1000L of air at STP was dissolved in water and required 2-5- 10 -5 moles of KMn O 4 for complete reaction of SO2 at pollutants- Thus- SO2 content in air is -

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Gay Lussac's Law

1000L of air ...

Question

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

1.4 p p m

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14 p p m

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2.8 p p m

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6.25 p p m

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Solution

The correct option is A $1.4 p p m$
Given:
1000L of air is dissolved at Standard Temperature Pressure into Water and then requires $2.5$ x $10^{- 5}$ moles of $K M n O_{4}$ for complete reaction of $S O_{2}$ which is dissolved in Water.

Solution:
Eqn 1:
$S O_{2}$ + $H_{2} O$ $\to$ $H_{2} S O_{3}$
Eqn2:
$2 M n O_{4}^{-} + 2 H_{2} S O_{3}^{-} \to 5 H_{2} S O_{4}^{-} + 2 M n^{2 +}$
It implies 2 moles of $M n O_{4}$ requires 5 moles of $S O_{2}$ .
Therefore, 2.5 x $10^{- 3}$ moles of $K M n O_{4}$ requires = $\frac{5}{2}$ x 2.5 x $10^{-} 3$ moles of $S O_{2}$ = 6.25 x $10^{- 3}$ moles

At STP, 6.25 x $10^{-} 3$ moles = 6.25 x 22.4 x $10^{- 3}$ = $1.4$ x $10^{- 3} L$

In 1000L, $\frac{1.4}{1000}$ x $10^{- 3}$ = 1.4 ppm.

So, the Correct option : A

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1000L of air at STP was dissolved in water and required 2.5×10−5 moles of KMnO4 for complete reaction of SO2 at pollutants. Thus, SO2 content in air is :

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