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Hydrogen Peroxide

To a 30 mL H ...

Question

To a $30 m L H_{2} O_{2}$ solution, excess of acidified solution of $K I$ was added. The iodine liberated required $20 m L$ of $0.3 N N a_{2} S_{2} O_{3} .$ Calculate the volume stregnth of $H_{2} O_{2}$ solution.

0.374 V

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2.544 V

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4.244 V

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1.12 V

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Solution

The correct option is D $1.12 V$
Miliequivalent of $25 m L H_{2} O_{2}$ = Miliequivalent of $I_{2}$ = Miliequivalent of $N a_{2} S_{2} O_{3}$ = $20 \times 0.3 = 6.$
Normality of $H_{2} O_{2} = \frac{6}{30} e q / L$
Let, the volume stregnth of $H_{2} O_{2}$ be $^{'} x V^{'} .$ Thus, $1 L$ of $H_{2} O_{2}$ produces $x L$ of $O_{2}$ at NTP.
Equivalent of $H_{2} O_{2} / L$ = Equivalent of $O_{2} = \frac{V (N T P)}{E q u i v a l e n t v o l . o f O_{2}} = \frac{x}{5.6}$

$\frac{x}{5.6} = \frac{6}{30} x = 1.12$
The volume stregnth of $H_{2} O_{2}$ is $1.12 V$

Alternate solution:
$Volume strength = Normality \times 5.6 Volume strength = \frac{6}{30} \times 5.6 = 1.12 V$

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