Redox and Equivalent ConceptHard
Question
KMnO4 reacts with oxalic acid according to the equation :
2MnO4- + 5C2O4- + 16H+ → 2Mn+2 + 10CO2 + 8H2O
Here 20 ml of 0.1 M KMnO4 is equivalent to :
2MnO4- + 5C2O4- + 16H+ → 2Mn+2 + 10CO2 + 8H2O
Here 20 ml of 0.1 M KMnO4 is equivalent to :
Options
A.20 mL of 0.5 M H2C2O4
B.50 mL of 0.5 M H2C2O4
C.50 mL of 0.1 M H2C2O4
D.20 mL of 0.1M H2C2O4
Solution
Meq of A = Meq of B.
Meq of KMnO4 = 20×0.5 = 10
Meq of 50 ml of 0.1 M
H2C2O4 = 50× 0.2 = 10
(0.1 M H2C2O4 = 0.2 N H2C2O4)
Meq of KMnO4 = 20×0.5 = 10
Meq of 50 ml of 0.1 M
H2C2O4 = 50× 0.2 = 10
(0.1 M H2C2O4 = 0.2 N H2C2O4)
Create a free account to view solution
View Solution FreeMore Redox and Equivalent Concept Questions
The oxidation states of Cr in [Cr(H2O)6]Cl3,[Cr(C6H6)2] and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are :...More number of oxidation states are exhibited by the actionoids than by the lanthanoids. The main reason for this is :-...A volume of 100 L of hard water requires 5.6 g of lime for removing temporary hardness. The temporary hardness in ppm of...In the reaction, P4 + NaOH + H2O → PH3 + NaH2PO2, the equivalent weight of P4 is...A quantity of 0.5 g of a metal nitrate gave 0.43 g of metal sulphate....