A solution of sucrose (molar mass =342) is prepared by dissolving 68.4 g in 1000 g of water. Given that the vapour pressure of water at 293 K is 0.023 atm, Kf for water =1.86Km−1 and Kb for water =0.52Km−1. What is the boiling point of solution?
A
272
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B
373.104
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C
400
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D
500
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Solution
The correct option is B373.104 The boiling point of solution can be calculated by using the relation,
ΔTb=Kb×1000×WBWA×MwB
Here Kb=0.52Km−1
∴ΔTb=0.52×1000×68.41000×342=0.104oC
∴ Boiling point of solution = Boiling point of pure water +ΔTb