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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. 0.4B. 0.6C. 0.3D. 0.7 |
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Answer» 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` |
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