A) 10.52 cal/(mol K)
B) 21.04 cal/(mol K)
C) 5.260 cal/ (mol K)
D) 0.526 cal/(mol K)
Correct Answer: C
Solution :
Molar entropy change for the melting of ice, \[\Delta {{S}_{melt}}=\frac{\Delta {{H}_{fusion}}}{T}\] \[=\frac{1.435\,kcal\,/mol}{(0+273)K}\] \[=5.26\times {{10}^{-3}}kcal/mol\,K\] \[=5.26\,cal/mol\,K\]
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