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`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 K_(2)Cr_(2)O_(7)` used after reducing the solution mixture with `Zn+H_(2)SO_(4)` is :A. `4.64 mL`B. `5.46 mL`C. `3.46 mL`D. `2.64 mL` |
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Answer» Meq. Of `Kr_(2)Cr_(2)O_(7)=Meq. Of Fe^(2+)` [from `Fe+` from `Fe_(2)(SO_(4))_(3)`] `V=4.64 mL` |
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