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Question

If Hvaporisation of substance X(l) (molar mass: 30 g/mol) is 300 J/g at it's boiling point 300 K, then molar entropy change for reversible condensation process is____________.

A
30 J/mol.K
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B
300 J/mol.K
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C
30 J/mol.K
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D
None of these
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Solution

The correct option is A 30 J/mol.K
Solution:- (A) 30J/molK
Given:-
ΔHvap.=300J/g
Molar mass of X=30g/mol
ΔHmolar=300×30=9000J/mol
Boiling point of substance X,(T)=300K
As we know that,
Molar entropy change, ΔSmolar=ΔHmolarT=9000300=30J/molK
Hence the molar entropy change for the given process will be 30J/molK.

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