Question

calculate the normality of oxalic acid solution containing 6.3 grams in 500 ml of solution​

Answer 1

The equivalent weight of H2C2O4.2H2O is 63

Thus normality is

N = (6.3 x 1000)/(63 x 500)

= 0.2

Answer 2

Answer:0.2

Explanation:

Normality = weight/G.M.W × 1000/volume in ml

Oxalic ac id weight is 63 grams

Given weight of oxalic acid is 6.3

Volume is 500 ml

Now N = 6.3/63 × 1000/500

N = 1/10 × 2

N = 0.2