The volume of $10$ volume H $_{2}$ O $_{2}$ solution that decolourises $200$ ml of $2$ N KMnO $_{4}$ solution in acidic medium is:

112

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336

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220

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224

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Solution

The correct option is A $224$ ml
$2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$

10 volume $H_{2} O_{2}$ means $1$ L of $H_{2} O_{2}$ will liberate $10$ L of $O_{2}$ at STP

Since $1$ mole oxygen gas is liberated by $2$ moles $H_{2} O_{2}$

$\Rightarrow 22.4 L$ of oxygen gas is liberated by $2$ moles $H_{2} O_{2}$

$\Rightarrow 10 L$ of oxygen gas is liberated by $\frac{2}{22.4} \times 10 = 0.8928$ moles $H_{2} O_{2}$

Mass of $H_{2} O_{2} = 0.8928 \times 34 = 30.357 g$

$N = \frac{w e i g h t \times 1 L}{e q u i v a l e n t w e i g h t} = \frac{30.357 \times 1}{17} = 1.78 g / L$

Using the relation $N_{1} V_{1} = N_{2} V_{2}$ ,

$\Rightarrow 1.78 \times V_{1} = 2 \times 200$

$\Rightarrow V_{1} = 224 m l$

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