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Secondary Cells

What volume o...

Question

What volume of $0.2 M$ $K M n O_{4}$ is required to react with $1.58 g$ of hypo solution ( $N a_{2} S_{2} O_{3}$ ) in acidic medium?

A

20

mL

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B

10

mL

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C

16.6

mL

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D

50

mL

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Solution

The correct option is B $10$ mL
Equivalent weight of $N a_{2} S_{2} O_{3} = \frac{158}{n} (n = 1)$
$2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 e^{-} (n = \frac{2}{2} = 1)$
$M n O_{4}^{⊝} = S_{2} O_{3}^{2 -}$
mEq. of $M n O_{4}^{-}$ = mEq. of $S_{2} O_{3}^{2 -}$
$0.2 M \times 5 (n - f a c t o r) \times V = \frac{1.58}{158} \times 10^{3}$
Therefore, $V = 10$ mL.

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