10 mL of $H_{2} O_{2}$ solution on treatment with KI released $I_{2}$ which required 10 mL of 1N hypo. Thus the concentration of $H_{2} O_{2}$ is ______.

1 N

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5.6 volume

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17 g

L^{- 1}

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All of these

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Solution

The correct option is D All of these
$H_{2} O_{2} + 2 I^{-} \to I_{2}$

$I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{-}$

$H_{2} O_{2} \equiv I_{2} \equiv 2 S_{2} O_{3}^{2 -}$

$N_{1} V_{1} = N_{2} V_{2}$

$H_{2} O_{2}$ hypo

$N_{1} (H_{2} O_{2}) = \frac{10 \times 1}{10} = 1 N$

Conc. $= N \times E = 17 g / L i t r e$

Volume strength = 5.6 $\times$ Normality

$= 5.6 \times 1$

$= 5.6$ volume strength

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