See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

The redox changes are 7 2 Mn 5e Mn       3 4 2 C 2C 2e      meq. of oxalate ion = meq. of KMnO4 (w/E) x 1000 = 90 x (1/20) w 9 x 1000 88/2 2  2 2 4 C O 88 E 2          2 2 4 C O w = 0.198g  % of oxalate in sample = (0.198 x 100)/0.3 = 66%