A solution of urea (mol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf, and Kb, for water are 1.86 and 0.512 K kg kg mol-1 respectively, the above solution will freeze at 

  1. A

    0.654°C 

  2. B

    —0.654°C 

  3. C

    654°C 

  4. D

    —654°C 

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    Solution:

    ΔTf=Kf.m ΔTb=Kb.m

    if the value of m is same, then
    Hence,  ΔTfKf=ΔTbKb
    or ΔTf=ΔTbKfKb
    ΔTb=100.18100=0.18C
    Tf=0.18×1.860.512=0.65C
    As the Freezing Point of pure water is 0°C, 
    ΔTf=T1(temperature of pure solvent)T2 (when solute is added) 0.654°C=0T2 T2=0.654°C

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