Question

${Br{O}_3}^{-}$ ion reacts with $B{r}^{-}$ to form $B{r}_2$, in acid medium. The equivalent mass of $B{r}_2$ in this reaction is:

• A. $\frac{4M.w.}{6}$
• B. $\frac{3M.w.}{5}$
• C. $\frac{5M.w.}{3}$
• D. $\frac{5M.w.}{8}$

Answer 1

Answer: (B)

$\frac{3M.w.}{5}$

$2Br{O}_3^{-}+12H^{+}+10e^{-}\to B{r}_2+6H_{2}O{n}_1=10$
$2B{r}^{-}\to B{r}_2+6e^{-}{n}_2=2$
$n-factor=\frac{n_1\times n_2}{n_1+n_2}=\frac{10\times 2}{10+2}=\frac{20}{12}=\frac{5}{3}$

$E.w.=\frac{M.w.}{5/3}=\frac{3M.w.}{5}$.