Question

Early evidence for the existence of noble gases was the discrepancy between the apparent molecular mass of nitrogen obtained from the atmosphere (after removal of oxygen, carbon dioxide, and water vapor) and that obtained from the decomposition of ammonium nitrate to nitrogen and water. Results: 28.01 g/mol (pure N2) 28.15 g/mol (N2 contaminated with Ar) If the discrepancy is solely due to argon (MM = 39.95 g mol-1), what is the nAr/nN2 in the atmosphere?

Answer 1

Let there be 'x' mole of Ar per mol of N₂ in the atmosphere.

Mol. weight of N₂ considered pure = 28.01

Mol. weight of N₂ when Ar is mixed = 28.15

Considering a mixture of 1 mol of N₂ and 'x' mole of Ar,

Mol. weight of the mixture = x₁M₁₊x₂M₂

where x1 and x2 are the mol fractions of N2 and Ar in the mixture respectively and given that mol weight of the mixture = 28.15

(1/1+x)28.01+(x/1+x)29.95 = 28.15

28.01+39.95x = 28.15 + 28.15x

0.14 = 11.8x

x = 0.0118

This is relevant with the experimental data