KMnO4 reacts with oxalic acid according to the equation2MnO4- +5C2O4^2- +16OH- -2MnO2+ + 10CO2 + 8H20here 20 mL of 0 - Chemistry - The Solid State

Question

KMnO4 reacts with oxalic acid according to the equation2MnO4-
+5C2O4^2- +16OH- -2MnO2+ + 10CO2 + 8H20here 20 mL of 0 1 M KMnO4 is
equivalent to
1 50mL of 0 5 M C2H2O4
2 20 mL of 0 1 M C2H2O4
3 20 mL of 0 5 M C2H2O4
4 50 mL of O 1 M C2H2O4

Anuradha Puri · Accepted Answer

Now, Normality = molarity x change in oxidation number So,

And

Also Number of milli equivalents (meq)= Normality x volume in ml Therefore,

And

Hence, 20 mL of 0 1 M KMnO4 is equivalent to 50 mL of 0 1 M C2H2O4

KMnO4 reacts with oxalic acid according to the equation2MnO4- +5C2O4^2- +16OH- -2MnO2+ + 10CO2 + 8H20here 20 mL of 0.1 M KMnO4 is equivalent to..... 1. 50mL of 0.5 M C2H2O4 2. 20 mL of 0.1 M C2H2O4 3. 20 mL of 0.5 M C2H2O4 4. 50 mL of O.1 M C2H2O4

KMnO4 reacts with oxalic acid according to the equation2MnO4- +5C2O4^2- +16OH- -2MnO2+ + 10CO2 + 8H20here 20 mL of 0.1 M KMnO4 is equivalent to.....

1. 50mL of 0.5 M C2H2O4

2. 20 mL of 0.1 M C2H2O4

3. 20 mL of 0.5 M C2H2O4

4. 50 mL of O.1 M C2H2O4