The density of air is 0.001293gm per c.c. it's vapour density will be at STP
Vapour density = (Molecular weight) / 2
Now 1 mole of any gas under standard Temperature ans pressure (STP) occupies 22.4 litres
That is : 1 mole of gas at STP = 22.4 L = 22400 ml
Now,
The density of air is given to us as 0.001293g/ml
Volume of gas = 22400 ml
Now we know that :-
Density = mass / volume
therefore, mass = density volume
Therefore,
Mass of air = 0.001293 22400 = 28.96 g
And from the above formula given at top we get:-
Vapour density = 28.96 / 2=14.48 gm
Therefore, the vapour density of air is = 14.48 gm
Hello,
Here is the solution for your question:
At STP,22.4 litres of gas is equal to 1 mole .
Given density =0.001293 g/cm^3
1 mole has 0.001293 gram
22.4 has 0.001293*22.4/10^-3
Molecular mass= 28.97
Vapour density=Molecular mass/2= 14.485
Hope this helps!