Chemistry - Redox Reactions Question with Solution | TestHub

To refine the solution, we need to correctly apply the concept of normality and equivalents.

The reaction involves BrO₃⁻ changing to Br⁻.

The change in oxidation state for Br is from +5 to -1, which is a change of 6 electrons.

Therefore, the n-factor for KBrO₃ is 6.

Number of equivalents of KBrO₃ = (mass / molar mass) * n-factor

$$=\frac{0.1262g}{167g/mol}\times 6$$

$$=0.004536equivalents$$

For standardization, equivalents of Na₂S₂O₃ = equivalents of KBrO₃.

Normality (N) = equivalents / Volume (L)

$$N\times 0.045L=0.004536equivalents$$

$$N=\frac{0.004536}{0.045}=0.1008N$$

Rounding to one significant figure, the normality is 0.1 N.

The final answer is $\boxed{0.1}$.