Some Basic Concepts of Chemistry 2 Question 38
38. A $5.0 \mathrm{~cm}^{3}$ solution of $\mathrm{H}{2} \mathrm{O}{2}$ liberates $0.508 \mathrm{~g}$ of iodine from an acidified KI solution. Calculate the strength of $\mathrm{H}{2} \mathrm{O}{2}$ solution in terms of volume strength at STP.
(1995, 3M)
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Solution:
- The redox reaction involved is :
$$ \mathrm{H}{2} \mathrm{O}{2}+2 \mathrm{I}^{-}+2 \mathrm{H}^{+} \longrightarrow 2 \mathrm{H}{2} \mathrm{O}+\mathrm{I}{2} $$
If $M$ is molarity of $\mathrm{H}{2} \mathrm{O}{2}$ solution, then
$$ \begin{aligned} & 5 M=\frac{0.508 \times 1000}{254}\left(\because 1 \text { mole }{2} \mathrm{O}{2} \equiv 1 \text { mole }_{2}\right) \ & \Rightarrow \quad M=0.4 \end{aligned} $$
Also, $n$-factor of $\mathrm{H}{2} \mathrm{O}{2}$ is 2, therefore normality of $\mathrm{H}{2} \mathrm{O}{2}$ solution is $0.8 \mathrm{~N}$.
$\Rightarrow$ Volume strength $=$ Normality $\times 5.6=0.8 \times 5.6=4.48 \mathrm{~V}$
