A commercial sample of H_{2}O_{2} is labeled | vedclass

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(A) $10 	ext{ V } H_{2}O_{2}$ means $1 	ext{ L}$ of this solution will produce $10 	ext{ L}$ of $O_{2}$ at $STP$.$2H_{2}O_{2} ightarrow 2H_{2}O +

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$3$

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(A) $10 	ext{ V } H_{2}O_{2}$ means $1 	ext{ L}$ of this solution will produce $10 	ext{ L}$ of $O_{2}$ at $STP$.$2H_{2}O_{2} ightarrow 2H_{2}O + O_{2}$$68 	ext{ g}$ of $H_{2}O_{2}$ produces $22.4 	ext{ L}$ of $O_{2}$ at $STP$.Therefore, $22.4 	ext{ L}$ of $O_{2}$ is obtained from $68 	ext{ g}$ of $H_{2}O_{2}$.$10 	ext{ L}$ of $O_{2}$ will be obtained from $H_{2}O_{2} = \frac{68}{22.4} 	imes 10 = 30.36 	ext{ g}$.$1000 	ext{ mL}$ of the given solution contains $30.36 	ext{ g}$ of $H_{2}O_{2}$.$100 	ext{ mL}$ of the given solution contains $\frac{30.36 	imes 100}{1000} = 3.036 	ext{ g}$ of $H_{2}O_{2}$.Thus, the percentage strength of $H_{2}O_{2}$ is approximately $3$.

$A$ commercial sample of $H_{2}O_{2}$ is labeled as $10 \text{ V}$. Its percentage strength is nearly:

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