Question

When $Br{O}_3^{-}$ ion reacts with $B{r}^{-}$ in acid medium, $B{r}_2$ is liberated. If the molecular weight of $B{r}_2$ is $M$, then the equivalent weight of $B{r}_2$ in this reaction is:

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

Answer 1

The correct option is C $\frac{3M}{5}$

$2Br{O}_3^{-}+12H^{+}+10B{r}^{-}\to 6B{r}_2+6H_{2}O$
$10$ moles electron required for formation of $6$ moles of $B{r}_2$.
$\therefore \text{n--factor of}B{r}_2=\frac{10}{6}=\frac{5}{3}$
$\text{Equivalent wt.}=\frac{\text{Molecular wt.}}{n}=\frac{M}{5/3}=\frac{3M}{5}$