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Byju's Answer

Standard XII

Chemistry

Concentration Terms

For estimatin...

Question

For estimating ozone in the air, a certain volume of air is passed through an acidified potassium iodide solution. Oxygen is evolved and iodide is then oxidized to give iodine. When such a solution is acidified, free iodine is evolved which can be titrated with standard sodium thio-sulphate solution. In an experiment $10 l i t r e$ of air at $1$ atm and $27^{o} C$ were passed through an alkaline $K I$ solution, at the end, the iodine entrapped in the solution on titration as above required $1.5 l i t r e$ of $0.01 N N a_{2} S_{2} O_{3}$ solution. Calculate volume percent of $0. 3$ in sample.

1.8

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4.6

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3.2

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Solution

The correct option is D $1.8$
$2 K I + O_{3} + H_{2} O ⟶ 2 K O H + I_{2} + O_{2}$
Meq. of $I_{2} =$ Meq. of $N a_{2} S_{2} O_{3}$ $= 150 \times 0.01 = 15$ (valence factor of $I_{2} = 2$ )
$∴$ mM of $I_{2} = \frac{15}{2} = 7.5$
Also, mM of $O_{3} =$ mM of $I_{2}$ (mole ratio 1:1)
$∴$ mM of $O_{3} = 7.5$
or mole of $O_{3} = 7.5 \times 10^{- 3}$
$∴ P_{O_{3}} = \frac{n R T}{V} = \frac{7.5 \times 10^{- 3} \times 0.0821 \times 300}{10}$ $= 184.725 \times 10^{- 4}$ atm
$∴ V o l % o f O_{3} = \frac{184.725 \times 10^{- 4} \times 100}{1} = 1.847 %$

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