The solubility of nitrogen in water is mol/L at when the pressure above water is atm. Calculate the Henry’s law constant for in units of for Henry’s law in the form , where is the gas concentration in mol/L. Calculate the solubility of in water when the partial pressure of nitrogen above water is atm at 0°C.

The solubility of nitrogen in water is 8.21 x 10 -4 mol/L at 0°C when the N2 pressure above water is 0.790 atm. Calculate the Henry’s law constant for N2 in units of mol/L • atm for Henry’s law in the form C = kP, where C is the gas concentration in mol/L. Calculate the solubility of N2 in water when the partial pressure of nitrogen above water is 1.10 atm at 0°C.

The answer to the question is

1.143 × 10⁻³ mol/L

Explanation:

Henry's law states that the concentration of a gas in a solution is directly proportional to the partial pressure of the same gas existing above the solution at constant temperature

or c ∝ P which gives c = k × P

where c = concentration

k = constant of proportionality

P = Partial pressure of the same gas above the solution

The solubility of nitrogen = 8.21×10⁻⁴ M/L

The pressure above water = 0.790 atm

Therefore from C = kP

8.21×10⁻⁴ M/L = k×0.790 atm

k = 1.03924×10⁻³mol/(L·atm)

Hence when partial pressure = 1.10 atm we have

C = k×P = 1.03924×10⁻³mol/(L·atm) × 1.10 atm = 1.143×10⁻³ mol/L

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i think   the answer is c hope i'm right