Find out the % of oxalate ion in given sample of oxalate salt of which 0 . 3 g is present in 100 mL of solution required 90 m L of N / 20 K M n O 4 for complete oxidation.

Step-by-step solution

The redox changes are Mn + 7 + 5 e − → Mn 2 + C 3 + 2 → 2 C 4 + + 2 e − ∵ meq. of oxalate ion=meq . of KMnO 4 w / E × 1000 = 90 × 1 / 20 Or w 88 / 2 × 1000 = 9 2 ∵ E C 2 O 4 − 2 = 88 2 or w C 2 O 4 − 2 = 0.198 g ∴ % of oxalate in sample = ( 0.198 × 100 ) / 0.3 = 66 %