In an oxidation-reduction reaction, ${M n O}_{4}^{-}$ is converted to ${M n}^{+ 2}$ . What is the number of equivalents of ${K M n O}_{4}$ (mol. wt = 158) present in 250 ml of 0.04 M ${K M n O}_{4}$ solution?

0.02

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0.05

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0.04

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0.07

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Solution

The correct option is B $0.05$
$M n O_{4}^{-}$
$x - 8 = - 1 \Rightarrow x = + 7 \Rightarrow M n^{+ 7}$ is converted to $M n^{+ 2}$
So, valency factor, $n = 5$
$N u m b e r o f e q u i v a l e n t s = n \times M \times V$

$= 5 \times 0.04 \times 0.250$

$= 0.05$

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Similar questions

Q. In an oxidation reduction reaction,

M n O_{4}^{-}

ion is converted to

M n^{2 +}

. What is the number of equivalents of

K M n O_{4}

(mol. wt.

= 158

) present in

250 m L

0.04 N

K M n O_{4}

solution?

What is the normality of a $K M n O_{4}$ solution in which $15.8 g$ of $K M n O_{4}$ is present in $250 m L$ solution and the solution is used as an oxidant, where ${M n O_{4}}^{-}$ is reduced to ${M n}^{2 +}$ ? (Given: molar mass of $K M n O_{4} = 158 g m o l^{- 1}$ )