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Q.

The amount of energy released by the neutralization of $10 ml of 10 M$ aqueous $KOH$ with excess diluted sulphuric acid is $Q$ . Then the enthalpy of neutralization is

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a

$- Q$

b

Q

c

$10 Q$

d

$- 10 Q$

answer is C.

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Detailed Solution

$10 ml of 10 M of enthalpy = Q$

$mole of KOH = \frac{MV}{100} = \frac{10 \times 10}{1000} = 0.1$

Neutralization enthalpy $=$

$1 mol KOH = 10 P$

Hence the correct answer is (C) $10 Q$ .

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The amount of energy released by the neutralization of 10ml of 10M aqueous KOH with excess diluted sulphuric acid is Q. Then the enthalpy of neutralization is