Question

a solution has been prepared by dissolving 0.49 gm of sulphuric acid in 100 ml of the solution calculate normality if the solution

Answer 1

answer for this question is 240N

Answer 2

Answer:

Explanation:

N= (wt.in gm*1000)/(eq.wt*vol.of solution in ml)

EQ.wt=mol. Wt/ factor

=98/2

=49

So,N=(0.49*1000)/49*100

=1/10

=0.1 ans