Question

A gas mixture has a denstiy of 1.0628g/l at 30C and 740 torr ? What is the possible component of the Mixture ? Answer: Ne-Ar ? What is the fraction of the lighter gas in the mixture ? Answer: 0.64

Answer 1

p*V=n*R*T
p*V=(mass/molar mass)*R*T
p*molar mass=(mass/V)*R*T
p*molar mass=density*R*T

p= 740 torr= 740/760 atm = 0.974 atm
density = 1.0628 g/L
R = 0.0821 atm-L/mol-K
T = 30 oC = (30+ 273) K = 303 K
p*molar mass=density*R*T
0.974*molar mass = 1.0628 * 0.0821 * 303
molar mass = 27.144 g/mol
Since molar mass is between Neon and Argon.
Its Ne-Ar mixture

Molar mass of Ne = 20 g/mol
Let fraction of Ne be x
Molar mass of Ar = 40 g/mol
Let fraction of Ar be 1-x

Net molar mass = x*20 + (1-x)*40
27.144 = 20*x +40 - 40*x
20*x = 12.856
x=0.64
Answer: 0.64