Vapour Pressure — Raoult's law, ideal solutions
Solutions
10
JEE Qs
8%
Hard
75
min
Always clearly differentiate between mole fractions in the liquid phase (x) for Raoult's Law and in the vapor phase (y) for Dalton's Law to avoid errors in multi-step problems.
🧮 Key Formulas
✅ Key Points for JEE
- 1Raoult's Law states that for a volatile component in a solution, its partial vapor pressure is proportional to its mole fraction in the liquid phase and its pure vapor pressure.
- 2For a solution containing a non-volatile solute, the vapor pressure of the solution is solely due to the solvent, following `P_solution = x_solvent * P_solvent^0`.
- 3The composition of the vapor phase is generally different from the liquid phase and is determined using Dalton's Law of Partial Pressures, where `y_i = P_i / P_total`.
- 4Ideal solutions are characterized by zero enthalpy of mixing (`ΔH_mix = 0`) and zero volume change on mixing (`ΔV_mix = 0`), meaning intermolecular forces between A-A, B-B, and A-B are comparable.
⚠️ Common Mistakes
- ✕Confusing mole fraction in the liquid phase (xᵢ) with mole fraction in the vapor phase (yᵢ) during calculations.
- ✕Incorrectly applying Raoult's Law for a non-volatile solute, leading to attempts to calculate its partial vapor pressure.
- ✕Assuming total vapor pressure is always the sum of pure component vapor pressures, ignoring mole fractions in the solution phase.
📝 Practice Questions
See allQ73.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).
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 .
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)
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
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
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)
NCERT Chapters
- Class 12 Chemistry Part 1 Ch 2: Solutions