Chemistry, asked by sandip6951, 10 months ago

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?

Answers

Answered by antiochus
18

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%

Answered by dharamrajktr6
24

here's the answer and hope this helps you

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