Equivalent wt.of \(KMnO_{4}\) in neutral medium -\(\frac{M.W}{x}\) the value of "x" is "

n-Factor of KMnO_(4) in neutral medium is

(a) Why do transition elements show catalytic properties? (b) Calculate equivalent weight of KMnO_4 in neutral medium. ( c) What is the cause of lanthanoid contraction?

The equivalent weight of KMnO_(4) in (a) neutral medium, (b) acidic medium and (c ) alkaline medium is M//.. ( where M is mol.wt. of KMnO_(4) )

\(K_{4}[Fe(CN)_{6}]\rightarrow Fe^{+3}+CO_{2}+NO_{2}\) In the given chemical equation Equivalent weight of \(K_{4}[Fe(CN)_{6}]\) is \(\frac{M}{x}\). M is molecular weight of \(K_{4}[Fe(CN)_{6}]\). What is the value of \(\frac{x}{11}\)