A) 20 cal \[mo{{l}^{-1}}{{K}^{-1}}\]
B) 26 cal \[mo{{l}^{-1}}{{K}^{-1}}\]
C) 24 cal \[mo{{l}^{-1}}{{K}^{-1}}\]
D) 28 cal \[mo{{l}^{-1}}{{K}^{-1}}\]
Correct Answer: B
Solution :
Key Idea: Use the following formula to solve the problem \[\Delta S=\frac{\Delta H}{T}\] where \[\Delta S=\] entropy = ? \[\Delta H=\] enthalpy \[=9710\,cal\,mo{{l}^{-1}}\] T = temperature \[={{100}^{o}}C=100+273=373\,K\] \[\therefore \] \[\Delta S=\frac{9710}{373}\] \[=26.032\,cal\,mo{{l}^{-1}}{{K}^{-1}}\] \[\approx 26\,cal\,mo{{l}^{-1}}{{K}^{-1}}\]
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