Heat of neutralisation of `NaOH` and `HCl` is `-57.46 kJ //` equivalent. The heat of ionisation of water in `kJ //mol` is `:`A. `-57.3 kJ mol^(-1)`B. `-114.6 kJ mol^(-1)`C. `+57.3 kJ mol^(-1)`D. `+114.6 kJ mol^(-1)`

Heat of neutralisation of `NaOH` and `HCl` is `-57.46 kJ //` equivalen

1.	Heat of neutralisation of `NaOH` and `HCl` is `-57.46 kJ //` equivalent. The heat of ionisation of water in `kJ //mol` is `:`A. `-57.3 kJ mol^(-1)`B. `-114.6 kJ mol^(-1)`C. `+57.3 kJ mol^(-1)`D. `+114.6 kJ mol^(-1)`
Answer» Correct Answer - C `NaOH+HCltoNaCl+H_(2)O` or `H^(+)+OH^(-)toH_(2)O` , `DeltaH=-57.3 kJ` ` :. H_(2)O to H^(+)+OH^(-)` , `DeltaH=+57.3 kJ` This is according to Laplace law.

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Heat of neutralisation of `NaOH` and `HCl` is `-57.46 kJ //` equivalent. The heat of ionisation of water in `kJ //mol` is `:`A. `-57.3 kJ mol^(-1)`B. `-114.6 kJ mol^(-1)`C. `+57.3 kJ mol^(-1)`D. `+114.6 kJ mol^(-1)`

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