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ChemistryMediumClass 12

Colligative Properties — ΔTb, ΔTf, π (osmosis)

Solutions

10

JEE Qs

8%

Hard

75

min

Always correctly identify the nature of the solute (electrolyte/non-electrolyte) to determine the Van't Hoff factor (i) and ensure consistency in units (molality vs. molarity, Kelvin vs. Celsius) for accurate calculations.

🧮 Key Formulas

ΔT_b = i * K_b * m
ΔT_f = i * K_f * m
π = i * C * R * T
i = 1 + (n - 1)α (for dissociation)
i = 1 + (1/n - 1)α (for association)

✅ Key Points for JEE

  • 1Colligative properties depend solely on the number of solute particles, not their chemical identity, in a given amount of solvent.
  • 2The Van't Hoff factor (i) is crucial for electrolytes (ionic compounds) to account for dissociation (i > 1) or association (i < 1) of solute particles in solution. For non-electrolytes, i=1.
  • 3ΔT_b and ΔT_f equations use molality (m, mol/kg) as the concentration unit, while osmotic pressure (π) uses molarity (C, mol/L).
  • 4Temperature (T) in the osmotic pressure formula (π = iCRT) must always be in Kelvin (K).
  • 5Colligative properties are inversely proportional to the molar mass of the solute, making them useful for molar mass determination.

⚠️ Common Mistakes

  • Forgetting to apply the Van't Hoff factor (i) for electrolytes, or incorrectly calculating 'i' based on the degree of dissociation/association.
  • Using molarity instead of molality for elevation in boiling point (ΔT_b) and depression in freezing point (ΔT_f) calculations.
  • Using temperature in Celsius instead of Kelvin for osmotic pressure (π) calculations.
  • Confusing the roles or values of cryoscopic constant (K_f) and ebullioscopic constant (K_b) for a given solvent.

📝 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