A) 881.1kcal
B) 771.4 kcal
C) 981.1 kcal
D) 871.2 kcal
Correct Answer: B
Solution :
Heat capacity, \[{{C}_{V}}=18.94\times 0.998\times {{10}^{3}}\] \[=18.902\,\times {{10}^{3}}\,cal/\deg \] Moles of benzoic acid \[=\frac{1.89}{1.22}\] \[{{C}_{p}}={{C}_{v}}+R\] \[=18.904\times {{10}^{3}}+2\] \[=18.904\times {{10}^{3}}\,\text{cal}\] Heat of combustion \[=\frac{{{C}_{p}}\times \Delta t}{moles}\] \[=\frac{18.904\times {{10}^{3}}\times 0.632\times 122}{1.89}\] \[=771.2\times {{10}^{3}}\,cal\] \[=771.2\,\text{kcal}\]You need to login to perform this action.
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