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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?

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The n- factor must be obviously wrong because two electrons are involved in the reduction of peroxide to oxygen. Write the half cell yourself. Indian chemistry textbooks should have pity on the world and stop teaching these equivalents to young students. This is a thing of the 18th century. Those who invented this concept moved on.

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  • $\begingroup$ Don't come to such a conclusion of "Indian textbooks" abruptly. How many "Indian" textbooks have you read? And if you have read about equivalents properly, you would come to know that it's a very powerful tool. $\endgroup$ – user600016 Dec 18 '19 at 11:06
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    $\begingroup$ @user600016 This might come out harsh, but the concept of equivalence has no use today as we are already able to determine molecular formulas and formula units. An equivalent and its derivative quantities such as gram-equivalent and normality are based on archaic representation. It's only powerful to those who don't know or don't want to know better. I'm not familiar with Indian literature, but judging from the questions posted by other users quoting Indian textbooks, there are way too many erroneous or outdated concepts still being taught (not to mention odd formatting and typos). $\endgroup$ – andselisk Dec 18 '19 at 11:45
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    $\begingroup$ user600016, I have seen more Indian textbooks and older British books teaching the concept of equivalents than you can imagine. No modern book ever mentions equivalents. Some analytical chemistry books aimed for technicians use it because equivalents make calculations easier in terms of N1V1=N2V2, which can be easily memorized by those who don't want to think about mole ratios. Yes, it was a powerful tool but in the 18th century. $\endgroup$ – M. Farooq Dec 18 '19 at 14:21
  • $\begingroup$ I did get the n factor wrong. $\endgroup$ – Abu Shahid Dec 18 '19 at 15:50
  • $\begingroup$ 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 $$\ce{H2O2 -> H2O + O2}$$? $\endgroup$ – Ashish Ahuja Feb 26 at 14:45

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