Question

1.80 g of impure sample of oxalate was dissolved in water and solution is made to 250 mL. On titration 20 mL of this solution required 30 mL of M/50 KMnO₄ solution. Calculate the percentage purity of the sample.

Answer 1

The balanced chemical equation for the reaction is: 

2MnO₄ + 5C₂O₂ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O 

Applying molarity equation, 

(M₁V₁/n₁)KMnO₁ = (M1⁄2V1⁄2/n₂)C₂O₂^-4

1/50 × 30/2 = M₂ × 20/5 

M₂ = 30 × 5/50 × 20 × 2 = 0.075 M 

Molecular mass of C₂O₂^-4 = 88 

he Mass of C₂O₂^-4 = 0.075 × 88 = 6.6 g 

In 250 mL = 6.6/1000 × 250 = 1.65 g 

% purity = 1.65/1.80 x 100 = 91.7%