Equivalent mass of $K M n O_{4}$ in acidic, basic and neutral are in the ratio of:

3 : 5 : 15

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5 : 3 : 1

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5 : 1 : 3

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3 : 15 : 5

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Solution

The correct option is D $3 : 15 : 5$
In acidic medium: ${M n O}_{4}^{-} + {8 H}^{+} + 5 e^{-} \to {M n}^{+ 2} + 4 H_{2} O$
In basic medium: ${M n O}_{4}^{-} + e^{-} \to {M n O}_{4}^{2 -}$
In neutral medium: ${M n O}_{4}^{-} + {4 H}^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O$
Equivalent mass is molecular mass divided by number of electrons gained.

In acidic medium, equivalent weight $=$ $\frac{M o l e c u l a r w e i g h t}{5}$

In basic medium, equivalent weight $=$ $\frac{M o l e c u l a r w e i g h t}{1}$

In neutal medium, equivalent weight $=$ $\frac{M o l e c u l a r w e i g h t}{3}$

Ration of equivalent mass $=$ $\frac{M o l e c u l a r w e i g h t}{5}$ : $\frac{M o l e c u l a r w e i g h t}{1}$ : $\frac{M o l e c u l a r w e i g h t}{3}$

Ratio of equivalent mass $= 3 : 15 : 5$

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