`2.68 xx 10^(-3)` moles of a solution containing an ion `A^(n+)` require `1.61 xx 10^(-1)` moles of `MnO_(4)^(-)` ion for the oxidation of `A^(n+)` to `AO_(3)^(-)` in acidic medium. What is the value of n?

Explanation

Since in acidic solution
`MnO_(4)^(-) + 8H^(+) + 5e^(-) rarr Mn^(2+) + 4H_(2)O`
`MnO_(4)^(-) "gains" 5 e^(-)`
So, its equivalent wt = molecular weight/5
It means `1.61 xx 10^(-3)` moles of `MnO_(4)^(-) = 5 xx 1.61 xx 10^(-3)` equivalents
`A^(n+) rarr AO^(3-)` (oxidation number of A in `AO_(3)^(-) = 5 `
Loss of electron = (5-n)
`2.68 xx 10^(-3)` moles of the solution containing `A^(+) "ions" = (5-n)xx2.68 xx 10^(-3)` equivalents
Equivalents of oxidised 'A' = Equivalents of reduced A
`(1.61 xx 10^(-3)) xx 5 = (5-n)xx2.68 xx 10^(-3)`
`(5-n) = 1.61 xx 10^(-3)xx5/(2.68 xx 10^(-3)) = 3 (5-n) = 3`
n = 5 - 3 = 2