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ChemistryMediumClass 12

Abnormal Molar Mass — Van't Hoff factor, association, dissociation

Solutions

10

JEE Qs

8%

Hard

60

min

Always first determine if the solute is undergoing dissociation, association, or neither, then correctly calculate the Van't Hoff factor 'i' before applying it to the colligative property formulas.

🧮 Key Formulas

i = (Observed Colligative Property) / (Calculated Colligative Property)
i = (Normal Molar Mass) / (Abnormal Molar Mass)
For dissociation: i = 1 + \alpha(n-1)
For association: i = 1 + \alpha(1/n - 1)
\Delta P/P^0 = i \cdot X_{solute}
\Delta T_b = i \cdot K_b \cdot m
\Delta T_f = i \cdot K_f \cdot m
\pi = i \cdot C \cdot R \cdot T

✅ Key Points for JEE

  • 1The Van't Hoff factor (i) quantifies the extent of dissociation or association of a solute, indicating how the number of particles in solution changes from initial expectations.
  • 2For dissociation, i > 1 as the number of particles increases; for association, i < 1 as the number of particles decreases; for non-electrolytes, i = 1.
  • 3Abnormal molar mass is observed when the calculated molar mass from colligative properties differs from the normal molar mass, which is a direct consequence of i \neq 1.
  • 4Always identify if the solute is an electrolyte (dissociation) or tends to associate (e.g., carboxylic acids in non-polar solvents) to determine the applicability of 'i'.
  • 5The degree of dissociation (\alpha) or association is critical for calculating 'i' when the process is not 100% complete (e.g., weak electrolytes).

⚠️ Common Mistakes

  • Incorrectly determining 'n' (number of particles formed) for dissociation, especially for complex salts or coordination compounds (e.g., K₄[Fe(CN)₆] gives 5 ions).
  • Confusing the formulas for degree of dissociation and degree of association, leading to sign errors in 'i' calculation.
  • Forgetting to multiply the Van't Hoff factor 'i' into the standard colligative property formulas, assuming ideal behavior.
  • Using molarity instead of molality (or vice versa) in colligative property calculations where specific concentration units are required.

📝 Practice Questions

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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).

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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

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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