Question

5.00g/dm3 of impure sodium carbonate were made up of to make 1dm3 of solution. 25cm3 of this required 18.00cm3 of HCl for neutralization. Calculate;
a. Percentage purity of the base
b. Percentage impurity of the base.

Answer 1

Answer:

Explanation:

Answer needs normality of acid

If normality of acid is n

Then answer is as follows

Equivalent = 18n

Equivalent of 1 ml is 18n/25

Moles × 2000 = 18n/25

Moles = 18n/50000

Mass = 0.03816n

Mass present in question = 5gm

Percentage purity = 3.186n/5

Impurity = 100-3.186n/5