Question

The mixing ratio of dioxygen in the atmosphere is 20.95%. Convert this to an atmospheric concentration into the following

(Use: Universal gas constant, R = 0.08206 atm·L·mol-1·K-1)

a.ppmv

b.mol L-1

c.gm-3 at P° (101325 Pa, 1.00 atm) and 25°C

Answer 1

Let the (v/v)% of dioxygen in atmosphere is =20.95%

(Volume of dioxygen/volume of atmospheric solution)*100

=20.95

Therefore the total volume of atmospheric gaseous solution=100 litre

The volume of dioxygen in atmosphere =20.95 litre

As per mole concept

22.4 litre volume is occupied by =1 mole gas

1 litre volume is occupied by=(1/22.4) mole gas

20.95 litre volume occupied by =(1/22.4)*20.95 mole gas

=0.93 mole of dioxygen

(a)The concentration of dioxygen in mole/litre=

(no of moles of dioxygen/volume of atmospheric gas solution)

=(0.93 mole dioxygen/100 litre solution)

=0.0093 mole/ litre

(b) The concentration of dioxygen in ppm

we know 1 ppm=1mg/L

Mass of 1 mole of dioxygen=32 g

Mass of 0.0093 mole of dioxygen=32*0.0093=0.29 g

=0.29*1000=290 mg

( As I g =1000 mg)

Therefore the concentration in ppm=290 mg/Litre

=290 ppm

(c) The concentration of dioxygen in g/m3

The pressure P=1 atm

Volume of gas =?

No of moles of dioxygen=0.93 mole

R=0.082 atm/mol/K

T=25+273K=298K

Volume=nRT/P(PV=nRT)

0.93 mole*0.082*298/1 atm=22.7 litre

The concentration in g/m3=

Mass of 0.93 moles of dioxygen/22.7 litre

=29.76 g/22.7 litre

(mass of 1 mole dioxygen=32g,mass of 0.93 mole dioxygen

=0.93*32 g=29.76 g)

Therefore the concentration is =

1.31g/L=1.31g/0.001 m3=1310g/m3

   ( 1 L=0.001 m3)