If air has an average density of 1.29 g/l and Co is reported at 2.0 ppm, what volume of air will contain 1.00 mole of Co?

There will have to be 1/(2*10^-6) = 500,000 times as much mass of air as CO. The mass of the CO is 28.0 g (1.00 mole), so that requires 14*10^6 g or air. Divided that by the density (1.29 g/l) for the volume in liters.