1.

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 "________"

Answer»

0.4
0.6
0.3
0.7

Solution :`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`


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