In class 11th in chapter of mole concept of chemistry subject we study a formula for valour density of gases which is :-

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!