Calculate the strength of $5$ volumes $H_2O_2$ solution.
(D) By definition,$5$ volumes $H_2O_2$ solution means that $1 \ L$ of this solution produces $5 \ L$ of $O_2$ gas at $STP$ upon decomposition.
The decomposition reaction is: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.
According to the stoichiometry,$2 \times 34 \ g$ $(68 \ g)$ of $H_2O_2$ produces $22.7 \ L$ of $O_2$ at $STP$.
Therefore,$22.7 \ L$ of $O_2$ at $STP$ is produced by $68 \ g$ of $H_2O_2$.
So,$5 \ L$ of $O_2$ at $STP$ will be produced by: $\frac{68 \times 5}{22.7} \ g \approx 14.98 \ g$.
The strength of the solution is approximately $15 \ g/L$.
The decomposition reaction is: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.
According to the stoichiometry,$2 \times 34 \ g$ $(68 \ g)$ of $H_2O_2$ produces $22.7 \ L$ of $O_2$ at $STP$.
Therefore,$22.7 \ L$ of $O_2$ at $STP$ is produced by $68 \ g$ of $H_2O_2$.
So,$5 \ L$ of $O_2$ at $STP$ will be produced by: $\frac{68 \times 5}{22.7} \ g \approx 14.98 \ g$.
The strength of the solution is approximately $15 \ g/L$.
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