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Latent heat in fusion of ice at 0^(@)C i...

Latent heat in fusion of ice at `0^(@)C` is 6025.24`J*mol^(-1)`. Calculate molar entropy of the process at `0^(@)C`.

Text Solution

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The change in entropy due to melting of 1 mol of ice
`=("molar latent (or enthalpy) of fusion of ice")/("Melting point of ice")`
`=(6025.24J*mol^(-1))/((273+0)K)=22.07J*K^(-1)*mol^(-1)`
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