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:
$H_{2} O_{2} (a q) \overset{}{\to} H_{2} O (l) + \frac{1}{2} O_{2} (g)$
$H_{2} O_{2}$ 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.
$∴$ Normality of $H_{2} O_{2}$ solution $= 2 \times$ Molarity of $H_{2} O_{2}$ solution

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