`25ml` of `0.017H_(2)SO_(4)` in strongly acidic medium required `16.9mL` of `0.01M KMnO_(4)` and in neutral medium required `28.6mL` of `0.01MKMnO_(4)` for complete conversion fo `SO_(3)^(2-)` to `SO_(4)^(2-)` .Assign the oxidation no of `Mn` in the product formed in each case.

Explanation

`SO_(3)^(2-)rarrSO_(4)^(2-)`
`S^(4+)rarrS^(6+)+2e`
`therefore ` Valence factor of `SO_(3)=2`
In acid medium: Meq.of `SO_(3)^(2-)="Meq.of" MnO_(4)^(-)`
`25xx0.017xx2=16.9xx0.01xxn_(1)`
`therefore n_(1)=5`
`therefore Mn^(7+)+5erarrMn^(4+)`
In neutral medium: Meq.of `SO_(3)^(2-)=` Meq.of `MnO_(4)^(-)`
`25xx0.017xx2=28.6xx0.01xxn_(2)`
`therefore n_(2)=3` ltbr. `therefore Mn^(7+)+3e rarrMn^(4+)`