86 = 2.15 mol/kg. 

Grams of KCl = 2.15 x 74.5 = 160.2 g/kg.