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`H_(2)SO_(4)+2NaOH rarr Na_(2)SO_(4) +2H_(2)O` In this equation the equivalent weight of sulphuric acid is..........

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`H_(2)SO_(4)+2NaOH rarr Na_(2)SO_(4) +2H_(2)O` In this equation the equivalent weight of sulphuric acid is..........
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Correct Answer - `98//2 = 49`
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Here change in oxidation no. is +2

So n factor =+2

Now equivalent weight= molecular weight/ n factor

EW= 98/2

EW = 49
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