Chemistry · Solutions · NEET
For a solution of two volatile liquids A and B, each liquid gives off its own vapour. Raoult's Law says the partial vapour pressure of A equals the vapour pressure of pure A times the mole fraction of A in the liquid. In symbols: p_A = p°_A × x_A and p_B = p°_B × x_B. Here p°_A is the pressure pure A would give alone, and x_A is how much of the liquid is A. The same rule holds for B.
You just add the two partial pressures. Total pressure P = p_A + p_B = p°_A x_A + p°_B x_B. Because x_B = 1 − x_A, you can also write P = p°_B + (p°_A − p°_B) x_A. This is a straight line: the total pressure changes smoothly from p°_B (pure B) to p°_A (pure A) as you change the composition. This straight-line behaviour is the sign of an ideal solution.
They are usually NOT the same. The mole fraction in the liquid (x) is used INSIDE Raoult's Law to find partial pressures. The mole fraction in the vapour (y) is found afterwards using Dalton's Law: y_A = p_A / P_total. The vapour is always richer in the more volatile liquid — the one with the higher pure vapour pressure. So if benzene has a higher p°, the vapour has more benzene than the liquid does. NEET loves this point.
It is the same core idea but written differently. For two volatile liquids, both components follow p = p° x, so both appear in the total pressure. When the solute is non-volatile (like sugar or urea), only the solvent gives vapour, so p_solution = p°_solvent × x_solvent. That special case leads to 'relative lowering of vapour pressure'. Both come from the one law: partial pressure of any component = its pure vapour pressure × its liquid mole fraction.
Because the total pressure is a weighted average of the two pure pressures. If you add more of the low-p° liquid, you increase its mole fraction and decrease the other one, pulling the average down. The molecules of the low-volatility liquid escape less easily, so fewer molecules enter the vapour. This is why the total pressure of an ideal mixture always lies between the two pure vapour pressures.
Find the vapour pressure of a solution at 45 °C with benzene to octane in molar ratio 3 : 2. At 45 °C the vapour pressure of pure benzene is 280 mm Hg and that of pure octane is 420 mm Hg. Assume an ideal solution.
Which statement about the composition of the vapour over an ideal 1:1 molar mixture of benzene and toluene at 25 °C is correct? Given p°(benzene) = 12.8 kPa and p°(toluene) = 3.85 kPa.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
p_A = p°_A × x_A and p_B = p°_B × x_B, where p° is the pure liquid's vapour pressure and x is its mole fraction in the liquid. Total pressure P = p_A + p_B.
Yes, for an ideal solution the total pressure is always between p°_A and p°_B, because it is a weighted average of the two using their mole fractions.
The liquid with the higher pure vapour pressure escapes more easily, so it contributes more to the vapour. Using y = p/P_total, its vapour mole fraction comes out higher than its liquid mole fraction.
It fails for non-ideal solutions, where A–B attractions differ from A–A and B–B attractions. This gives positive or negative deviation. It holds well when the two liquids are similar, like benzene and toluene.
With a non-volatile solute, only the solvent gives vapour, so p = p°_solvent × x_solvent. With two volatile liquids, both contribute vapour and both terms appear in the total pressure.
