Question

What will be the value of the equivalent weight of $KBr{O}_3$ in the following ionic equation?
$2Br{O}_3^{-}+12H^{+}+10e^{-}\to B{r}_2+6H_{2}O$
(Given: The molecular weight of $KBr{O}_3$ is $M$.)

• A. $\frac{M}{4}$
• B. $\frac{M}{6}$
• C. $\frac{M}{10}$
• D. $\frac{M}{5}$

Answer 1

The correct option is D $\frac{M}{5}$

$2Br{O}_3^{-}+12H^{+}+10e^{-}\to B{r}_2+6H_{2}O$
$E_{KBr{O}_3}=\frac{\text{molecular weight}}{\text{number of electrons gained}}=\frac{2Br{O}_3^{-}}{10}\Rightarrow \frac{Br{O}_3^{-}}{5}=\frac{KBr{O}_3}{5}=\frac{M}{5}$
Where $M$ stands for molecular weight of $KBr{O}_3$.