`1g` pure iron is dissolved in excess of `H_(2)SO_(4)`. The clear filtrate is made up `100 mL. 10mL` of this solution is treated with `0.1 M KMnO_(4)` solution till whole of the `Fe^(2+)` ions are oxidised to `Fe^(3+)` ions. Now `0.2 g Fe_(2)(SO_(4))_(3)` is dissolved in it. the solution is now treated with `Zn` and `H_(2)SO_(4)`.
The volume of `0.1 M KMnO_(4)` used after reducing the solution mixture with `Zn+H_(2)SO_(4)` is:

Options

A. `5.572 mL`
B. `3.572 mL`
C. `4.572 mL`
D. `6.572 mL`

Explanation

Meq. of `Fe^(2+)` [form `Fe`]+Meq. of `Fe^(2+)`
[from `Fe_(2)(SO_(4))_(3))`]=Meq of `KMnO_(4)`
`1/56xx1000xx10/100+0.2/(400/2)xx1000=0.1xx5xxV`
`:. V=([1.786+1])/0.5=5.572 mL`