The strength of $20$ volume of $H_{2} O_{2}$ is:

13.6 g / l i t r e

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60.7 g / l i t r e

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160 g / l i t r e

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20.2 g / l i t r e

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Solution

The correct option is A $60.7 g / l i t r e$
$2 H_{2} O_{2} ⟶ 2 H_{2} O + O_{2}$

$2 \times 34 = 68 g m$ $22.4 L$ at N.T.P

$22.4 L$ of $O_{2}$ at NTP is produced from $68 g m H_{2} O_{2}$ .

$20 L$ of $O_{2}$ at N.T.P is produced from $H_{2} O_{2}$ $\Rightarrow \frac{68 \times 20}{22.4} = 60.7$

$∴$ Strength of $H_{2} O_{2}$ in $20$ volume $H_{2} O_{2} = 60.7 g / L$

Hence, option B is correct.

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