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Equivalent Mass

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Question

Calculate the equivalent mass of $K M n O_{4}$ in the below reaction.
$M n O_{4}^{-} (+ 7) \to M n O_{4}^{- 2} (+ 6)$ (alkaline medium)

A

158

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B

316

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C

79

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D

157

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Solution

The correct option is A 158
$M n O_{4}^{-} (+ 7) \to M n O_{4}^{- 2} (+ 6)$
$E q u i v a l e n t w e i g h t o f K M n O_{4} = \frac{M o l e c u l a r w e i g h t}{N u m b e r o f e l e c t r o n s i n v o l v e d i n t h e r e a c t i o n}$
Molecular weight of $K M n O_{4}$ = 158
In the given reaction, only one electron is gained, so
$E q u i v a l e n t w e i g h t = \frac{M o l e c u l a r w e i g h t o f K M n O_{4}}{1} = 158$
Note: It is important to note that $K M n O_{4}$ acts as an oxidant in every medium although with different strength which follows the order,

acidic medium > neutral medium > alkaline medium

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Similar questions

Q. Calculate the equivalent mass of

K M n O_{4}

in the below reaction.

M n O_{4}^{-} (+ 7) \to M n O_{4}^{- 2} (+ 6)

(alkaline medium)

Q. Find the equivalent weight of

K M n O_{4}

in the given reaction, if the molar mass of

M n O_{4}^{-}

M n O_{4}^{-} \to M n^{4 +}

(neutral medium)

Q. Find the equivalent weight of

K M n O_{4}

in the given reaction, if the molar mass of

M n O_{4}^{-}

K M n O_{4} \to M n^{2 +}

(acidic medium)

M n O_{4}^{-}

ions can be reduced in strong alkaline medium to give

M n O_{4}^{2 -}

M n O_{2}

M n O_{4}^{-}

in acidic medium. Calculate equivalent weight of

M n O_{4}^{2 -}

.

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Equivalent Mass

Standard XIII Chemistry

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