For a liquid, enthalpy of fusion is $1.435 k c a l {m o l}^{- 1}$ and molar entropy change is $5.26 c a l {m o l}^{- 1} K^{- 1}$ . The freezing point of liquid in degree celcius will be:

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Solution

The correct option is A 0
$△ S_{f u s i o n} =$ $^{△} H$ $\frac{f u s i o n}{T_{f}}$
$5.26 =$ $1.435 * 10^{3} T_{f}$
$T_{f} =$ $0.00366 ≃ O^{o}$

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