Question

A mixture of helium and neon gases has a density of 0.4430 g/L at 31.8°C and 606 torr. What is the mole fraction of neon in this mixture?

Answer 1

T = 31.8 + 273 = 304.8 K

Pressure (P) = 606 torr = 606 / 760 = 0.797 atm

density (d) = 0.4430 g /L

Volume (V) = 1 Litre

R = universal gas constant = 0.0821 L-atm / mol K

PV = nRT

0.797 x 1 = n x 0.0821 x 304.8

n = 0.03185

mass =0.4430

let mass of He = x

mass of Ne = 0.4430 -x

moles of He = x / 4

moles of Ne = 0.4430 -x / 20.18

total moles = x /4 + 0.4430 -x / 20.18

x /4 + 0.4430 -x / 20.18   = 0.03185

20.18 x + 1.772 -4x = 2.571

16.18 x = 0.7989

x = 0.04938

mass of He = 0.04938

mass of Ne = 0.4430 -x= 0.4430 - 0.04938 = 0.3936 g

moles of He = 0.04938 / 4 = 0.012345

moles of Ne = 0.3936 / 20.18 = 0.0195

mole fraction of Ne = moles of Ne / total moles

                               = 0.0195 / 0.03185

                              = 0.6122

mole fraction of Ne = 0.6122

or 61.22%