In alkaline solution, $KMnO _{4}$ is reduced to $MnO _{2}$ (colourless).
$\frac{2 KMnO _{4}+2 H _{2} O \longrightarrow 2 MnO _{2}+2 KOH+3[ O ] KI +3[ O ] \longrightarrow KIO _{3}}{2 KMnO _{4}+2 H _{2} O + Kl \longrightarrow 2 MnO _{2}+2 KOH + KlO _{3}}$
Hence, two moles of $KMnO _{4}$ are reduced by one mola of $KI$