Enthalpy of fusion of water is $6.01 k J {m o l}^{- 1}$ . The entropy change of $1$ mole of ice at its melting point will be:

22 k J {m o l}^{- 1}

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109 k J {m o l}^{- 1}

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44 k J {m o l}^{- 1}

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11 k J {m o l}^{- 1}

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Solution

The correct option is B $22 k J {m o l}^{- 1}$
$Δ S_{f u s i o n} = \frac{Δ H_{f u s i o n}}{T_{m p}} = \frac{6.01 \times 1000}{T_{m p}} = 22 k J {m o l}^{- 1}$

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