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When `50cm^(3)` of `0.2N H_(2)SO_(4)` is mixed with `50cm^(3)` of `1 N KOH`, the heat liberated isA. 11.46 kJB. 57.3 kJC. 573 kJD. 573 J |
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Answer» Correct Answer - D For complete neutralization of strong acid and strong base energy released is 57.32 KJ/mol No. of mole of `H_(2)SO_(4)=(0.2xx50)/(1000)=10^(2)` No. of mole of KOH = `(1)/(1000)xx50=5xx10^(-2)` `"So = "57.32xx10^(-2)=0.5732 kJ = 573.2" Joule"`. |
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