Calculate the amount of CaCl2 (molar mass =111gmol−1) which must be added to 500 g of water to lower its freezing point by 2K, assuming CaCl2 is completely dissociated. (Kf for water =1.86Kkgmol−1).
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Solution
CaCl2→Ca2++2Cl−
∴i=3
ΔTf=i×Kf×m
2=3×1.86×m
∴m=23×1.86=25.58=0.358
∴ Amount of CaCl2 to be added to
500 g of water =0.358×5001000×111