Question

At Nikel, Russia, the annual average concentration of sulfur dioxide is observed to be 50 mu g/m^3 at 15 degree C and 1 atm. What is this concentration of SO_2 in parts per billion?

Answer 1

It is known that

1 m³ = 1000 L.

Concentration of SO₂ = 50 μg/m³

Volume of air = 1000 L; temperature of air = 15ºC = (15 + 273) K = 288 K; pressure of air = 1 atm.

Molar mass of SO₂ = 64 g/mol.

Determine moles of SO₂ corresponding to 50 μg/m³.

Moles SO₂ = (volume of air)*(50 μg/m³)

= (1 m³)*(50 μg/m³)

= 50 μg

= [(50 μg)*(1 g)/(1.0*10⁶ μg)*(1 mol)/(64 g)]

= 7.8125*10^-7 mole.

Moles air = PV/RT

= (1 atm)*(1000 L)/(0.082 L-atm/mol.K)*(288 K)

= 42.3442 mole.

Mixing ratio = (moles SO₂)/(moles air)

= (7.8125*10^-7 mole)/(42.3442 mole)

= 1.8450*10^-8

≈ 18.45*10^-9

This simply means that there are 18.45*10^-9 moles SO₂ in 1 mole air.

A concentration of 1 mole SO₂ in 1.0*10⁹ moles air or more clearly, a concentration of 1.0*10^-9 moles SO₂ in 1 mole air is defined as a concentration of 1 ppb.

Therefore, concentration of SO₂ = 18.45 ppb (ans).