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Question

# Calculate the molal elevation constant, Kb for water and the boiling of 0.1 molal urea solution. Latent heat of vaporisation of water 9.72 kcal mol1 at 373.15K.

A
Kb=0.512kgmolK1,Tb=373.20K
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B
Kb=05.12kgmolK1,Tb=378.20K
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C
Kb=1.02kgmolK1,Tb=383.20K
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D
Kb=0.512kgmolK1,Tb=385.70K
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Solution

## The correct option is A Kb=0.512kgmolK−1,Tb=373.20KKb=R×M1×T2b1000×ΔvapHM1=18g, Tb=373.15K ΔvapH=9.72kcal/mol=40629.6J/molKb=0.512kgmolK−1ΔTb=Kb×m=0.512×0.1=0.0512KHence, Tb=373.15+0.0512=373.20KHence, option A is correct

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