A solution of urea (mol. mass 60 g mol-1) boils at 100.18 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
- 0.654
- -0.654
- 6.54
- -6.54
Answer
(2) -0.654
Solution
Tb = 100.18 - 100 = 0.18 K
Tb = Kb m and Tf = Kf m (same molality m)
Tf Tb = KfKb
Tf = 0.18 1.860.512
= 0.654 K
Freezing point = 0 - 0.654 = -0.654
(The molar mass of urea is not needed.)
Answer: (2)