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ChemistryMediumClass 12

Vapour Pressure + Raoult's Law

Solutions

36

JEE Qs

8%

Hard

75

min

Master the calculation of mole fractions in both liquid and vapour phases, and deeply understand how intermolecular forces dictate deviations from Raoult's Law.

🧮 Key Formulas

P_A = x_A * P_A^0 (Raoult's Law for component A in a solution)
P_B = x_B * P_B^0 (Raoult's Law for component B in a solution)
P_Total = P_A + P_B = x_A * P_A^0 + x_B * P_B^0 (Dalton's Law + Raoult's Law for binary solution)
y_A = P_A / P_Total (Mole fraction of A in vapour phase)
y_B = P_B / P_Total (Mole fraction of B in vapour phase)
(P^0 - P_s) / P^0 = x_solute (Relative Lowering of Vapour Pressure for non-volatile solute)

✅ Key Points for JEE

  • 1Vapour pressure of a liquid increases with temperature due to increased kinetic energy of molecules and is independent of the amount of liquid or surface area.
  • 2Raoult's Law states that the partial vapour pressure of each volatile component in a solution is directly proportional to its mole fraction in the liquid phase and its vapour pressure in the pure state.
  • 3Ideal solutions strictly obey Raoult's Law over the entire range of concentrations, exhibiting similar intermolecular forces between all components (A-A, B-B, A-B).
  • 4Non-ideal solutions show deviations from Raoult's Law: positive deviation (A-B forces < A-A, B-B, leading to higher VP) or negative deviation (A-B forces > A-A, B-B, leading to lower VP).
  • 5For a solution of a non-volatile solute, only the solvent contributes to the vapour pressure. The relative lowering of vapour pressure is a colligative property, equal to the mole fraction of the non-volatile solute.

⚠️ Common Mistakes

  • Confusing mole fractions in the liquid phase (x) with those in the vapour phase (y) or incorrectly assuming they are equal.
  • Misapplying Raoult's Law for non-volatile solutes, often attempting to calculate the 'vapour pressure' of the non-volatile solute.
  • Not correctly identifying the pure component vapour pressures (P_A^0, P_B^0) versus the partial pressures in the solution (P_A, P_B).

📝 Practice Questions

See all

Q73.Consider the following cases of standard enthalpy of reaction (ΔH∘r in kJmol−1) 7 C2H6( g) + O2( g) →2CO2( g) + 3H2O(l)ΔH∘1 = −1550 2 C (graphite) + O2( g) →CO2( g) ΔH∘2 = −393.5 The magnitude of ΔH∘fC2H6( g) is _______ 1 H2( g) + O2( g) →H2O(l) ΔH∘3 = −286 2 kJmol−1 (Nearest integer).

2025·NumericalEasy

Q70.When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg . The mole fraction of the solute in the solution is 0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ? (1) 0.8 (2) 0.4 (3) 0.2 (4) 0.6 Q71.0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion produced 0.9 gH2O. Molar mass of (X) is _____ gmol−1 .

2025·MCQMedium

Q69.Arrange the following solutions in order of their increasing boiling points. (i) 10−4M NaCl (ii) 10−4M Urea (iii) 10−3M NaCl (iv) 10−2M NaCl (1) (i) < (ii) < (iii) < (iv) (2) (iv) < (iii) < (i) < (ii) (3) (ii) < (i ) ≡(iii) < (iv) (4) (ii) < (i) < (iii) < (iv)

2025·MCQMedium

Q52.Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is : (1) 1.79 m (2) 2.79 m (3) 2 m (4) 3 m

2025·MCQMedium

Q64.Consider a binary solution of two volatile liquid components 1 and 2. x1 and y1 are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of 1 vs 1 x1 y1 are given respectively as: P02−P01 (1) P0 P02−P01 (2) P0 2 , 1 , P01 P02 P02 P02 P01−P02 (3) P0 P01−P02 (4) P0 1 , 2 , P02 P02 P01 P02

2025·MCQMedium

Q69.Consider the given plots of vapour pressure (VP) vs temperature(T/K). Which amongst the following options is correct graphical representation showing ΔTf , depression in the freezing point of a solvent in a solution? (1) (2) (3) (4)

2025·Graph basedMedium

NCERT Chapters

  • Class 12 Chemistry Ch 2: Solutions