Question

The normality of 0.98 w/v. H2so4solution is

Answer 1

HELLO THERE!

First of all:

In a solution of 0.98 (w/v),

0.98 grams of Sulphuric acid is present in 100 mL of solution.

Now,

$N = \frac{1000\times W}{EV}$

where W is the mass of sulphuric acid, E is the equivalent weight and V is the volume of solution.

We know, Sulphuric acid is a dibasic acid, hence, its Equivalent weight = Molecular weight/ 2 = 49.

So, Normality = (0.98 x 1000) / (49 x 100)

= 0.2 N

THANKS!