$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to 2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$
Here $20\, mL$ of $0.1\, M\, KMnO_4$ is equivalent to
- A$20\, mL$ of $0.5\, M\, H_2C_2O_4$
- B$50\, mL$ of $0.5\, M\, H_2C_2O_4$
- ✓$50\, mL$ of $0.1\, M\, H_2C_2O_4$
- D$20\, mL$ of $0.1\, M\, H_2C_2O_4$
