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Mole fraction of B in the vapour phase in equilibrium with an ideal solution

Type the stem's data into a dark equilibrium bench and run it through five stages: weigh out a kilogram of A and add B by molality, build the liquid mole fraction from the molecules on screen, let the vessel reach equilibrium and split the pressure gauge into p(A) and p(B) by Raoult's law, then put twenty-five molecules of each phase side by side and count - five are B in the liquid and only four in the vapour, which is the whole question.

Sub-topics covered

• Molality to mole fraction, and why the solvent's molar mass is needed
• Raoult's law for a binary ideal solution
• Partial pressure of the second component from the measured total
• Dalton's law of partial pressures in the vapour
• Liquid-phase x against vapour-phase y
• Relative volatility and enrichment of the vapour
• Recognising data in a stem that the question does not need

What it teaches

• Molality counts moles of B per kilogram of A while mole fraction counts them against every mole present, so the molality has to be converted before Raoult's law can be used - and the molar mass of A is given for exactly that
• Raoult's law uses the LIQUID composition: each component's partial pressure is its own pure vapour pressure scaled by its own mole fraction in the liquid
• Once the total pressure is known and one partial pressure has been calculated, the other is simply what is left over - no second application of Raoult's law is needed
• Dalton's law governs the vapour: in an ideal gas mixture each component's partial pressure is its mole fraction IN THE VAPOUR times the total, so the vapour composition is just the pressure share
• x and y are different numbers for the same vessel at the same moment, and the question is built around a student handing in x
• Combining the two laws gives y(i) = x(i)p(i)deg/p(total), so a component less volatile than the mixture as a whole is depleted in the vapour and one more volatile is enriched
• Here p(B)deg works out lower than p(A)deg, so B is the less volatile liquid and the vapour ends up poorer in B than the liquid it came from - the enrichment factor is exactly p(B)deg/p(total)
• The vapour being richer in the more volatile component is the entire basis of fractional distillation

Simulation ID

• ADV-2026-P2-CHE-Q16 · revision 1

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