Question

What is the residence time of a pollutant XY if its atmospheric concentration is 6.0 ppm and 300 million tons are released into the atmospheric yearly? XY has a molecular weight of 50 g per mole. The total amount of air in the atmosphere is 1.8×10²⁰ moles.

Answer 1

Number of moles of gas in the atmosphere = 1.8*10²⁰ moles

Mass of XY released = 300 million tons/year = 300*10⁶ tons/year

Conver tons into kgs

1 ton = 907.2 kg

So,  300*10⁶ tons/year = 907.2* 300*10⁶ tons/year = 2721*10⁸ kg/year

Concentration of XY in Atmosphere = 6.0 ppm = 6.0*10^-6

Moles of XY = (6.0*10^-6)*(1.8*10²⁰) = 10.8*10¹⁴ mol

molar mass of XY = 50 g/mol

So, mass of XY present = moles*molar mass = (10.8*10¹⁴ )*50 = 540*10¹⁴ g = 540*10¹¹ kg

Residence time = mass of XY present/mass of XY released per year

= (540*10¹¹)/(2721*10⁸) = 198 years