Question

A mixture of Ar and N2 gases has a density of 1.477 g/L at STP. What is the mole fraction of each gas?

Answer 1

consider; X = mole fraction Ar

then, 1-X = mole fraction N2.

Now find the weight per liter of each gas at STP

For Ar, Molecular wt over 22.4 liters/mole
= 39.94 /22.4 liters per mole = 1.78 g/liter at STP

for N2 = 28g/ 22.4 liters per mole = 1.25 g/liter

Total density = 1.477

this can be represented as, (density = mass/volume)
(mole fraction Ar) wt./liter Ar) + (mole fraction N2) wt./liter N2= 1.477 g/liter of mixture.

so X (1.78g/Liter Ar) + (1-X) (1.25g/liter N2) = 1.477 g/liter mixture

now solve for X and 1-x

1.78X + 1.25-1.25X = 1.477/liter

gather terms .53X + 1.25 = 1.477

and 0.53X = 0.227

X = 0.4283

So mole fraction Ar = 0.4283 and mole fraction N2 = 1-0.4283 which is = 0.5717