An aqueous solution freezes at -0.186°C (Kf = 1.86°; Kb= 0.512°).What is the elevation in boiling point?

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0.186

0.512

0.5121.86

0.0512

answer is D.

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Detailed Solution

∆Tb=Kb×Molality ∆Tf=Kf×Molality ∴∆Tb∆Tf=KbKf⇒∆Tb=∆Tf×KbKf=0.186×0.5121.86=0.0512

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