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)` .

Options

A. `6.08 JK^(-1)`
B. `109.5JK^(-1)`
C. `21.89JK^(-1)`
D. `-21.89JK^(-1)`

Answer

3

Explanation

`DeltaW_(vap)=40850J mol^(-1),T_(b)=373K`
`DeltaS_(vap)=(DeltaH_(vap))/(T_(b))=(40850J mol^(-1))/(373K)=109.5K^(-1)mol^(-1)`
`DeltaS_(vap)` per gram`=(505.5JK^(-1)mol^(-1))/(18gmol^(-1))=6.083JK^(-1)g^(-1)`
Entropy change for `3.6g` water `=6.083JK^(-1)g^(-1)xx3.5g=21.69JK^(-1)`