The correct options are
B 50mL of 0.10 M
H2C2O4 C 25mL of 0.20 M
H2C2O42MnO−4+5C2O2−4+16H+⟶2Mn2++10CO2+8H2O20 ml of 0.1 M KMnO4 contains 0.1×0.020 moles of KMnO4 i.e, 0.002 moles of KMnO4
(A) 120 mL of 0.25M H2C2O4 contains 0.25×0.12 moles of H2C2O4 i.e, 0.03 moles of H2C2O4
(B) 50 mL of 0.10 M H2C2O4 contains 0.05×0.1 moles of H2C2O4 i.e, 0.005 moles of H2C2O4
(C) 25 mL of 0.20 M H2C2O4 contains 0.005 moles of H2C2O4
(D) 50 mL of 0.20 M H2C2O4 contains 0.01 moles of H2C2O4
By stiochimetry, 2 moles of KMnO4 is equivalent to 5 moles of H2C2O4
Therefore, 0.002 moles of KMnO4 is equivalent to 0.005 moles of H2C2O4.
Therefore, (B) and (C) are correct options.