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Mole Concept

The molarity ...

Question

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

0.2 M

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0.02 M

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0.4 M

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0.04 M

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Solution

The correct option is B $0.02 M$
$M n O_{4}^{⊝} (a c i d i c) \equiv (2 S_{2} O_{3}^{2 -} \to S_{4} O_{3}^{2 -} + 2 e^{-})$
$^{'} n^{'}$ factor $= 5$ $^{'} n^{'}$ factor $= 1$ $(n = \frac{2}{2} = 1)$
As Mn(VII) is reduced to Mn(II), therefore, n=5
$100 m L \times N \equiv 100 m L \times 0.1 \times 1$ ( $n$ factor)
$N = 0.1$
$M = \frac{0.1}{n f a c t o r} = \frac{0.1}{5} = 0.02 M$

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