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:

1g pure iron is dissolved in excess of H_(2)SO_(4). The clear filtrate

1.	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:
Answer» `5.572 mL` `3.572 mL` `4.572 mL` `6.572 mL` SOLUTION :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`