`A^(n+1)` is maximum oxidised by acidified `KMnO_(4)` solution into `AO_(3)^(-)`. If `2.68` m moles of `A^(+(n+1))` requires `32.16 mL` of a `0.05M` acidified `KMnO_(4)` solution for complete oxidation,value of n is

Answer

1

Explanation

Change in oxidation no. of `A = 4n`
Change in oxidation no. of `Mn = 5`
`(4n)x 2.68 = 5 xx 32.16 xx 0.05 :. N = 1`