Question

A sample of air (containing 20% 02and 80% N2) is dissolved in water when total pressure of air is 1atm The solubility of N2and 02( in terms of mole fraction) is x and y. This sample is taken at height of H. above sea level where pressure of air is 0.4 atm and temperature is 300 K and solubility of N2 and O2

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Air contains O

2

and N

2

in the ratio of 1:4. Calculate the ratio of solubilities in terms of mole fraction of O

2

and N

2

dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O

2

and N

2

are 3.30×10

7

torr and 6.60×10

7

torr respectively.

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Answer

Given,

The ratio of the O

2

&N

2

=1:4

So, pO

2

=K

H

XO

2

1=K

H

2

XO

2

XO

2

=

K

H

2

1

pN

2

=K

H

XN

2

1=KH

N

2

XN

2

XN

2

=

K

H

N

2

1

Thus the ratio is:

K

H

N

2

1

K

H

2

1

XO

2

:XN

2

=

3.33×10

7

1

:

6.6×10

7

1

1:2