Some Basic Concepts of Chemistry - Result Question 36
38. A $5.0 \mathrm{~cm}^3$ solution of $\mathrm{H}_2 \mathrm{O}_2$ liberates 0.508 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{array}{rlrl} & & 5 M & =\frac{0.508 \times 1000}{254}\left(\because 1 \text { mole } \mathrm{H}_2 \mathrm{O}_2 \equiv 1 \mathrm{~mole} \ \mathrm{H}_2 \mathrm{O}_2\right) \\ \Rightarrow \quad & M & =0.4 $\begin{array}
Also, $n$-factor of $\mathrm{H}_2 \mathrm{O}_2$ is 2, therefore normality of $\mathrm{H}_2 \mathrm{O}_2$ solution is 1.6 N .
$\Rightarrow$ Volume strength $=$ Normality $\times 5.6=0.8 \times 5.6=4.48 \mathrm{~V}$
BETA
