`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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For the oxidation of `A^(n+)` as : `A^(n+) rarr AO_(3)^(-)` n-factor `= 5-n` `rArr` Gram equivalent of `A^(n+) = 2.68 xx 10^(3)(5-n)` Now equating the above gram equivalent with gram equivalent of `KMnO_(4)` : `2.68 xx 10^(-3) (5-n) = 16.1 xx 10^(-3) xx 5` `rArr n = +2`
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