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Question

If ΔHvap of pure water at 100oC is 40.627 kJmol1, then the value of ΔSvap is:

A
108.91 kJmol
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B
108.91 JK1 mol1
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C
606.27 JK1mol1
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D
808.27 JK1 mol1
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Solution

The correct option is C 108.91 JK1 mol1
For vapourisation of water at 100 C,

ΔH=TΔS
ΔS=ΔHvapT
ΔS=40.627373=0.1089kJK1mol1=108.91 JK1mol1

Option B is correct.

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