Question

A mixture of He and Ne has a density of `1.36 xx 10^(-3) gm //ml` at `0^(@)C` and 2.24 atm. Then mole fraction of neon in this mixture is `"________"`
A. 0.4
B. 0.6
C. 0.3
D. 0.7

Answer 1

Correct Answer - B
`d = 1.36 xx 10^(-3) gm//ml = 1.36 gm// "litre"`
`X_(He) = a`
`X_(Ne) = 1 - a`
`M_(avg) = x_(1)m_(1) + x_(2) m_(2)`
`M_(avg) = 4a + 20 (1 - a)`
`d = (PM_(avg))/(RT)`
`M_(avg) = (dRT)/(P) = (1.36 xx 0.0821 xx 273)/(2.24)`
`= 13.6 gm//mol`
`M_(avg) = 13.6 = 4a + 20 (1 - a)`
`a = 0.4`
`x_(He) = 0.4`
`x_(Ne) = 0.6`