An aqueous solution freezes at -2.55°C. What is its boiling point KbH2O = 0. 52 K m-1 ; Kf (H2O)= 1.86 K m-1

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107.0° C

100.6° C

100.1° C

100.7° C

answer is D.

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Detailed Solution

∆Tf=2.55°C=Kf.m ∴m=2.55Kf ∆Tb=Kbm⇒∆Tb = 2.55×KbKf=2.55×0.521.86°C=0.7°C ⇒Tb=100+0.7=100.7°C

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