The solubility of pure Nitrogen gas at 25C and 1atm is 6.8*10^-4mol/L. What is the concentration of Nitrogen dissolved in water under atmospheric conditions? The partial pressure of Nitrogen gas in atmosphere is 0.78atm
Asked by hilhew7 | 6th Jun, 2015, 07:20: PM
Expert Answer:
Since in this problem ‘k’ is not given so we have to find out.
Applying Henry’s equation,S = k P
k= S/P = 6.8 X 10-4 mol L-1 = 6.8 X 10-4 mol L-1
1
So the solubility of N2 at 0.78 atm:
S= (6.8 X 10-4 mol L-1 ) (0.78 atm) = 5.3 x10-4 M
Since in this problem ‘k’ is not given so we have to find out.
S = k P
k= S/P = 6.8 X 10-4 mol L-1 = 6.8 X 10-4 mol L-1
1
So the solubility of N2 at 0.78 atm:
S= (6.8 X 10-4 mol L-1 ) (0.78 atm) = 5.3 x10-4 M
Since in this problem ‘k’ is not given so we have to find out.
Applying Henry’s equation,S = k P
k= S/P = 6.8 X 10-4 mol L-1 = 6.8 X 10-4 mol L-1
1
So the solubility of N2 at 0.78 atm:
S= (6.8 X 10-4 mol L-1 ) (0.78 atm) = 5.3 x10-4 M
Since in this problem ‘k’ is not given so we have to find out.
Applying Henry’s equation,S = k P
k= S/P = 6.8 X 10-4 mol L-1 = 6.8 X 10-4 mol L-1
1
So the solubility of N2 at 0.78 atm:
S= (6.8 X 10-4 mol L-1 ) (0.78 atm) = 5.3 x10-4 M
Since in this problem ‘k’ is not given so we have to find out.
S = k P
k= S/P = 6.8 X 10-4 mol L-1 = 6.8 X 10-4 mol L-1
1
So the solubility of N2 at 0.78 atm:
S= (6.8 X 10-4 mol L-1 ) (0.78 atm) = 5.3 x10-4 M
Since in this problem ‘k’ is not given so we have to find out.
Applying Henry’s equation,S = k P
k= S/P = 6.8 X 10-4 mol L-1 = 6.8 X 10-4 mol L-1
1
So the solubility of N2 at 0.78 atm:
S= (6.8 X 10-4 mol L-1 ) (0.78 atm) = 5.3 x10-4 M
Answered by Vaibhav Chavan | 7th Jun, 2015, 01:03: AM
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