Question

A mixture of helium and neon gases has a density of 0.4175 g/L at 36.8°C and 730 torr. What is the mole fraction of neon in this mixture?

Answer 1

PV = nRT

R = 0.082057 L atm K-1 mol-1

P = 760 torr = 1 atm
V = 1.00 L
T = 309.8 K
n = PV/RT = [1 x 1] / [0.082057×309.8] = 1/ 25.4212 = 0.03934 mol

Consider, no. of moles of He = x/4.0026;

no. of mol of Ne = (0.4175-x)/20.1798
Hence (x/4.0026) + (0.4175-x)/20.1798 = 0.03934

20.1798x + 1.6711 -4.0026 x = 3.1775

16.177 x = 1.5064

x = 0.09312

therefore;

weight of He = 0.09312
moles of He = 0.02326 mol
mol of Ne = (0.4175 – 0.09312)/ 20.1798 = 0.016074 mol
mol fraction of Ne = 0.016074/0.039334 = 0.408 or 40.8 %