Question

The atmospheric pressure is 1.0 atm and Henry’s law constant for O2 is 1.66 x 10^−6 M/mm Hg at 25 °C. Assume air contains 21% oxygen. Calculate the partial pressure of oxygen. (21% of air is oxygen and the mole fraction of O2 is 0.21).

(a) 180 mm Hg

(b) 130 mm Hg

(c) 120 mm Hg

(d) 160 mm Hg

Answer 1

Correct answer is (d) 160 mm Hg

Easiest explanation: The partial pressure of oxygen is:

P (O2) = (1.0 atm) ((760 mm Hg)/(1 atm)) (0.21) = 160 mm Hg.