A 5 mL solution of ${\mathrm{H}}_2{\mathrm{O}}_2$ liberates 0.508 g of Iodine from an acidified KI solution. What is the volume strength of the ${\mathrm{H}}_2{\mathrm{O}}_2$ solution ?

$\begin{array}{l}\mathrm{Meq}.\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2=\mathrm{Meq}.\mathrm{of}{\mathrm{I}}_2\\ \Rightarrow \frac{{\mathrm{W}}_{{\mathrm{H}}_2{\mathrm{O}}_2}}{34}\times 2\times 1000=\frac{0.508}{254}\times 2\times 1000\\ \therefore {\mathrm{W}}_{{\mathrm{H}}_2{\mathrm{O}}_2}=0.068\mathrm{g}\\ {\mathrm{H}}_2{\mathrm{O}}_2\stackrel{ }{\to }{\mathrm{H}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2\\ \because 34\mathrm{g}{\mathrm{H}}_2{\mathrm{O}}_2\mathrm{gives}11.2\mathrm{litres}\mathrm{of}{\mathrm{O}}_2\mathrm{at}\mathrm{STP}\\ \therefore 0.068\mathrm{g}\mathrm{gives}=\frac{11.2}{34}\times 0.068=22.4\mathrm{ml}{\mathrm{O}}_2\\ 5\mathrm{ml} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2(0.068\mathrm{g})\to 22.4\mathrm{mi} {\mathrm{O}}_2\\ 1\mathrm{ml} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \to ?\\ \therefore \mathrm{Volume}\mathrm{Strength}\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2=\frac{22.4}{5}=4.48\mathrm{volume}\end{array}$