In alkaline medium KMnO4 can be converted into MnO₂. The decrease in oxidation state in ‘n’. So
equivalent mass of ${\mathrm{KMnO}}_4=\frac{\text{ Molar mass }}{\mathrm{n}}$. What is ‘n’ ?

In alkaline solution, KMnO₄​ is first reduced to manganate and then to insoluble manganese dioxide.

$2{\mathrm{KMnO}}_4+{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{alkaline} \mathrm{medium}}{\to }2{\mathrm{MnO}}_2+2{\mathrm{OH}}^{⊝}+3\left[\mathrm{O}\right]$

$\stackrel{+7}{{\mathrm{MnO}}_4^{-}}⟶\stackrel{+4}{{\mathrm{MnO}}_2}$