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In an oxidation-reduction, `MnO_(4)^(-)` ion is converted to `Mn^(2+)`, what is the number of equivalents of `KMnO_(4)` (mol. Wt.=158) present in 250 mL of 0.04 M `KMnO_(4)` solution ?A. 0.02B. 0.05C. 0.04D. 0.07 |
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Answer» Correct Answer - B In redox-reaction : `8H^(+)+MnO_(4)^(-)+5e^(-) to Mn^(2+)+4H_(2)O` Change in oxidation state of `MnO_(4)^(-) =(+7)-(+2)=+5` `therefore N_(KMnO_(4))=M_(KMnO_(4))xx5` `=0.04xx5=0.20` Number of equivalents `=(NV)/(1000)=(0.2xx250)/(1000)=0.05` |
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