Explanation:
\({\text{2K}}\mathop {{\text{Mn}}}\limits^{{\text{ + 7}}} {{\text{O}}_{\text{4}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\xrightarrow{{{\text{Alkaline}}}}\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\mathop {{\text{Mn}}}\limits^{ + 4} {{\text{O}}_2}{\text{ + 2KOH + 3}}[{\text{O}}]\)
\({\rm{(As}}\,{\rm{change}}\,{\mkern 1mu} {\rm{in}}\,{\rm{oxidation}}{\mkern 1mu} \,{\rm{number}}\,{\rm{of}}\,{\mkern 1mu} {\rm{Mn}}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} {\rm{3)}}\)
So, equivalent mass = \(\frac{{{\rm{molar}}\,{\rm{mass}}}}{3}\)
Hence, the equivalent weight of \({\rm{KMn}}{{\rm{O}}_{\rm{4}}}\)
in alkaline solution is equal to\(\frac{1}{3}{\rm{rd}}\) of the molar mass of \({\rm{KMn}}{{\rm{O}}_{\rm{4}}}\)