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 amount of K_(2)Cr_(2)O_(7) to be dissolved to prepare V mL of K_(2)Cr_(2)O_(7), which is just sufficient to completely oxidised 10 mL of above FeSO_(4) solution ?

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 amount of K_(2)Cr_(2)O_(7) to be dissolved to prepare V mL of K_(2)Cr_(2)O_(7), which is just sufficient to completely oxidised 10 mL of above FeSO_(4) solution ?
Answer» `0.0875 G` `0.875 g` `8.75 g` `0.0087g` Solution :Meq. Of `Kr_(2)Cr_(2)O_(7)` needed for `FeSO_(4)` =meq of `Fe` in `10 mL` `w/49xx1000/VxxV=1/56xx1000x10/100` `:. W=0.0875 g`