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Raoult's Law and Vapour Composition
Concepts tested here
- Composition of the vapour phase
All Questions
2005 AIPMT 1 question
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A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at $20^\circ C$ are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be -Mole ratio pentane : hexane = 1 : 4, so in the liquid $x_{pentane} = \dfrac{1}{5} = 0.2$ and $x_{hexane} = \dfrac{4}{5} = 0.8$.
Partial pressure of pentane: $p_{pentane} = 0.2 \times 440 = 88$ mm Hg
Partial pressure of hexane: $p_{hexane} = 0.8 \times 120 = 96$ mm Hg
Total vapour pressure $= 88 + 96 = 184$ mm Hg
By Dalton's law, the mole fraction of pentane in the vapour $= \dfrac{p_{pentane}}{P_{total}} = \dfrac{88}{184}$
$= 0.478$
The vapour is richer in pentane (0.478) than the liquid (0.200), because pentane is the more volatile component.