In: Chemistry
Compare the number of moles of nitrogen-oxygen species in air containing 300 ppbv nitric oxide, with fog consisting of 10,000 droplets per cm^3, having an average diameter of 2 um, and containing nitrate ion at a concentration of 3 * 10^5 moles L-1
1 m3 = 1000 L contains
10 000 droplets/ cm3 x 1000 cm3/L x 1000 L/ m3 x (4/3) x 3.14 x 1 x 10-18 m3 /doplet =
= (4/3) x 3.14 x 10-8 m3 droplet volume / m3 air
= 4.18 x 10-5 L droplets with [NO3-] = 3 x 10-5 mol/L
The number of NO3- mol in 1 m3 air is
3 x 10-5 mol/L x 4.18 x 10-5 L = 1.25 x 10-9 NO3- mol
300 ppb (volume) = 300 x 10-9 L NO/ L air
300 x 10-9 L NO/ L air x 1000 L air = 3.00 x 10-4L ( 0.3 mL)
3.00 x 10-4L / 22.4 L/mol = 1.34x10-5 mol NO in 1 m3 air
1.34x10-5 mol NO in 1 m3 air / 1.25 x 10-9 mol NO3- in 1 m3 air = 1.07 x 104
There are about 10 700 more NO mol in air than NO3- mol in the fog