At 100 the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be
- 102
- 103
- 101
- 100
Answer
(3) 101
Solution
At 100 , vapour pressure of pure water p = 760 mm, ps = 732 mm.
Relative lowering of vapour pressure:
p - psps = n2n1
n2n1 = 760 - 732732 = 28732 = 0.0383
Molality (moles of solute per 1000 g water, n1 = 100018):
m = 0.0383 100018 = 2.13 mol kg-1
Tb = Kb m = 0.52 2.13 = 1.1 K
∴ Tb = 100 + 1.1 = 101.1 101
Answer: (3)