`MnO_(4)^(2-)` (`1` mole) in neutral aqueous medium is disproportionate toA. `(2)/(3) " mol of "MnO_(4)^(ɵ) and (1)/(3) " mol of "MnO_(2)`B. `(1)/(3) " mol of "MnO_(4)^(ɵ) and (2)/(3) " mol of "MnO_(2)`C. `(1)/(3) " mol of "Mn_(2)O_(7) and (2)/(3) " mol of "MnO_(2)`D. `(2)/(3) " mol of "Mn_(2)O_(7) and (1)/(3) " mol of "MnO_(2)`

`MnO_(4)^(2-)` (`1` mole) in neutral aqueous medium is disproportionat

1.	`MnO_(4)^(2-)` (`1` mole) in neutral aqueous medium is disproportionate toA. `(2)/(3) " mol of "MnO_(4)^(ɵ) and (1)/(3) " mol of "MnO_(2)`B. `(1)/(3) " mol of "MnO_(4)^(ɵ) and (2)/(3) " mol of "MnO_(2)`C. `(1)/(3) " mol of "Mn_(2)O_(7) and (2)/(3) " mol of "MnO_(2)`D. `(2)/(3) " mol of "Mn_(2)O_(7) and (1)/(3) " mol of "MnO_(2)`
Answer» Correct Answer - A Remember it as fact: in nautral medium `K_(2)MnO_(4)` (potassium manganate) disproportionate as follows: `K_(2)MnO_(4)toKMnO_(4)+MnO_(2)` On balancing we get: `3MnO_(4)^(2-)+2H_(2)Oto2MnO_(4)^(ɵ)-=(1)/(3)molMnO_(2)`

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The purity of `H_2O_2` in a given sample is `85%.` Calculate the weight of impure sample of `H_2O_2` which requires `10mL` of `M//5 KMnO_4` solution in a titration in acidic mediumA. 2 gB. 0.2 gC. 0.17 gD. 0.15 g
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