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
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