W g of benzoic acid is completely burnt in bomb calorimeter in excess of $O_{2}$ so that Q KJ heat is released at 298K. Internal energy change for combustion of 1 mole benzoic acid (molecular weight = M) will be (in KJ)

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-Q

$\frac{- (QM)}{W}$

$\frac{- (QW)}{M}$

answer is C.

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Detailed Solution

When W g of of benzoic acid is completely burnt in bomb calorimeter in excess of $O_{2}$ then the amount of heat released is Q KJ.

Now this heat will be in negative because the heat is released by the system, i.e., -Q KJ.

1 mole of benzoic acid will be equal Molecular weight (M).

So, the internal energy will be:

$\frac{(- QM)}{W}$

Therefore, the correct option is C.

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