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Ratio of the molar volume of pure B as a vapour to its molar volume as a liquid in an ideal solution

Type the stem's data into a dark neon bench, press run, and watch it count 1000 g of A into a sealed container mole by mole, drop B in by molality and build the mole fraction from the dots, boil the solution until the gauge reaches the measured total and solve Raoult's law live for the vapour pressure of pure B, then fill two chambers with one mole of B each - a tiny liquid cube beside a vapour cube drawn at the honest edge ratio - before spraying the ratio across the screen.

Sub-topics covered

• Molality to mole fraction, and why the solvent's molar mass is needed
• Raoult's law for a binary ideal solution
• Vapour pressure of a pure component recovered from a measured total
• Molar volume of a liquid from its density
• Molar volume of an ideal gas at a stated pressure
• Converting mm Hg to atm to match the units of R
• Dimensionless ratios of two molar quantities

What it teaches

• Molality counts moles of solute per kilogram of solvent while mole fraction counts them against every mole present, so a molality always has to be converted before Raoult's law can be used - and the conversion is what the molar mass of A is given for
• The mass of solvent taken is a free choice: x(B) = m / (m + 1000/M(A)) contains no w(A), so the answer is the same for 500 g or 2500 g of A
• For an ideal solution each component's partial pressure is its pure vapour pressure times its own mole fraction, and the total is a straight line between p(A) and p(B)
• With p(total), p(A) and both mole fractions known, Raoult's law has exactly one unknown left, which is what makes the first half of this question a one-line solve
• A liquid's molar volume comes from density alone - M/rho - with no temperature and no pressure in it, because its molecules are already touching
• Molar volume of pure B in the vapour phase means one mole of B alone at its own vapour pressure p(B), not at the mixture's total pressure and not at 1 atm - this is the single decision the whole question turns on
• R quoted in L atm K-1 mol-1 forces the pressure into atmospheres, which is why the stem also hands over 1 atm = 760 mm Hg
• Both molar volumes are per mole, so the ratio is a pure number: RTrho x 1000 / (M x p)

Simulation ID

• ADV-2026-P2-CHE-Q15 · revision 1

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