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Oxides of Sulphur

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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Similar questions

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 :

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:

Q. A sample of

F e S O_{4}

and

F e C_{2} O_{4}

is dissolved in

H_{2} S O_{4}

. The complete oxidation of sample required 8/3 eqv. of

K M n O_{4}

. After oxidation, the reaction mixture was reduced by

Z n

. On further oxidation by

K M n O_{4}

5/3 eqv were used. The moles of

F e C_{2} O_{4}

in the mixture is

Q. How is

S O_{2}

an air pollutant?

Q. Henry's law constant for the solubility of

N_{2}

gas in water at

298

K is

1.0 \times 10^{5}

atm. The mole fraction of

N_{2}

in air is

0.6

. The number of moles of

N_{2}

from air dissolved in

10

moles of water at

298

K and

5

atm pressure is:

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

$1000 L$ 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: