Let, solubility of AB2(1:2 type electrolyte) is pure water =5molL−1=5M ⇒Ksp=4S3=2.56×10−4M3 (given) ∴S=0.04M=0.04m (molal) under standard condition of the solution. Depression of freezing point, ∆Tf=Kf×m×i=1.8×0.04×3=0.216K [∵ Assuming complete dissociation of AB2 van't Hoff factor, i=3]