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02/10/2020 3:34 pm
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A solution of urea (mol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at
(a) 0.654°C
(b) -0.654°C
(c) 6.54°C
(d) -6.54°C
1 Answer
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02/10/2020 3:45 pm
(b) -0.654°C
Explanation:
As ΔTf = Kf.m
ΔTb = Kb.m
Hence, we have m = ΔTf /Kf
or ΔTf = ΔTb Kf/Kb
=> [ΔTb = 100.18 - 100 = 0.18°C]
As the Freezing Point of pure water is 0°C,
ΔTf = 0 - Tf
0.654 = 0 - Tf
∴ Tf = - 0.654
Thus the freezing point of solution will be -0.654°C.