34 g of hydrogen peroxide is present in 1120 ml of solution. This solution is called

$\begin{array}{l}\mathrm{Amount}\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2\mathrm{in}\mathrm{one}\mathrm{ml}=\frac{34}{1120}\mathrm{g}\\ {\mathrm{H}}_2{\mathrm{O}}_2\stackrel{ }{\to }{\mathrm{H}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2\\ \therefore 34\mathrm{g}\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2\mathrm{gives}16\mathrm{g}\mathrm{or}11200\mathrm{ml}\mathrm{of}{\mathrm{O}}_2\\ \therefore \frac{34}{1120}\mathrm{g}{\mathrm{H}}_2{\mathrm{O}}_2=\frac{11200}{34}\times \frac{34}{1120}=10\mathrm{ml}{\mathrm{O}}_2\\ \therefore \mathrm{Volume}\mathrm{strength}\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2=10\mathrm{volume}\end{array}$