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What is the molarity of `H_(2)SO_(4)` solution if 25 ml of exactly neutralised with 32.63 ml of 0.164M, NaoHA. 0.107 MB. 0.126 MC. 0.214 MD. `-0.428M` |
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Answer» Correct Answer - A 0.164 M NaOH`-=`0.164 N NaOH We know, `N_(1)V_(1)=N_(2)V_(2),N_(1)xx25=0.164xx32.63` 0.214N `H_(2)SO_(4)-=(0.214)/(2)"M "H_(2)SO_(4)` (`because"basicity of "H_(2)SO_(4)` is 2) `-=0.107" M "H_(2)SO_(4)` |
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