On electrolysis of a sample of acidified water, 22.4 ml of hydrogen was obtained. The volume of oxygen in ml obtained is:

22.4

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44.8

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11.2

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2.24

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Solution

The correct option is C 11.2
22.4 litre of $H_{2}$ at STP equals to 1 mole of $H_{2}$ .

$2 H_{2} O \to 2 H_{2} + O_{2}$

2 mole of $H_{2}$ has 1 mole $O_{2}$ on electrolysis.
$∴$ 1 mole of $H_{2}$ has 0.5 mole of $O_{2}$

Volume of 0.5 mole $O_{2} = 0.5 \times 22.4$

$= 11.2 l$

Hence, option C is correct.

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On electrolysis of a sample of acidified water, 22.4 ml of hydrogen was obtained. The volume of oxygen in ml obtained is:

The correct option is C 11.222.4 litre of H2 at STP equals to 1 mole of H2.2H2O→2H2+O22 mole of H2 has 1 mole O2 on electrolysis.∴ 1 mole of H2 has 0.5 mole of O2Volume of 0.5 mole O2=0.5×22.4 =11.2 lHence, option C is correct.