{\KMnO}_{4} reacts with oxalic acid according to the equation 2 {\MnO}_{4}^{-}+5 {\C}_{2} {\O}_{4}^{2-}+16 {\H}^{+} \to 2 {\Mn}^{2+}+10 {\CO}_{2}+8 {\H}_{2} {\O} Here 20{\mL} of 0.1 {\M} {\KMnO}_{4} is equivalent to (1996)

Correct Answer is : 50mL of 0.1M C2H2O4

50 {\mL} of 0.5 {\M} {\C}_{2} {\H}_{2} {\O}_{4}

20 {\mL} of 0.1 {\M} {\C}_{2} {\H}_{2} {\O}_{4}

20 {\mL} of 0.5 {\M} {\C}_{2} {\H}_{2} {\O}_{4}

50 {\mL} of 0.1 {\M} {\C}_{2} {\H}_{2} {\O}_{4}

Class 12 NEET Chemistry D And F Block Elements Questions

Question : 98 of 110

Marks: +1, -0

KMnO4 reacts with oxalic acid according to the equation
2MnO4−+5C2O42−+16H+→ 2Mn2++10CO2+8H2O
Here 20‌mL of 0.1M KMnO4 is equivalent to

(1996)

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