Question

the normality of 0.098%(w/v) h2so4 solution is
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Answer 1

Answer:

Explanation:

10 grams of  H2SO4  is present in 100 ml aqueous solution.

So that means 100 grams of  H2SO4  is present in 1 litre solution

Also molecular weight of  H2SO4  is 98 .

Hence we can find out total no of moles of  H2SO4  which comes out to be 100/98 = 1.02041

So Molarity of the solution is no of moles per litre of solution

So M = 1.02041/1 = 1.02041

Now as sulphuric acid is a dibasic acid i.e it gives 2  H+  ions per molecule on dissociation hence its valency factor is 2

Now Normality can be calculated as  N=Z×M  where

N is normality

Z is valency factor

M is the Molarity

So Normality of our solution will be 2×1.02041= 2.04082N