1.

Calculate the equivalent weight of KMnO_(4) in (i) Acidic medium (in) Basic medium (iii) Neutral medium

Answer»

Solution :(i) `"Equivalent weight of" KMnO_(4) "in acidic medium" = ("Molecular weight of" KMnO_(4) )/("No. of moles of ELECTRONS TRANSFERRED")= (158)/(5)= 31.6 `
(ii) `"Equivalent weight of" KMnO_(4) "in basic medium" = ("Molecular weight of" KMnO_(4) )/("No. of moles of electrons transferred")= (158)/(1)= 158`
(iii) `"Equivalent weight of" KMnO_(4) "in neutral medium"= ("Molecular weight of" KMnO_(4))/("No. of moles of electrons transferred")= (158)/(3)= 52.67`


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