When $Br{O_3}^{-}$ Ion react with $B{r}^{-}$ in acidic medium $B{r}_2$ is liberated. The equivalent mass of $B{r}_2$ in this reaction is

 $2Br\stackrel{-}{o_3}+12H^{+}+10\stackrel{-}{Br}\to 6B{r}_{2}6H_{2}O$
10 mole $\stackrel{-}{e}$ are required for the formation of 6 moles of  $B{r}_2$
n.factor of  $B{r}_2=\frac{10}{6}=\frac{5}{3}$
eq.wt  $\frac{m.w}{n_f}=\frac{M}{5/3}=\frac{3M}{5}$