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 ratio of equivalent of KMnO_(4) and K_(2)Cr_(2)O_(7) used for reducing the solution 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 ratio of equivalent of KMnO_(4) and K_(2)Cr_(2)O_(7) used for reducing the solution is:
Answer» `5//6` `6//5` `1` `2` SOLUTION :EQ. of `KMnO_(4)`=Eq. of `K_(2)Cr_(2)O_(7)`=Eq. of `FE^(2+)`