Find the percentage strength of ${\mathrm{H}}_2{\mathrm{O}}_2$ in a ample marked "10 volume"

$10\text{ volume }{\mathrm{H}}_2{\mathrm{O}}_2$ mean that 1ml of ${\mathrm{H}}_2{\mathrm{O}}_2$ liberates 10ml oxygen 

Reaction :         $\underset{2\left[\right(2\times 1)+(2\times 16\left)\right]=68\mathrm{g}}{2{\mathrm{H}}_2{\mathrm{O}}_2}\to \underset{ }{2{\mathrm{H}}_2\mathrm{O}}+\underset{22,400\mathrm{ml}}{{\mathrm{O}}_2}$

$22400{\mathrm{mLO}}_2$is liberated from ${\mathrm{H}}_2{\mathrm{O}}_2=68\mathrm{g}$

$\therefore 10{\mathrm{mLO}}_2$ is liberated 

from ${\mathrm{H}}_2{\mathrm{O}}_2=\frac{68\mathrm{g}}{22400\mathrm{mL}}\times 10\mathrm{mL}=\frac{68}{2240}\mathrm{g}$

Thus :  $1{\mathrm{mLH}}_2{\mathrm{O}}_2\text{ contains }{\mathrm{H}}_2{\mathrm{O}}_2=\frac{68}{2240}\mathrm{g}$

$100{\mathrm{mLH}}_2{\mathrm{O}}_2$ contains 

${\mathrm{H}}_2{\mathrm{O}}_2=\frac{68}{2240}\times 100=\mathbf{3}\mathbf{.03}\mathbf{\%}$