$100 mL$ of $H_{2} O_{2}$ solution having volume strength $11.35 V$ is mixed with $500 mL$ of $0.5 M$ KI solution to liberate $I_{2}$ gas such that equilibria gets established. All the $I_{2}$ gas liberated is dissolved to form $500 mL$ . solution. $200 ml$ of the solution required $50 mL$ of $\frac{2}{3} M$ hypo solution. Calculate volume strength of remaining $H_{2} O_{2}$ mixture. Round off your answer.

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Detailed Solution

The reaction is given as,

$\begin{array}{l} H_{2} O_{2} + KI ⟶ I_{2} \\ n f = 2 n f = 1 n f = 2 \end{array}$

$\begin{array}{l} I_{2} + {Na}_{2} S_{2} O_{3} ⟶ {Na}_{2} S_{4} O_{6} \\ n f = 2 n f = 2 \end{array}$

$2 \times {mmI}_{2} = 1 \times mm {Na}_{2} S_{2} O_{3}$

${mmI}_{2} = 1 \times \frac{50}{2} \times \frac{2}{3} = \frac{50}{3} mm$

_In $500 ml$ _{of solution} $mm$ _of $I_{2} = \frac{50}{3} \times \frac{5}{2} = \frac{125}{3} mm$

$mm of H_{2} O_{2} \times 2 = 2 \times {mmI}_{2}$

$mm of H_{2} O_{2} = \frac{125}{3} reacted$

$mm of H_{2} O_{2} left = 100 - \frac{125}{3} = \frac{175}{3} mm$

$\begin{array}{l} M = \frac{175}{3 \times 600} = 0.097 \\ V S = 0.097 \times 11.35 = 1.1 \end{array}$

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