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`2.68xx10^(-3)` moles of solution containing anion `A^(n+)` require `1.61xx10^(-3)` moles of `MnO_(4)^(-)` for oxidation of `A^(n+)` to `AO_(3)^(-)` in acidic medium. What is the value of `n`? |
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Answer» Correct Answer - 2 Redox changes are: For `KMnO_(4):Mn^(7+)+5erarrMn^(2+)` For `A^(n+): A^(n+)rarrA^(5+)+(5-n)e` Now meq.of `A^(n+)="meq.of" KMnO_(4)` `( becausemeq.="mole"xx"valencyfactor"xx1000)` `2.68xx10^(-3)xx(5-n)xx1000=1.61xx10^(-3)xx5xx1000` `therefore n=1.99approx 2` |
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