What is the percentage strength ( $w e i g h t / v o l u m e$ ) of $11.2 V$ $H_{2} O_{2}$ ?

1.7

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3.4

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34

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None of these

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Solution

The correct option is B $3.4$
$1$ mL $H_{2} O_{2} = \frac{68 x}{22400} g H_{2} O_{2}$ .
Here, $x$ is the volume strength of $H_{2} O_{2}$ which is equal to $11.2 L$ .
Hence, $1$ mL $H_{2} O_{2} = \frac{68 \times 11.2}{22400} = 0.034 g H_{2} O_{2}$ .
The percentage strength of $H_{2} O_{2}$ is $0.034 \times 100 = 3.4$ % ( $w / v$ ).

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