The volume of 10 volume $H_{2} O_{2}$ solution that decolorises 200 mL of 2 N $K M n O_{4}$ solution in an acidic medium is:

112 mL

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336 mL

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200 mL

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236 mL

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Solution

The correct option is D 236 mL
$2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$

10 volume is 1L of $H_{2} O_{2}$ which can liberate 10 L of $O_{2}$ at STP
1 mole oxygen gas is liberated by 2 moles $H_{2} O_{2}$

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

10 L of oxygen gas is liberated by 22.42×10=0.8928 moles $H_{2} O_{2}$

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

N= $e q u i v a l e n t w e i g h t \times 1 L = 1730.35 \times 1 = 1.78 g / L$

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

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

$V_{1} = 224 m L$

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Similar questions

H_{2} O_{2}

K M n O_{4}

10 mL of $H_{2} O_{2}$ solution (volume strength = x) requires 10 mL of N/0.56 $M n O_{4}^{-}$ solution in acidic medium. Hence x is:

H_{2} O_{2}

K M n O_{4}

H_{2} O_{2}

10

H_{2} O_{2}

= x

10

\frac{N}{0.56}

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