Q. 393.6( 14 Votes )

# 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 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 10^{7} mm and 6.51 × 10^{7} mm respectively, calculate the composition of these gases in water.

Answer :

Given-

K_{H} for O_{2} = 3.30 × 10^{7} mm Hg,

K_{H} for N_{2} = 6.51 × 10^{7} mm Hg

Percentage of oxygen (O_{2}) = 20 %

Percentage of nitrogen (N_{2}) = 79%

Total pressure = 10 atm

Using Henry’s law,

→

where, p is the partial pressure of gas in the solution and K_{H} is Henry’s constant.

Now, to determine the mole fraction of oxygen in solution, , we use

To determine the mole fraction of nitrogen in solution, ,

__Thus, the__ __mole fraction of oxygen in solution, x _{oxy} = 4.61x10^{-5}__

__and the__ __mole fraction of nitrogen in solution, x _{nit} is 9.22x10^{-5}__

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