How many molecules of `CO_(2)` are present in one litres of air containing 0.03 % volume of `CO_(2)` at N.T.P ?

Explanation

100 litres air contain `CO_(2)` at N.T.P = 0.03 L
1 litre of air contains `CO_(2)` at N.T.P `= (0.03)/(100)L= 3xx10^(-4)L`
Now, `22.4 L` of `CO_(2)` at N.T.P contain molecules `= 6.022 xx 10^(23)`
`:. 3 xx 10^(-4) L` of `CO_(2)` at N.T.P will contain molecules `= (6.022xx10^(23))/((22.4L))xx(3xx10^(-4)L)`
`= 8.07 xx 10^(18)` molecules.