CBSE Class 12-science Answered
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
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