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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
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