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Solutions - Abnormal Molar Masses (van't Hoff Factor)

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Abnormal Molar Mass occurs when the solute undergoes association (e.g., dimerization of CH3COOHCH_3COOH in benzene) or dissociation (e.g., NaClNaCl in H2OH_2O) in a solution.

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The van't Hoff factor (ii) is defined as the ratio of the observed value of a colligative property to the theoretically calculated value assuming the solute is a non-electrolyte.

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If i>1i > 1, the solute undergoes dissociation (the number of particles increases).

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If i<1i < 1, the solute undergoes association (the number of particles decreases).

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If i=1i = 1, the solute undergoes neither association nor dissociation.

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The degree of dissociation (α\alpha) is the fraction of the total substance that undergoes dissociation: α=i−1n−1\alpha = \frac{i - 1}{n - 1}, where nn is the number of ions produced per formula unit.

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The degree of association (α\alpha) is the fraction of the total substance that undergoes association: α=1−i1−1/n\alpha = \frac{1 - i}{1 - 1/n}, where nn molecules associate to form one aggregate.

📐Formulae

i=Normal Molar MassAbnormal (Observed) Molar Massi = \frac{\text{Normal Molar Mass}}{\text{Abnormal (Observed) Molar Mass}}

i=Observed Colligative PropertyCalculated Colligative Propertyi = \frac{\text{Observed Colligative Property}}{\text{Calculated Colligative Property}}

i=Total moles of particles after association/dissociationTotal moles of particles before association/dissociationi = \frac{\text{Total moles of particles after association/dissociation}}{\text{Total moles of particles before association/dissociation}}

Relative Lowering of Vapour Pressure: P1∘−P1P1∘=i⋅χ2\text{Relative Lowering of Vapour Pressure: } \frac{P_1^\circ - P_1}{P_1^\circ} = i \cdot \chi_2

Elevation of Boiling Point: ΔTb=i⋅Kb⋅m\text{Elevation of Boiling Point: } \Delta T_b = i \cdot K_b \cdot m

Depression of Freezing Point: ΔTf=i⋅Kf⋅m\text{Depression of Freezing Point: } \Delta T_f = i \cdot K_f \cdot m

Osmotic Pressure: π=i⋅C⋅R⋅T\text{Osmotic Pressure: } \pi = i \cdot C \cdot R \cdot T

💡Examples

Problem 1:

Calculate the van't Hoff factor (ii) for a 0.1 m0.1\, m aqueous solution of K2SO4K_2SO_4 if it is 90%90\% dissociated.

Solution:

  1. Write the dissociation equation: K2SO4→2K++SO42−K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}.
  2. Here, n=3n = 3 (2 potassium ions + 1 sulfate ion).
  3. Degree of dissociation α=0.90\alpha = 0.90.
  4. Use the formula: i=1+(n−1)αi = 1 + (n - 1)\alpha.
  5. i=1+(3−1)×0.90=1+2×0.90=1+1.80=2.80i = 1 + (3 - 1) \times 0.90 = 1 + 2 \times 0.90 = 1 + 1.80 = 2.80.

Explanation:

Since the salt dissociates into 3 particles, if 100%100\% dissociated, ii would be 33. At 90%90\%, the observed number of particles is 2.802.80 times the original.

Problem 2:

2 g2\,g of benzoic acid (C6H5COOHC_6H_5COOH) dissolved in 25 g25\,g of benzene shows a depression in freezing point equal to 1.62 K1.62\,K. Molal depression constant (KfK_f) for benzene is 4.9 K kg mol−14.9\,K \, kg \, mol^{-1}. What is the percentage association of acid if it forms a dimer in solution?

Solution:

  1. Calculated Molar Mass of C6H5COOH=122 g mol−1C_6H_5COOH = 122\,g\,mol^{-1}.
  2. Observed Molar Mass M2=1000⋅Kf⋅w2w1⋅ΔTf=1000×4.9×225×1.62≈241.98 g mol−1M_2 = \frac{1000 \cdot K_f \cdot w_2}{w_1 \cdot \Delta T_f} = \frac{1000 \times 4.9 \times 2}{25 \times 1.62} \approx 241.98\,g\,mol^{-1}.
  3. i=Normal Molar MassObserved Molar Mass=122241.98≈0.504i = \frac{\text{Normal Molar Mass}}{\text{Observed Molar Mass}} = \frac{122}{241.98} \approx 0.504.
  4. For dimerization, n=2n=2. α=1−i1−1/n=1−0.5041−1/2=0.4960.5=0.992\alpha = \frac{1-i}{1-1/n} = \frac{1-0.504}{1-1/2} = \frac{0.496}{0.5} = 0.992.
  5. Percentage association = 99.2%99.2\%.

Explanation:

Benzoic acid undergoes association (dimerization) in benzene, leading to a higher observed molar mass and a van't Hoff factor less than 1.

Abnormal Molar Masses (van't Hoff Factor) Revision - Class 12 Chemistry CBSE