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$1.0 g$ Sample containing $K O_{2}$ and some inert impurity is dissolved in excess of aqueous $H I$ solution and finally diluted to $100 m L .$ The solution is filtered off and $20 m L$ of filtrate required $15 m L$ $0.4 M N a_{2} S_{2} O_{3}$ solution to reduce the liberated iodine. The mass % of $K O_{2}$ in the original sample would be

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Detailed Solution

The balanced redox reaction is :

$2 K O_{2} + 6 H l \to 2 K O H + 2 H_{2} O + 3 l_{2}$

meq of hypo $= 15 \times 0.4 = 6 =$ meq of $l_{2}$

$\Rightarrow$ Total $I_{2}$ liberated $= 15 m m o l$

$\Rightarrow$ m mol of $K O_{2}$ in original sample $= 10$

$\Rightarrow$ m% $= 10 \times 10^{- 3} \times 71 \times 100 = 71 %$

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