`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 ?A. `0.0875 g`B. `0.875 g`C. `8.75 g`D. `0.0087g`

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

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 ?A. `0.0875 g`B. `0.875 g`C. `8.75 g`D. `0.0087g`
Answer» 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`

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