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

The correct answer is M/5 (Option 1).

Understanding the Problem:
We need to find the equivalent weight of KBrO₃ using the given ionic equation:

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

Step 1: Recall the formula for Equivalent Weight
The equivalent weight of a substance in a redox reaction is defined as:

$\text{Equivalent Weight}=\frac{\text{Molar Mass (M)}}{\text{n-factor}}$

where the n-factor is the number of electrons gained or lost per molecule of the substance.

Step 2: Determine the change in oxidation state of Bromine
On the reactant side, in KBrO₃, the oxidation state of Br can be calculated:
K is +1, each O is −2, so: +1 + x + 3(−2) = 0 ⇒ x = +5.
Therefore, Br is in the +5 oxidation state in KBrO₃.

On the product side, in Br₂, bromine is in its elemental form, so its oxidation state is 0.

Step 3: Calculate the number of electrons transferred per molecule of KBrO₃
The change in oxidation state per bromine atom = +5 → 0 = 5 electrons gained per Br atom.

Each molecule of KBrO₃ contains one Br atom, so:

n-factor of KBrO₃ = 5

Step 4: Verify using the ionic equation
The equation shows that 2 moles of KBrO₃ accept 10 electrons in total.
So, per mole of KBrO₃: 10 ÷ 2 = 5 electrons.
This confirms the n-factor = 5.

Step 5: Calculate the Equivalent Weight

$\text{Equivalent Weight}=\frac{M}{5}$

Therefore, the equivalent weight of KBrO₃ is M/5.