1000 L of air...

Question

$1000 L$ of air at STP was dissolved in water and it required $2.5 \times 10^{- 5}$ moles of $K M n O_{4}$ for complete reaction with the pollutant $S O_{2}$ . Thus, the $S O_{2}$ content in air is:

1.4 ppm

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14 ppm

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2.8 ppm

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6.25 ppm

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Solution

The correct option is A 1.4 ppm
$S O_{2} + H_{2} O \to H_{2} S O_{3}$
$2 M n O_{4}^{-} + 5 H_{2} S O_{3} \to 2 M n^{2 +} + 3 H_{2} O + 5 S O_{4}^{2 -} + 4 H^{+}$
$2 m o l e s M n O_{4}^{-} = 5 m o l e s o f H_{2} S O_{3} o r S O_{2}$
$2.5 \times 10^{- 5} m o l e s M n O_{4}^{-} = \frac{5}{2} \times 2.5 \times 10^{- 5} m o l e s S O_{2}$
$= 6.25 \times 10^{- 5} m o l e s S O_{2}$
At S.T.P, $6.25 \times 10^{- 5} m o l e s = 6.25 \times 10^{- 5} \times 22.4 L S O_{2}$
$= 1.4 \times 10^{- 3} L$
In 1000 L air, $S O_{2} = 1.4 \times 10^{- 3} L$
$I n 10^{6} L, S O_{2} = \frac{1.4 \times 10^{- 3} \times 10^{6}}{10^{- 3}} = 1.4 p p m$

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