Respuesta :
Answer:
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
