The density of air is 0.001293gm per c.c. it's vapour density will be at STP
Answers (2)
16th Sep, 2020
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
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
Comments (0)
16th Sep, 2020
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!
Comments (0)
