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Equivalent Mass

Find the equi...

Question

Find the equivalent weight of $K M n O_{4}$ in the given reaction, if the molar mass of $M n O_{4}^{-}$ is 'M' ?
$K M n O_{4} \to M n^{2 +}$ (acidic medium)

41.6

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21.6

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11.6

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31.6

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Solution

The correct option is D $31.6$
In acidic medium
$(+ 7) K M n O_{4} \to (+ 2) M n^{2 +}$
$n_{f} = | 2 - 7 | \times 1 = 5$
Here 5 electrons are taken so
$equivalent weight = \frac{M o l . w t . o f K M n O_{4}}{5} = \frac{158}{5} = 31.6$

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Similar questions

K M n O_{4}

M n O_{4}^{-}

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

K M n O_{4}

M n O_{4}^{-}

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

K M n O_{4}

M n O_{4}^{-}

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

K M n O_{4} + 8 H^{+} + 5 e^{-} \to K^{+} + M n^{2 +} + 4 H_{2} O

K M n O_{4}

K M n O_{4} = 158

K M n O_{4}

31.6

M n O_{4}^{-}

M n^{2 +}

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