What is the % strength of 22.4 volume of H_2O | vedclass

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(D) $22.4$ volume $H_2O_2$ means $1 \ mL$ of $H_2O_2$ solution gives $22.4 \ mL$ of $O_2$ at $NTP$.$2H_2O_2 \longrightarrow 2H_2O + O_2$$2 \ mol \long

Vedclass · Accepted Answer

$6.8$

Vedclass · Answer

(D) $22.4$ volume $H_2O_2$ means $1 \ mL$ of $H_2O_2$ solution gives $22.4 \ mL$ of $O_2$ at $NTP$.$2H_2O_2 \longrightarrow 2H_2O + O_2$$2 \ mol \longrightarrow 1 \ mol$$68 \ g$$22400 \ mL$ at $NTP$$	herefore 22400 \ mL$ of $O_2$ is obtained by $68 \ g$ of $H_2O_2$.$	herefore 22.4 \ mL$ of $O_2$ is obtained by $\frac{68 	imes 22.4}{22400} = 0.068 \ g$ of $H_2O_2$.Since $1 \ mL$ of $H_2O_2$ solution contains $0.068 \ g$ of $H_2O_2$,$	herefore 100 \ mL$ of $H_2O_2$ solution contains $0.068 	imes 100 = 6.8 \ g$ of $H_2O_2$.Thus,the strength is $6.8 \%$.

What is the $\%$ strength of $22.4$ volume of $H_2O_2$ solution (in $\%$)?

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