A) 4.8 kcal
B) 7.2 kcal
C) 8.2 kcal
D) 2.4 kcal
Correct Answer: D
Solution :
[d] Number of moles \[=\frac{1}{35.5}\] Given, \[1eV=23.06\text{ }kcal\text{ }mo{{l}^{-1}}\] \[3.7\text{ }eV=3.7\times 23.06\text{ }kcal\text{ }mo{{l}^{-1}}\] i.e. 1 mole release energy \[=3.7\times 23.06\text{ }kcal\] \[\therefore \] Energy released \[=\frac{1}{35.5}\times 3.7\times 23.06\text{ }kcal=2.4\text{ }kcal\]You need to login to perform this action.
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