Question

${{\mathrm{BrO}}_3}^{-}$ ion reacts with ${\mathrm{Br}}^{-}$to form ${\mathrm{Br}}_2$ in an acid medium. The equivalent mass of ${\mathrm{Br}}_2$ in this reaction is:

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

Answer 1

The correct option is A

$\frac{3\mathrm{M}.\mathrm{W}}{5}$

Explanation for correct option:

A. $\frac{3\mathrm{M}.\mathrm{W}}{5}$

Step 1: Formation of Bromine:

• ${{\mathrm{BrO}}_3}^{-}$ion reacts with ${\mathrm{Br}}^{-}$to form ${\mathrm{Br}}_2$ in an acid medium.
• The chemical reaction can be depicted as:
• $$\begin{gathered} 2{{\mathrm{BrO}}_3}^{-}+12{\mathrm{H}}^{+}+10{\mathrm{e}}^{-}\to {\mathrm{Br}}_2+6{\mathrm{H}}_2\mathrm{O}{\mathrm{n}}_1=10 \\ 2{\mathrm{Br}}^{-}\to {\mathrm{Br}}_2+6{\mathrm{e}}^{-}{\mathrm{n}}_2=2 \end{gathered}$$

Step 2: Calculation of n-factor:

$\mathrm{n}-\mathrm{factor}=\frac{{\mathrm{n}}_1\times {\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}=\frac{10\times 2}{10+2}=\frac{20}{12}=\frac{5}{3}$

Step 3: Calculation for equivalent weight:

$$\begin{gathered} \mathrm{E}.\mathrm{W}=\frac{\mathrm{M}.\mathrm{W}}{5/3} \\ \mathrm{E}.\mathrm{W}=\frac{3\mathrm{M}.\mathrm{W}}{5} \end{gathered}$$

Thus, the equivalent weight of Bromine$\left({\mathrm{Br}}_2\right)$ is $\frac{3\mathrm{M}.\mathrm{W}}{5}$.

Explanation for incorrect options:

The equivalent mass of ${\mathrm{Br}}_2$ in this reaction is $\frac{3\mathrm{M}.\mathrm{W}}{5}$. Thus options B, C, and D are incorrect.

Hence, the correct option is A i.e. $\frac{3\mathrm{M}.\mathrm{W}}{5}$.