The strength of $H_2{O}_2$ is expressed in several ways like molarity, normality, % (w/V), volume strength, etc. The strength of “10 V” means 1 volume of $H_2{O}_2$ on decomposition gives 10volumes of oxygen at 1atm and 273K or 1 litre of $H_2{O}_2$ gives 10 litre of $O_2$ at 1 atm and 273K. The decomposition of $H_2{O}_2$ is shown as under: 
$H_2{O}_2\left(aq\right)\stackrel{ }{\to }{H}_{2}O\left(l\right)+\frac{1}{2}{O}_2\left(g\right)$   
$H_2{O}_2$ can acts as oxidising as well as reducing agent. At oxidizing agent $H_2{O}_2$ is converted into $H_{2}O$ and as reducing agent $H_2{O}_2$ is converted into $O_2.$ For both cases its $n-$ factor is 2.
$\therefore$ Normality of $H_2{O}_2$ solution $=2\times$ Molarity of $H_2{O}_2$ solution