Question

A solution of ${Na}_2{S}_2{O}_3$ is standardized iodometrically against $0.167g$ of ${KBrO}_3$. This process requires $50mL$ of the ${Na}_2{S}_2{O}_3$ solution. What is the normality of the ${Na}_2{S}_2{O}_3$ ?

• A. $0.2N$
• B. $0.12N$
• C. $0.72N$
• D. $0.02N$

Answer 1

The correct option is B $0.12N$

$\begin{array}{c}\text{Here, n-factor of}{KBrO}_3\text{is}6\text{.}\\ \text{So, Equivalents will be balanced.}\\ \text{Equivalent of}{Na}_2{S}_2{O}_3=\text{Equivalent of}{KBrO}_3\\ \Rightarrow N\times \frac{50}{1000}=\frac{0.167}{167}\times 6.\end{array}$

$\therefore N=\frac{0.167}{167}\times 6\times \frac{1000}{50}=0.12N$

So, correct option is (B)