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`150 mL` of `0.5N HCl` solution at `25^(@)C` was mixed with `150 mL` of `0.5 N NaOH` solution at same temperature. Calculate the heat of neutralizatio

Monjur Alahi

`150 mL` of `0.5N HCl` solution at `25^(@)C` was mixed with `150 mL` of `0.5 N NaOH` solution at same temperature. Calculate the heat of neutralization of `HCl` with `NaOH`, if find temperature was recorded to be `29^(@)C`.
`(rho_(H_(2)O)=1g//mL)`


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Salim Ahmed Kawsar

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Correct Answer - `-16 kcal`.

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Correct Answer - c `HCl+NaOHto NaCl+H_(2)O` at t=0, number of moles = `(500xx0.1)/1000 = (200xx0.2)/1000 ` =0.05 =0.04 during neutralisation of 1 mole of NaOH by 1 mole of HCl, heat...

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2 Answers 1 views
Assertion(A) : Heat of neutralisation of `HCl` and `NaOH` is same as that of `H_(2)SO_(4)` with `NaOH`. Reason(R ) : `HCl`, `H_(2)SO_(4)` and `NaOH` a

Correct Answer - A Heat of neutralisation of strong acid with strong base is contant `(-137 Kcal//mol^(-1))`

2 Answers 1 views
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2 Answers 7 views
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2 Answers 1 views
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2 Answers 1 views
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2 Answers 1 views

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