Question

Example. Calculate the normality of a solution
containing 6.3 g oxalic acid dissolved in 250ml. of
solution.​

Answer 1

Answer:

Normality=

1 liter of solution

number of equivalent of solute

=

eq.weight

Weight

×

V(ml)

1000

By putting all the given values, we get-

0⋅1=

2

126

6⋅3

×

V

1000

V=1000ml=1litre

Volume of water=Total Volume - Volume of oxalic acid

=1−⋅25

=0⋅75 litre

Answer 2

Answer:

0.4 N

Explanation:

another formula is...

Normality×Equivalent Weight = Molarity× Molecular weight