Question

250ml of a solution contains 6.3grams of oxalic acid. What is the volume of water to be added to this solution to make it a 0.1N solution ​

Answer 1

your answer is 0.75

explanation


Eq. Wt. of oxalic acid = 126/2 = 63

Normality of oxalic acid = 6.3 ×1000/(63× 250)
= 0.4N

Equation for dilution
V 2= V1N1/N2
= (0.250×0.4)/0.1
=1 lit

Volume of water added = 1-0.25 = 0.75 lit

Answer 2

Answer:

Explanation:

your answer is 0.75

explanation

Eq. Wt. of oxalic acid = 126/2 = 63

Normality of oxalic acid = 6.3 ×1000/(63× 250)

= 0.4N

Equation for dilution

V 2= V1N1/N2

= (0.250×0.4)/0.1

=1 lit

Volume of water added = 1-0.25 = 0.75 lit