Amixture of helium and neon gases has a density of 0.6048 g/l at 48.7°c and 792 torr. what is the mole fraction of neon in this mixture?

The volume of one mole of gas is 22.4 liters at 0 °C and 760 Torr. The volume of the given condition can be calculated using:
(PV)/T = constant
(P₁V₁)/T₁ = (P₂V₂)/T₂
(760 * 22.4)/273 = (792 * V₂) / 321.7
V₂ = 25.3 Liters
The Mr of He = 4
The Mr of Ne = 20
Density of He = 4/25.3 = 0.1581 g/L
Density of Ne = 20/25.3 = 0.7905 g/L
Let x be the fraction of Ne
1 - x is the fraction of He
avg density = sum[(mole fraction * Mr)/V]
0.6048 = (20x + 4(1 - x))/25.3
x = 0.70

The answer is b metal denting

16- counting by fours