What is the equivalent weight of I₂ when it is converted into IO₄^– ?

Complete balanced chemical reaction:

$2{\mathrm{IO}}_4^{-}+16{\mathrm{H}}^{+}+14{\mathrm{e}}^{-}\to {\mathrm{I}}_2+8{\mathrm{H}}_2\mathrm{O}$

Oxidation state of I in ${\mathrm{IO}}_4^{-}$ is +7 and in I₂ is zero. Equivalent weight of I₂ is calculated as:

$$\begin{gathered} \mathrm{Equivalent} \mathrm{weight}=\frac{\mathrm{Molecular} \mathrm{weight}}{\mathrm{Number} \mathrm{of} \mathrm{electrons} \mathrm{transferred}} \\ =\frac{\mathrm{Molecular} \mathrm{weight}}{7} \end{gathered}$$