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.

To find the volume of air that will contain 1.00 mole of Co (Cobalt), we need to determine the amount of air in grams needed to have the same mass as 1.00 mole of Co.

Since the mass of 1.00 mole of Co is 28.0 grams, we need to find the mass of air that is 500,000 times as much as Co. This can be calculated by multiplying the mass of Co by 500,000:

Mass of air = Mass of Co × 500,000 = 28.0 g × 500,000 = 14,000,000 g

Now, we can use the density of air to convert the mass of air to volume in liters. The density of air is given as 1.29 g per liter.

Volume of air = Mass of air / Density of air = 14,000,000 g / 1.29 g/L ≈ 10,853,658.54 L

Therefore, the volume of air that will contain 1.00 mole of Co is approximately 10,853,658.54 liters.