A mixture of argon and nitrogen has a density of 1.40gL^-1 at STP. What is mole fraction of nitrogen.

At STP conditions 1mole of any gas occupies 22.4L of volume, so x moles of gas occupies 22.4x L of volume.

Let us take there are ''A'' moles of Argon and N moles of Nitrogen in the mixture.

Then the total mass of the mixture  = (40A+28N) gm.    (since molecular weights of Argon and Nitrogen are 40 & 28)

Total Volume of the mixture = 22.4(A+N) Liters.

So, density of the mixture = (40A+28N)/22.4(A+N)  gm/Liter

Given density of the mixture = 1.4 gm/Liter

Equating them we get  A = (8/27)N

   Adding N on both sides, we get

      A+N =(35/27)N  =>  [N/(A+N)] = 27/35

        So, mole fraction of Nitrogen in the above mixture is 27/35

let no. of moles of nitrogen per be x and that of argon be y.

density=1.4=(no. of moles of nitrogen*molar mass of nitrogen +no. of moles of argon*molar mass of argon)/volume

d=(28x + 40y)/[22.4*(x+y)]=1.4

3.36x=8.64y

x=2.57y

mole fraction of nitrogen=x/(x+y)

=2.57y/(2.57y+y)

=0.72