# What is the relation between volume strength and normality of H2O2? [closed]

I have been told that: $$\text{equivalent weight of }\ce{H2O2} =\text{molecular weight of } \ce{H2O2}$$

as it's $$n$$-factor is $$1$$ (from balanced equation).

But in my book it's given that:

Molarity $$= \frac {V }{11.2}$$

Normality $$= \frac {V} {5.6}$$

Why are they different if equivalent and molecular weights are equal?

• I just posted an answer in which I am indeed getting the n-factor for the overall decomposition of $\ce{H2O2}$ as $1$, which I believe is the same reaction using which volume strength is defined. The n-factor of $1$ seems to fit the law of chemical equivalence, so am I doing something wrong in my answer or is this answer wrong? Or maybe volume strength is not defined using the reaction $$\ce{H2O2 -> H2O + O2}$$? – Ashish Ahuja Feb 26 at 14:45