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The boiling point of an aqueous solution is found to be 100.280C. Then the freezing point of the same solution is –x0C. What is the value of ‘x’?

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detailed solution

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Tb=0.28=Kb×molarityKb=0.52 for water0.280.52=molarity  molarity=0.5384Tf=1.86×molarityTf=1.86×0.5384Tf=1.00Tf=10C

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