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# The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. - CBSE (Science) Class 12 - Chemistry

ConceptSolubility Solubility of a Gas in a Liquid

#### Question

The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 Kif the Henry’s law constants for oxygen and nitrogen are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.

#### Solution

Percentage of oxygen (O2) in air = 20 %

Percentage of nitrogen (N2) in air = 79%

Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, (10 × 760) mm Hg = 7600 mm Hg

Therefore,

Partial pressure of oxygen,  p_(o_(2)) = 20/100xx7600 mm Hg

= 1520 mm Hg

Partial pressure of oxygen,  p_(N_(2)) = 79/100xx7600 mm Hg

= 6004 mmHg

Now, according to Henry’s law:

KH.x

For oxygen:

p_(o_(2))  =K_H.x_(o_2)

=> x_(o_2) = p_(o_(2))/K_H

= (1520 "mm Hg")/3.30xx10^(7) mm Hg   (Given K_H = 3.30 xx 10^(7) mm Hg)

= 4.61 x 10-5

For nitrogen:

p_(N_(2)) = K_H.x_(N_(2))

=>x_(N_(2)) = p_(N_(2))/K_H

 = (6004 "mm Hg")/(6.51xx10^(7)"mm Hg")

= 9.22 x 10-5

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10−5and 9.22 × 10−5 respectively.

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Solution The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. Concept: Solubility - Solubility of a Gas in a Liquid.
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