DTf = iKf m
DTf = Depression in freezing point
Kf = Freezing Point Depression Constant
m = molality
i = van’t Hoff factor, for CaCl2 i = 3
Moles of CaCl2 = 5.7 g/110.98 g mol−1 = 0.0513 mol
Mass of water in kg = 18.6 kg
Hence molality of CaCl2 = 0.0513 mol/18.6 kg = 2.758×10-3 mol kg-1
Kf for water is =1.86 K kg mol–1
DTf = 3×1.86 K kg mol–1×2.758×10-3 mol kg-1
= 0.01539 K
Depression in freezing point 0.01539 K
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