Abnormal Molar Mass & van't Hoff Factor
Module 5 | Solutions Chapter Master Review Sheet
1. Concept of Abnormal Molar Mass
In the previous module, we assumed that the solute added to the solvent is a non-electrolyte (it does not dissociate or associate). However, what happens when we dissolve an electrolyte like NaCl in water?
NaCl dissociates into Na+ and Cl- ions. If we dissolve 1 mole of NaCl, we actually get 2 moles of particles in the solution. Since colligative properties depend strictly on the number of particles, the observed colligative property will be roughly double the expected value.
Similarly, some molecules, like acetic acid (CH3COOH) in benzene, undergo association (dimerization) due to hydrogen bonding. Two molecules join to form one. Here, the number of particles is halved, and the colligative property decreases.
2. The van't Hoff Factor (i)
To account for the extent of dissociation or association, Jacobus Henricus van't Hoff introduced a factor 'i', known as the van't Hoff factor.
i = Observed Colligative Property / Calculated Colligative Property
i = Total number of moles of particles after association/dissociation / Number of moles of particles before association/dissociation
| Solute Behavior | Value of 'i' | Effect on Colligative Property | Effect on Molar Mass |
|---|---|---|---|
| Dissociation (e.g., NaCl, KCl, K2SO4) | i > 1 | Increases | Decreases (Lower than normal) |
| Association (e.g., Acetic acid in benzene) | i < 1 | Decreases | Increases (Higher than normal) |
| Neither (e.g., Urea, Glucose, Sucrose) | i = 1 | Remains same | Normal Molar Mass |
3. Modified Equations for Colligative Properties
When solving numericals involving electrolytes (acids, bases, salts), we must multiply the previously learned formulas by the van't Hoff factor (i).
(p10 - p1) / p10 = i × x2 = i × [ (n2) / (n1 + n2) ]
2. Elevation of Boiling Point:
ΔTb = i × Kb × m
3. Depression of Freezing Point:
ΔTf = i × Kf × m
4. Osmotic Pressure:
Π = i × C × R × T
4. Degree of Dissociation (α)
The degree of dissociation is the fraction of total substance that undergoes dissociation into ions.
Consider an electrolyte An that dissociates into 'n' ions:
An → nA
Initial moles: 1 0
Moles at eq: 1 - α nα
Total particles = (1 - α) + nα = 1 + α(n - 1)
Since i = (Total particles after dissociation) / (Initial particles) = [1 + α(n - 1)] / 1
Where 'n' is the number of ions produced per formula unit. (e.g., for K2SO4, n = 3).
5. Degree of Association (α)
The degree of association is the fraction of total substance that associates (combines) to form larger molecules.
Consider 'n' molecules associating to form a single macromolecule:
nA → An
Initial moles: 1 0
Moles at eq: 1 - α α/n
Total particles = 1 - α + α/n = 1 + α(1/n - 1)
Since i = [1 + α(1/n - 1)] / 1
6. NCERT Solved Examples (Step-by-Step)
NCERT Example 2.12: 2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol-1. What is the percentage association of acid if it forms dimer in solution?
Step 1: Calculate the observed/abnormal molar mass (M2)
W2 = 2 g; W1 = 25 g; ΔTf = 1.62 K; Kf = 4.9 K kg mol-1
M2(observed) = (Kf × W2 × 1000) / (ΔTf × W1)
M2(observed) = (4.9 × 2 × 1000) / (1.62 × 25) = 241.98 g mol-1
Step 2: Calculate van't Hoff factor (i)
Normal molar mass of benzoic acid (C6H5COOH) = (7×12) + (6×1) + (2×16) = 122 g mol-1
i = Normal Molar Mass / Observed Molar Mass
i = 122 / 241.98 = 0.504
Step 3: Calculate Degree of Association (α)
Since it forms a dimer, n = 2.
α = (1 - i) / (1 - 1/n) = (1 - 0.504) / (1 - 1/2)
α = 0.496 / 0.5 = 0.992
Percentage association = 99.2%
7. Previous Year Questions (PYQs) & Exhaustive Question Bank
Part A: Conceptual & Assertion-Reason (1 Mark)
Q1. What is the expected value of van't Hoff factor for K3[Fe(CN)6] assuming complete dissociation?
K3[Fe(CN)6] → 3K+ + [Fe(CN)6]3-
It gives a total of 4 ions. Hence, for complete dissociation, i = 4.
Q2. Assertion (A): 0.1 M solution of NaCl has a higher boiling point than 0.1 M solution of glucose.
Reason (R): Elevation of boiling point is directly proportional to the number of particles in the solution.
Part B: Numerical Problems (3-5 Marks)
Q3. A 0.01 m aqueous solution of AlCl3 freezes at -0.068°C. Calculate the percentage of dissociation. (Given: Kf for water = 1.86 K kg mol-1).
Step 1: Find van't Hoff factor (i)
ΔTf = 0 - (-0.068) = 0.068 K
Using formula: ΔTf = i × Kf × m
0.068 = i × 1.86 × 0.01
i = 0.068 / 0.0186 = 3.65
Step 2: Calculate degree of dissociation (α)
AlCl3 dissociates as: AlCl3 → Al3+ + 3Cl-. So, n = 4 ions.
α = (i - 1) / (n - 1) = (3.65 - 1) / (4 - 1)
α = 2.65 / 3 = 0.883
Percentage dissociation = 88.3%
Q4. Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250 g of water. (Kb for water = 0.512 K kg mol-1, Molar mass of NaCl = 58.44 g). Assume complete dissociation of NaCl.
Given: W2 = 15.0 g; M2 = 58.44 g mol-1; W1 = 250 g = 0.25 kg.
Since NaCl completely dissociates (NaCl → Na+ + Cl-), i = 2.
Step 1: Calculate molality (m)
Moles of NaCl = 15.0 / 58.44 = 0.256 mol
m = Moles of solute / Mass of solvent in kg = 0.256 / 0.25 = 1.024 m
Step 2: Calculate Elevation in BP (ΔTb)
ΔTb = i × Kb × m
ΔTb = 2 × 0.512 × 1.024 = 1.05 K
Step 3: Calculate Boiling Point of Solution
Boiling point of pure water = 373.15 K (or 100°C)
Boiling point of solution = 373.15 + 1.05 = 374.20 K (or 101.05°C)
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