The volume of oxygen evolved at STP by decomposition of $0.68 g$ '20 volume' hydrogen peroxide solution is :

2.24 m L

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22.4 m L

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224 m L

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2240 m L

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Solution

The correct option is D $224 m L$
$2 H_{2} O_{2} ⟶ 2 H_{2} O + O_{2}$
$2 m o l$ $1 m o l$

$= 2 \times 34 = 68 g$ $22400 m L$

$∵$ At STP, $68 g H_{2} O_{2}$ produced $O_{2} = 22400 m L$

$∴$ AT STP, $0.68 g H_{2} O_{2}$ will produced $O_{2} = \frac{22400 \times 0.68}{68} = 224 m L$

Hence, the correct option is $C$ .

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0.68 g^{'} 20 v o l u m e^{'}

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V

m L

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