Answer
$-14.30\frac{J}{K.mol}$
Work Step by Step
We know that
$\Delta S=\frac{\Delta H_{fus}}{T}$
We plug in the known values to obtain:
$\Delta S=\frac{-69\frac{J}{g}}{298K}\times \frac{60.05g}{1mol}=-14.30\frac{J}{K.mol}$
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