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?
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Answer:
Volume strength of H2O2=5.6*Normality
56 volume strength of =5.6*Normality
Normality=56/5.6=10
In terms of normality
As Milli equivalents of reacted= Equivalent of
Moles of
Weight of
% purity of in the sample
=(0.1*375)/55
=(0.6818)*100
=68.18%
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