Question

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

Answer 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