Enthalpy of fusion of a liquid is $1.435 k c a l m o l^{- 1}$ and molar entropy change is $5.26 c a l m o l^{- 1} K^{- 1}$ , hence melting point of liquid is

100^{\circ} C

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0^{\circ} C

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373 K

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- 273^{\circ} C

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Solution

The correct option is B $0^{\circ} C$
Given:
$△ H_{f u s i o n} = 1.435 k c a l m o l^{- 1}$
$= 1435 c a l m o l^{- 1}$
$△ S = 5.26 c a l m o l^{- 1} k^{- 1}$
we know:
$△ S_{f u s i o n} = \frac{△ H_{f u s i o n}}{T} T = \frac{△ H_{f u s}}{△ S_{f u s}} = \frac{1435}{5.26} = 272.81 K = (272.81 - 273)^{o} C \approx 0^{\circ} C$

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