The heat is liberated when 1.89 ~g of benzoic acid is burnt in a...

Mass of benzoic acid $=1.89 \mathrm{~g}$ Temperature of bomb calorimeter $=25^{\circ} \mathrm{C}=298 \mathrm{~K}$ Mass of water $(m)=18.94 \mathrm{~kg}=18940 \mathrm{~g}$ [ncrease in temperature $(\Delta t)=0.632^{\circ} \mathrm{C}$ Specific heat of water $(s)=0.998 \mathrm{cal} / \mathrm{g}-\mathrm{deg}$ Heat gained by water or heat liberated by benzoic $\operatorname{acid}(Q)=m s \Delta t=18940 \times 0.998 \times 0.632=11946.14 \mathrm{cal}$ Since, $1.89 \mathrm{~g}$ of acid, liberates $11946.14 \mathrm{cal}$ of heat, therefore heat liberated by $122 \mathrm{~g}$ (mol. $$\begin{aligned} & \text{ wt. of benzoic acid) of acid }=\frac{11946.14 \times 122}{1.89} \\ & =771126.5 \mathrm{cal} \approx 771.1 \mathrm{kcal}\end{aligned}$$