Q11 of 26 Page 1

Calculate the amount of CaCl2 (molar mass = 111g/mol) which must be added to 500g of water to lower its freezing point by 2K assuming CaCl2 is completely disassociated. (Kf for water is = 1.86 Kg K/mol)

As we know the depression in freezing point is a colligative property. It can be calculated by the formula:

∆ Tb = iKb m (where, ∆ Tb =change in freezing point ; Kb = molal depression constant ; m = molality, I = vantHoff factor)


For completely disassociatingCaCl2 the i=3)


2 = 3× 1.86 ( n × 1000 / 500)


2= 3 × 1.86 ( w × 1000/ M × 500)


w= = 19.89g


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