Heat liberated with 100 ml of 1 N NaOH is neutralised by 300 ml of 1N HCl :

11.56 kJ

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5.73 kJ

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22.92 kJ

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17.19 kJ

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Solution

The correct option is A 5.73 kJ
The standard enthalpy change of neutralization for a strong acid and base is -57.3 kJ/mol.

Heat liberated with 100 ml of 1 N NaOH is neutralised by 300 ml of 1 N HCl is $\frac{57.3 \times 100 m l \times 1 N}{1000 m l / l} = 5.73 k J$

Hence the correct option is B.

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Similar questions

Heat liberated when 100 ml of 1 N NaOH is neutralized by 300 ml of 1N HCl is

When 100 ml of 1 N HCl is mixed with 100 ml of 1 N NaOH the heat evolved was found to be Q. The heat of neutralisation of HCl with NaOH is:

m o l^{- 1}

20 m L

0.5 M N a O H

100 m L

0.1 M H C l

x k J

k J m o l^{- 1}

100 m L

100 m L

4 M N a O H

H C l

= 1.2)

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