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# Kf for water is 1.86∘C kg mol−1. What is the molality of a solution which freezes at −0.192∘C? Assuming no change in the solute by raising the temperature, at what temperature will the solution boil? Kb for H2O=0.515∘C kg mol−1)

A
102.053oC
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B
100.053oC
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C
100.15oC
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D
101.153oC
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Solution

## The correct option is B 100.053oCFreezing point depression of a solution is given by, △Tf=Kf×m where, △Tf is the depression in freezing point. Kf is molal depression constant. m is molality of solution. Freezing point of the pure water is 0∘ Thus, the depression in freezing point of water is 0.192∘C Molality =△TfKf =0.1921.86=0.103 m Boiling point elevation of a solution is given by, △Tb=Kb×m where, △Tb is the elevation in boiling point. Kb is molal elevation constant. m is molality of solution. △Tb=0.515×0.103=0.053oC Assuming the boiling point of pure water to be 100∘C, the boiling point of the solution will be 100.053oC.  Suggest Corrections  1      Similar questions  Explore more