Question

`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`?

Answer 1

Correct Answer - 2
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`