posted by Dave .
When air pollution is high, ozone (O3) contents can reach 0.60 ppm (ie., .60 mol ozone per million mol air). How many molecules of ozone are present per liter of polluted air if the barometric pressure is 755 mm Hg and the temperature is 79 degrees F?
I converted temperature to K and pressure to atm but now I'm stuck. Where do i go from here?
You know T, P, and V (one liter)
Solve for n, the moles of air. PV= nRT
once you have the number of moles of air, multiply by .60*10^-6 moleozone/moleair.
That will give you moles of ozone.
Ok that makes sense, but why is it .60*10^-6 instead of .60*10^6?
0.6 ppm = 0.6/106 = 0.6 x 10-6.