Question

When $Br{O}_3^{-}$ ion reacts with ${Br}^{-}$ in acidic medium, ${Br}_2$ is liberated. The equivalent weight ${Br}_2$ 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 $\frac{3M}{5}$

$2{BrO}_3^{-}+12H^{+}+10{Br}^{-}\to 6{Br}_2+6H_{2}O$

$10\text{mole electron require for formation of}6\text{mole of}B{r}_2$

$\begin{array}{cc}\Rightarrow n\text{- factor of}B{r}_2=& \frac{10}{3}=\frac{5}{3}\\ \Rightarrow \text{eq.}\omega t=\frac{\text{mol.}\omega t.}{n}& =\frac{M}{5/3}.\\ & =\frac{3M}{5}\end{array}$