Question

30g KMnO4 sample containing inert impurity is completely reacted with 60ml of 56 volume of H2O2 in acidic medium then what is the percentage purity of KMnO4?

Answer 1

Answer:

Volume strength of H2O2=5.6*Normality

56 volume strength of $H_{2} O_{2}$=5.6*Normality

Normality=56/5.6=10

In terms of normality

As Milli equivalents of $KMnO_{4}$ reacted= Equivalent of $H_{2} O_{2}$

Moles of $KMnO_{4} =1/10=0.1$

Weight of $KMnO_{4} =0.1*375$

% purity of $KMnO_{4}$ in the sample

=(0.1*375)/55

=(0.6818)*100

=68.18%

Answer 2

here's the answer and hope this helps you