At `25^(@)C` the heat of combustion at constant volume of 1 mol of a compound is 5150 kJ. The temperature of a bomb calorimeter rises from `25^(@)C` to `30.5^(@)C` when a certain amount of the compound is burnt in it. If the heat capacity of the calorimeter is 9.76 `kJ*K^(-1)` then how much of the compound was taken for combustion. [Molar mass of the substance=128]

Explanation

As given in the question, the gram-molecular weight of that compound`=128g*mol^(-1)`
Therefore, the amount of heat that evolves in the combustion of 128 g of the substance=5150 kJ
Again, `DeltaT=[(273+30.5)-(273+25)]K=5.5K and C_(cal)=9.76kJ*K^(-1)`.
Thus, the amount of heat evolves when a certain amount of compound is burnt=`C_(cal)xxDeltaT=9.76xx5.5=53.68kJ`.
Now, 5150 kJ of heat is liberated due to the combustion of 128 g compound hence, the amount of compound required for the evolution of 53.68 kJ of heat`=(128)/(5150)xx53.68=1.334g`.