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Question

The entropy change involved in the conversion of 1 mole of liquid water at 373 K to vapour will be: 
Given:  Hvap=2.257 kJ/g


A
118.5 JK1mol1
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B
108.9 JK1mol1
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C
150 JK1mol1
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D
130.6 JK1mol1
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Solution

The correct option is D 108.9 JK1mol1
The enthalpy change of vaporisation for 1 mole of water can be calculated as follows:
Hvap= 2.257 kJ/g×18 g/mol=40.6 kJ mol1
so,
Svap=HvapT=40.6×103 J mol1373 K=108.9 J K1mol1

Chemistry

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