A commercial sample of $H_{2} O_{2}$ is labelled as $10 V$ . Its % strength is nearly :

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Solution

The correct option is A 3
10 volume $H_{2} O_{2}$ means 1 ml $H_{2} O_{2}$ solution will produce 10 ml of $O_{2}$ .

$2 H_{2} O_{2} ⟶ 2 H_{2} O + O_{2}$

2 mole 1 mole at S.T.P(22.4 L)
$2 \times 34 g = 68 g$

22400 ml of $O_{2}$ is produced from 68 g $H_{2} O_{2}$ .

10 ml $O_{2}$ is produced from $\frac{68 \times 10}{22400} g = 0.03036 g$

100 ml $H_{2} O_{2}$ contains $0.03 \times 100 = 3 g$

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