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Calculate the entropy change when 3.6g o...

Calculate the entropy change when `3.6g` of liquid water is completely converted into vapour at `100^(@)C` . The molar heat of vaporization is `40.85KJ mol^(-1)` .

A

`6.08JK^(-1)`

B

`109.5JK^(-1)`

C

`21.89JK^(-1)`

D

`-21.89JK^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`DeltaH_(vap)=40850Jmol^(-1),T_(b)=373K`
`DeltaS_("vap")=(DeltaH_("vap"))/(T_(b))=(40850Jmol^(-1))/(373K)=109.5K^(-1)mol^(-1)`
`DeltaS_(vap)` per gram `=(109.5JK^(-1)mol^(-1))/(18gmol^(-1))=6.-083JK^(-1)g^(-1)`
Entropy change for `3.6g` water `=6.083JK^(-1)g^(-1)xx3.6g`
`=21.89JK^(-1)`
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Knowledge Check

  • Entropy of vaporisation of water at 100^(@)C , if molar heat of vaporisation is 9710 cal mol^(-1) will be

    A
    `20 cal mol^(-1)K^(-1)`
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  • Entropy of vaporisation of water at 100^(@)C , if molar heat of vaporisation is 8710 cal mol^(-1) will be

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    `20 cal mol^(-1) K^(-1)`
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    `23.36 cal mol^(-1) K^(-1)`
    C
    `24 cal mol^(-1) K^(-1)`
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    `28.0 cal mol^(-1) K^(-1)`

  • Calculate the ebullioscopic constant for water. The heat of vaporization is 40.685 kJ mol^(-1)

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