Get Answers to all your Questions

header-bg qa

solve this question please 

Answers (1)

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 KMnO_{4} of  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%

Posted by

Satyajeet Kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE