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Question

The entropy of vaporization of benzene is 85JK1 mol1. When 117 benzene vaporizes at its normal boiling point, the entropy change of surrounding is:

A
85JK1
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B
85×1.5JK1
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C
85×1.5JK1
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D
None of these
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Solution

The correct option is B 85×1.5JK1
To calculate the moles, we use the formula,
Moles=GivenmassMolarmass
Given mass of Benzene=117g
Molar mass of Benzene=78.11,g/mol
We get moles=117g78.11g/mol=1.491.5mole
For 1 mole of Benzene, the entropy of vaporization is 85J/Kmol
So, for 1.49 moles of Benzene, the entropy of vaporization will be =1.5×85=127.5J/K
Now at, boiling point Benzene is in equilibrium with its vapour. Therefore, the two phases are in equilibrium and for such processes
Ssystem+Ssurrounding=0
Ssurrounding=Ssystem=127.5J/K

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