The heat liberated when 1.89 g of benzoic ac | vedclass

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Question

(A) $1$. Calculate the heat released $(q)$ by the combustion of $1.89 \ g$ of benzoic acid: $q = m 	imes c 	imes \Delta T$ $2$. $m = 18.94 \ kg = 18

Vedclass · Accepted Answer

$771.1$

Vedclass · Answer

(A) $1$. Calculate the heat released $(q)$ by the combustion of $1.89 \ g$ of benzoic acid: $q = m 	imes c 	imes \Delta T$ $2$. $m = 18.94 \ kg = 18940 \ g$,$c = 0.998 \ cal/g \cdot ^\circ C$,$\Delta T = 0.632 \ ^\circ C$ $3$. $q = 18940 \ g 	imes 0.998 \ cal/g \cdot ^\circ C 	imes 0.632 \ ^\circ C = 11942.5 \ cal = 11.9425 \ kcal$ $4$. Calculate the moles of benzoic acid: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{1.89 \ g}{122 \ g/mol} \approx 0.01549 \ mol$ $5$. Heat of combustion per mole = $\frac{q}{n} = \frac{11.9425 \ kcal}{0.01549 \ mol} \approx 771.1 \ kcal/mol$ $6$. Therefore,the correct option is $A$.

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