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When 100 ml of an aqueous solution of $H_{2} O_{2}$ is titrate with an excess of $K I$ solution in dilute $H_{2} O_{2} .$ The liberated $I_{2}$ required $50 m l$ of $0.1 M N a_{2} S_{2} O_{3}$ for complete reaction. The volume strength of $H_{2} O_{2}$ solution is _____V

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Detailed Solution

$m . e q o f H_{2} O_{2} = m . e q o f N a_{2} S_{2} O_{3}$

$\begin{array}{l} M_{1} \times n f \times V_{1} = M_{2} \times n f \times V_{\begin{array}{l} 2 \end{array}} \\ M_{1} \times 2 \times 100 = 0.1 \times 1 \times 50 \end{array}$

$M_{1} = \frac{0.1}{4} = \frac{1}{40}$

$V . S = 11.2 \times M o l a r i t y = 11.2 \times \frac{1}{40} = 0.28 V$

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