The molarity $(M)$ of $K M n O_{4}$ in aqueous medium is :

0.8 M

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0.08 M

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0.26 M

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0.026 M

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Solution

The correct option is D $0.26 M$
$M n O_{4}^{⊝} (a q u e o u s) \equiv S_{2} O_{3}^{2 -} (S_{2} O_{3}^{2 -} \to 2 H S O_{4}^{⊝} + 8 e^{-}) (n = 8)$
$n$ factor $= 3$ $n$ factor $= 8$
In aqueous medium, Mn(VII) is reduced to Mn(IV), n=3
$100 m L \times N = 100 m L \times 0.1 \times 8$
$N = 0.8$
$M = \frac{0.8}{3} = 0.26$

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