The equivalent weight of `KMnO_(4)` in (a) neutral medium, (b) acidic medium and (c ) alkaline medium is `M//..` ( where `M` is mol.wt. of `KMnO_(4)`)A. Molecular weightB. `("Molecular weight")/(2)`C. `("Molecular weight")/(3)`D. `("Molecular weight")/(5)`

The equivalent weight of `KMnO_(4)` in (a) neutral medium, (b) acidic

1.	The equivalent weight of `KMnO_(4)` in (a) neutral medium, (b) acidic medium and (c ) alkaline medium is `M//..` ( where `M` is mol.wt. of `KMnO_(4)`)A. Molecular weightB. `("Molecular weight")/(2)`C. `("Molecular weight")/(3)`D. `("Molecular weight")/(5)`
Answer» Correct Answer - C `KMnO_(4)` in neutral medium changes to `MnO_(2)` `underset((+7))(MnO_(4)^(-))to underset((+4))(MnO_(2))` Thus n-factor = change in O.N. = 3 `therefore "Eq. wt." = ("Mol.wt.")/(3)`

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The equivalent weight of `KMnO_(4)` in (a) neutral medium, (b) acidic medium and (c ) alkaline medium is `M//..` ( where `M` is mol.wt. of `KMnO_(4)`)A. Molecular weightB. `("Molecular weight")/(2)`C. `("Molecular weight")/(3)`D. `("Molecular weight")/(5)`

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