Question

The solubility of pure nitrogen gas at 25°c and 1 atm is 6.8× 10-⁴

Answer 1

Answer:

Explanation:

ince 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

Answer 2

Answer:

partial pressure of N2 at 0.78 atm is 5.304 x 10^-4 mol/L

Explanation:

Acc to Henry's law;

partial pressure is directly proportional to mole fraction(solubility)

X1/X2=P1/P2

6.8 x 10^-4/X2=1/0.78

X2=6.8 x 10^-4 x 0.78+

    = 5.304 x 10^-4

Hope you have understood