$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to 2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$
અહી $20\, mL$ of $0.1\, M\, KMnO_4$ એ કોના બરાબર હશે
$0.1\,M\,KMnO_4 = 0.5\,N\,KMnO_4$
$\therefore $ Meq of $KMnO_4 = 20 \times 0.5 = 10$ ($n$ factor $= 5$)
Meq of $50 \,ml$ of $0.1\,M\,H_2C_2O_4 = 50 \times 0.2 = 10$
$(0.1\,M\,H_2C_2O_4 = 0.2\,N\,H_2C_2O_4)$