Question

In the ionic equation $2K^{+}+Br{O}^{3-}+12H^{+}+10e^{-}\to B{r}_2+6H_{2}O+2K$ the equivalent weight of $KBr{O}_3$ will be:

• A. $M/5$
• B. $M/3$
• C. $M/6$
• D. $M/4$

Answer 1

Answer: (A)

$M/5$

$2K^{+}+Br{O}_3^{-}+12H^{+}+10e^{-}\to B{r}_2+6H_{2}O+2K$

$(+5)$ $0$

Since, it is a disproportionate reaction, so equivalent weight in this reaction is defined as the molar mass divided by n-factor.

n-factor is the number of electrons lost or gained.

So for above reaction, $n-factor=5$

As $1$ mole of $B{r}_2$ is produced in the reaction,

So, $MassofB{r}_2=M$

$\therefore Equivalentweight=\frac{M}{5}$