(C) Molar mass of benzoic acid $(C_6H_5COOH) = 122 \ g \ mol^{-1}$.Heat gained by water $(Q) = m \times s \times \Delta T$.$Q = 18.94 \times 10^3 \ g

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(C) Molar mass of benzoic acid $(C_6H_5COOH) = 122 \ g \ mol^{-1}$.Heat gained by water $(Q) = m \times s \times \Delta T$.$Q = 18.94 \times 10^3 \ g \times 0.998 \ cal \ g^{-1} \ ^{\circ}C^{-1} \times 0.632 \ ^{\circ}C$.$Q = 11946.3 \ cal = 11.946 \ kcal = 49.98 \ kJ$.Heat liberated by $1.89 \ g$ of benzoic acid $= 49.98 \ kJ$.Heat of combustion for $1 \ mol$ $(122 \ g)$ $= \frac{49.98 \ kJ}{1.89 \ g} \times 122 \ g \ mol^{-1} \approx 3226 \ kJ \ mol^{-1}$.The closest option is $3240 \ kJ \ mol^{-1}$.

Class 11 Chemistry Thermodynamics Mix Examples-Thermodynamics and Thermochemistry

Medium

The heat liberated when $1.89 \ g$ of benzoic acid is burnt in a bomb calorimeter at $25^{\circ} C$ increases the temperature of $18.94 \ kg$ of water by $0.632^{\circ} C$. If the specific heat of water at $25^{\circ} C$ is $0.998 \ cal / (g^{\circ} C)$,then find the heat of combustion of benzoic acid.

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A
$2540 \ kJ \ mol^{-1}$
B
$1975 \ kJ \ mol^{-1}$
C
$3240 \ kJ \ mol^{-1}$
D
$2825 \ kJ \ mol^{-1}$

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