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A solution of 122g of benzoic acid in 1000g of benzene shows a boiling point elevation of 1.4degree celsius. Assuming that the solute is dimerised to the extent of 80%, calculate the normal boiling point of benzene. Given molar enthalpy of vaporisation of benzene = 7.8Kcal/mol

A solution of 122g of benzoic acid in 1000g of benzene shows a boiling point elevation of 1.4degree celsius. Assuming that the solute is dimerised to the extent of 80%, calculate the normal boiling point of benzene. Given molar enthalpy of vaporisation of benzene = 7.8Kcal/mol

Grade:12

2 Answers

Ravleen Kaur
askIITians Faculty 1452 Points
4 years ago
Please go through revision notes and you will get your answer.

https://www.askiitians.com/iit-jee-solutions-colligative-properties/
Deepanshu
15 Points
2 years ago
WB=122g,MB=122g,WB=1000g
∆tb=1.4°C=1.4K,∆Hvap=7.8×103cal/mol
Hence,solution is dimerized,therefor,n=2
\alpha×100=80
\alpha=0.8
Hence solution is of hydrocarbon
It is a case of association
i=1-\alpha+\alpha/n=0.6​
∆tb=iKbm
Kb=7/3
•.•Kb=Rto2×MA/1000×∆Hvap
to=418.33K ans
 

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