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Enthalpy of Neutralisation

10 mL of H 2 ...

Question

10 mL of $H_{2} O_{2}$ solution on treatment with KI and titration of liberated $I_{2}$ , required 10 mL of 1N hypo. Thus $H_{2} O_{2}$ is -

1 N

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5.6 v o l u m e

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17 g L^{- 1}

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A l l a r e c o r r e c t

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Solution

The correct option is D $A l l a r e c o r r e c t$
$H_{2} O_{2} + 2 I^{-} \to I_{2}$
$I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{-} H_{2} O_{2} \equiv I_{2} \equiv 2 S_{2} O_{3}^{2 -}$
$N_{1} V_{1} = N_{2} V_{2} H_{2} O_{2} h y p o$
$N_{1} (H_{2} O_{2}) = \frac{10 \times 1}{10} = 1 N$
$C o n c . = N \times E = 17 g / l i t r e$
Volume strength $= 5.6 \times n o r m a l i t y$
$= 5.6 \times 1 = 5.6 v o l u m e$

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