Commercial sample of $\mathrm{H}_{2} \mathrm{O}_{2}$ is labeled as $10 \mathrm{V}$. Its % strength is nearly

10 V $\mathrm{H}_{2} \mathrm{O}_{2}$ means $1 \mathrm{L}$ of this solution will produce $10 \mathrm{LO}_{2}$ at STP. $$ \begin{array}{c} 2 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\quad \mathrm{O}_{2} \\ 68 \mathrm{g} \quad \quad \quad \quad 22.4 \text{L at STP} \end{array} $$ $\therefore 22.4 \mathrm{L}$ of $\mathrm{O}_{2}$ is obtained from $\mathrm{H}_{2} \mathrm{O}_{2}=68 \mathrm{g}$ 10 L of $O_2$ will be obtained from $$ \mathrm{H}_{2} \mathrm{O}_{2}=\frac{68}{22.4} \times 10=30.36 \mathrm{g} $$ $1000 \mathrm{mL}$ of the given solution contains $30.36 \mathrm{g}$ $\mathrm{H}_{2} \mathrm{O}_{2}$ and $100 \mathrm{mL}$ of the given solution contains $$ \frac{30.36 \times 100}{1000}=3.03 \mathrm{g} \mathrm{H}_{2} \mathrm{O}_{2} $$ Thus, $\%$ strength of $\mathrm{H}_{2} \mathrm{O}_{2}$ is $3.03(\approx 3)$