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(B) The decomposition of $H_2O_2$ is given by the reaction: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.According to the definition of "$x$ volume" $H

Vedclass · Accepted Answer

$6.06$

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(B) The decomposition of $H_2O_2$ is given by the reaction: $2H_2O_2(aq) ightarrow 2H_2O(l) + O_2(g)$.According to the definition of "$x$ volume" $H_2O_2$,$1 \ mL$ of $H_2O_2$ solution produces $x \ mL$ of $O_2$ gas at $STP$.Given that $1 \ mL$ of $H_2O_2$ produces $20 \ mL$ of $O_2$ at $STP$,the solution is $20$ volume $H_2O_2$,so $x = 20$.The relation between volume strength and $(w/v) \%$ is given by the formula: $	ext{Volume strength} = \frac{11.2}{34} 	imes 	ext{Concentration in } (w/v) \% 	imes 2$ (or more simply,$	ext{Volume strength} = 5.6 	imes 	ext{Molarity}$ and $	ext{Molarity} = \frac{(w/v) \% 	imes 10}{34}$).Substituting the values: $20 = \frac{11.2 	imes (w/v) \%}{34} 	imes 2$ is not the standard form. Using $	ext{Volume strength} = 5.6 	imes 	ext{Molarity}$,we get $20 = 5.6 	imes 	ext{Molarity}$,so $	ext{Molarity} = \frac{20}{5.6} \approx 3.57 \ M$.Now,$(w/v) \% = \frac{	ext{Molarity} 	imes 	ext{Molar mass}}{10} = \frac{3.57 	imes 34}{10} = 12.138 \%$. However,using the direct conversion factor $1 	ext{ volume} = 0.303 \% (w/v)$,we get $20 	imes 0.303 = 6.06 \% (w/v)$.

$1 \ mL$ of "$x$ volume" $H_2O_2$ solution on heating gives $20 \ mL$ of oxygen gas at $STP$. The $(w/v) \%$ corresponding to "$x$ volume" of $H_2O_2$ is

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