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Question

Calculate the enthalpy of vaporisation (in kJ/mol) for ethanol. Given, entropy change for the process is 109 J K1 mol1 and boiling point of ethanol is 78.5 oC.


A
50.29 kJ mol1
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B
12.58 kJ mol1
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C
63.52 kJ mol1
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D
38.31 kJ mol1
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Solution

The correct option is D 38.31 kJ mol1
We know that, 
Svapourisation=HvapourisationTb

Given:  Svapour=109 J K1 mol1
 Tb=78.5+273=351.5K

Substituting these values in above equation, we get 
109=Hvap351.5
Hvap=38313.5 J mol1
                 =38.31 kJ mol1

Chemistry

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