Abnormal Molar Masses | Chemistry Class 12 - NEET PDF Download

ABNORMAL MOLAR MASS

Molecular masses of a solute can be determined from colligative properties - relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. The relations between molecular mass and colligative properties are derived under certain assumptions:

• The solution is sufficiently dilute so that the solvent obeys Raoult's law.
• The solute neither dissociates nor associates in the solution.

When these assumptions are not valid, the experimentally determined molar mass may differ from its true value. Such a difference is called an abnormal molar mass. The two common causes of abnormal molar masses are association and dissociation of solute particles in solution.

CAUSES OF ABNORMAL MOLAR MASS

Association of solute particles

• Some solute molecules associate to form dimers, trimers, etc., in the solvent.
• Association reduces the number of solute particles present in the solution.
• Colligative properties depend on the number of solute particles; therefore they are smaller than expected for the same amount of solute.
• Since colligative properties are inversely proportional to molar mass (for a given concentration), the apparent molar mass appears larger than the true molar mass.
• Example: Ethanoic acid (CH₃COOH) forms hydrogen-bonded dimers in non-polar solvents such as benzene.

Image 2: Ethanoic acid dimer due to hydrogen bonding

Dissociation of solute particles

• Electrolytes and some other solutes dissociate into two or more ions when dissolved.
• Dissociation increases the number of solute particles in solution.
• Colligative properties increase for the same amount of substance when dissociation occurs.
• Consequently, the apparent molar mass determined from colligative properties is lower than the true molar mass.
• Example: Hydrofluoric acid (HF) partially dissociates in water: HF → H⁺ + F^-.

NaCl dissociate into two ions when dissolved in water

VAN'T HOFF FACTOR (i)

To account for association or dissociation when using colligative properties, Jacobus Henricus van't Hoff introduced the van't Hoff factor, denoted by i. It relates the observed colligative effect to the effect expected when no association or dissociation occurs.

The van't Hoff factor may be expressed as:

• i = Normal molar mass / Abnormal molar mass
• i = Observed colligative property / Calculated colligative property (for non-associating, non-dissociating solute)
• i = Total number of moles of particles after association/dissociation ÷ Total number of moles before association/dissociation

Possible values and their meanings:

• i = 1 : No association or dissociation (solute is normal).
• i > 1 : Dissociation has occurred (more particles formed).
• i < 1 : Association has occurred (fewer particles present).

Typical approximate values in solution:
Potassium chloride (KCl) in aqueous solution, i ≈ 2.
Barium chloride (BaCl₂) in aqueous solution, i ≈ 3.
Benzoic acid (C₆H₅COOH) in benzene, i ≈ 0.5.
Acetic acid (CH₃COOH) in benzene, i ≈ 0.5.

CALCULATION OF i (VAN'T HOFF FACTOR)

Case 1: Dissociation of an electrolyte

Consider an electrolyte A_xB_y that dissociates as:

A_xB_y → x A^y+ + y B^x-

Let c be the initial concentration (moles per litre) of the electrolyte and α the degree of dissociation (fraction dissociated).

Number of moles of undissociated A_xB_y at equilibrium = c(1 - α)

Number of moles of A-ions at equilibrium = c x α

Number of moles of B-ions at equilibrium = c y α

Total number of moles of particles at equilibrium = c(1 - α) + c x α + c y α

Therefore,

i = (total number of moles of particles at equilibrium) / (initial number of moles of particles)

i = [c(1 - α) + c x α + c y α] / c

i = 1 - α + x α + y α

Let n = x + y (total particles produced per formula unit on complete dissociation).

i = 1 + (n - 1) α

Case 2: Association of molecules

Let n molecules associate to form one associated particle:

n A → A_n

Let the initial concentration be C and the degree of association be α (fraction of molecules that associate).

At equilibrium, number of monomer molecules = C(1 - α)

Number of associated species = C α / n

Total number of particles at equilibrium = C(1 - α) + C α / n

Therefore,

i = (total number of particles at equilibrium) / (initial number of particles)

i = [1 - α + α / n]

i = 1 - [(n - 1)/n] α

MODIFIED FORMULAE FOR COLLIGATIVE PROPERTIES

When association or dissociation occurs, insert the van't Hoff factor i in the colligative-property equations:

• Boiling point elevation: ΔT_b = i K_b m
• Freezing point depression: ΔT_f = i K_f m
• Osmotic pressure: π = i C R T
• Lowering of vapour pressure: ΔP = i X_solute P° (for dilute solutions with nonvolatile solute)

EXAMPLES

Example 1. Acetic acid (CH₃COOH) associates in benzene to form dimers. 1.65 g of acetic acid when dissolved in 100 g of benzene raised the boiling point by 0.36°C. Calculate the van't Hoff factor and the degree of association of acetic acid in benzene. (K_b = 2.57 K kg mol^-1).

Solution:  Calculate molality:
Molar mass of CH₃COOH = 60 g mol^-1
Moles of acetic acid = 1.65 / 60 = 0.0275 mol
Mass of benzene = 100 g = 0.100 kg
Molality m = 0.0275 / 0.100 = 0.275 m
For observed boiling point elevation, ΔT_b(obs) = 0.36°C
If no association, predicted ΔT_b = K_b m = 2.57 × 0.275 = 0.70675°C
Van't Hoff factor i = ΔT_b(obs) / (K_b m)
i = 0.36 / 0.70675 = 0.509 ≈ 0.509
Since acetic acid dimerises (n = 2),
i = 1 - [(n - 1)/n] α = 1 - (1/2) α
0.509 = 1 - 0.5 α
0.5 α = 1 - 0.509 = 0.491
α = 0.982 ≈ 0.981 (degree of association)

Example 2. Derive a relationship between K_b, T_b and ΔH_vap for a liquid.

Solution: Consider a pure solvent whose vapour pressure at its boiling point T_b is P° and equals the external pressure.
When a non-volatile solute is dissolved, the vapour pressure at a temperature slightly above T_b is lowered. The boiling point of the solution T_b + ΔT_b is reached when the vapour pressure of the solution becomes equal to the external pressure.
From Clausius-Clapeyron relation:

Example 3. To 500 cm³ of water, 3.0 × 10^-3 kg of acetic acid is added. If 23% of acetic acid is dissociated, what will be the depression in freezing point? K_f and density of water are 1.86 K kg mol^-1 and 0.997 g cm^-3 respectively.

Solution: Mass of acetic acid = 3.0 × 10^-3 kg = 3.0 g
Moles of acetic acid = 3.0 / 60 = 0.05 mol
Volume of water = 500 cm³; density = 0.997 g cm^-3
Mass of solvent (water) = 500 × 0.997 = 498.5 g = 0.4985 kg
Molality m = 0.05 / 0.4985 = 0.1003 m
Degree of dissociation α = 0.23
Effective molality = m (1 + α) = 0.1003 × (1 + 0.23) = 0.1003 × 1.23 = 0.123369 m
Freezing point depression ΔT_f = K_f × (effective molality)
ΔT_f = 1.86 × 0.123369 = 0.2294°C ≈ 0.23°C

Example 4. At 300 K, the vapour pressure of an ideal solution containing one mole of A and three moles of B is 500 mm of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Calculate the vapour pressures of A and B in their pure states.

Solution: Let P°_A and P°_B be the vapour pressures of pure A and B respectively at 300 K.
Initial composition: 1 mole A and 3 moles B → X_A = 1/4, X_B = 3/4.
Using Raoult's law for ideal solutions:
(1/4) P°_A + (3/4) P°_B = 500 mm
On adding one mole of B, composition becomes 1 mole A and 4 moles B → X_A = 1/5, X_B = 4/5.
New vapour pressure = 500 + 10 = 510 mm. Thus:
(1/5) P°_A + (4/5) P°_B = 510 mm
Multiply the first equation by 4: P°_A + 3 P°_B = 2000
Multiply the second equation by 5: P°_A + 4 P°_B = 2550
Subtract the first from the second:

P°_B = 550 mm

Substitute back to find P°_A:

P°_A + 3(550) = 2000

P°_A = 2000 - 1650 = 350 mm

Therefore, P°_A = 350 mm and P°_B = 550 mm.

Example 5. A decinormal solution of NaCl exerts an osmotic pressure of 4.6 atm at 300 K. Calculate its degree of dissociation. (R = 0.082 L atm K^-1 mol^-1).

Solution: Decinormal solution means concentration C = 0.1 mol L^-1.
For NaCl, 0.1 N = 0.1 M
For non-dissociated NaCl, expected osmotic pressure:
π_normal = C R T = 0.1 × 0.082 × 300 = 2.46 atm
Observed osmotic pressure π_obs = 4.6 atm
Van't Hoff factor i = π_obs / (C R T)
i = 4.6 / 2.46 = 1.8699 ≈ 1.87
For NaCl, on dissociation n = 2 so i = 1 + α
α = i - 1 = 1.87 - 1 = 0.87 = 87%

Example 6. Vapour pressure of C₆H₆ (benzene) and C₇H₈ (toluene) mixture at 50°C is given by: P = 179 X_B + 92, where X_B is mole fraction of benzene. Calculate:

(A) Vapour pressures of pure liquids.

(B) Vapour pressure of the liquid mixture obtained by mixing 936 g C₆H₆ and 736 g toluene.

(C) If the vapours are removed and condensed into liquid and again brought to 50°C, what would be the mole fraction of C₆H₆ in the vapour state?

Solution: (A) For pure benzene (X_B = 1): P°_B = 179 × 1 + 92 = 271 mm
For pure toluene (X_B = 0): P°_T = 92 mm

(B) Moles of benzene = 936 / 78 = 12
Moles of toluene = 736 / 92 = 8
Total moles = 20; X_B = 12/20 = 0.6; X_T = 8/20 = 0.4
Vapour pressure of mixture PM = P°_B X_B + P°_T X_T
PM = 271 × 0.6 + 92 × 0.4 = 162.6 + 36.8 = 199.4 mm

(C) Mole fraction of benzene in vapour (y_B) above the initial liquid mixture:
y_B = (X_B P°_B) / PM = (0.6 × 271) / 199.4 = 162.6 / 199.4 = 0.815
y_T = 1 - 0.815 = 0.185
These vapours are condensed; the condensed liquid has composition X′_B = 0.815 and X′_T = 0.185. On heating again to 50°C the new total vapour pressure is:

P′ = X′_B P°_B + X′_T P°_T = 0.815 × 271 + 0.185 × 92 = 220.865 + 17.02 = 237.885 mm
New mole fraction of benzene in vapour:
y′_B = (X′_B P°_B) / P′ = 220.865 / 237.885 = 0.928
New mole fraction of toluene in vapour = 0.072

Example 7. The freezing point depression of an HF (aq) solution with molality 0.10 m is -0.201°C. What is the van't Hoff factor? Take K_f = 1.86°C m^-1.

Solution: HF partially dissociates: HF ⇌ H⁺ + F^-
Given ΔT_f = 0.201°C (magnitude), m = 0.10 m, K_f = 1.86°C m^-1
ΔT_f = i K_f m
i = ΔT_f / (K_f m)
i = 0.201 / (1.86 × 0.10) = 0.201 / 0.186 = 1.0806 ≈ 1.08

Example 8. A millimolar solution of potassium ferricyanide is 70% dissociated at 27°C. Find the osmotic pressure of the solution.

Solution: Concentration C = 1 millimolar = 1 × 10^-3 M

Potassium ferricyanide dissociates as K₃[Fe(CN)₆] → 3 K⁺ + [Fe(CN)₆]^3-
n = total ions produced = 4
Degree of dissociation α = 0.70
Van't Hoff factor i = 1 + (n - 1) α = 1 + 3 × 0.70 = 1 + 2.1 = 3.1
Osmotic pressure π = i C R T
π = 3.1 × 1 × 10^-3 × 0.0821 × 300
π = 0.07635 atm ≈ 0.0764 atm

TRY YOURSELF!

Q.1. A 0.5 percent aqueous solution of Potassium Chloride was found to freeze at 0.24°C. Calculate the Van't Hoff factor and degree of dissociation of the solute at this concentration (K_f for water is 1.86).

Solution: Assume 0.5% w/w KCl means 0.5 g KCl in 100 g solution. Approximate solvent mass as 100 g (for the given solution procedure).
Moles of KCl = 0.5 / 74.5 = 0.006711 mol
Mass of solvent ≈ 100 g = 0.100 kg
Molality m ≈ 0.006711 / 0.100 = 0.06711 m
Observed freezing point depression ΔT_f = 0.24°C
i = ΔT_f / (K_f m) = 0.24 / (1.86 × 0.06711) = 0.24 / 0.124866 ≈ 1.923
For KCl, i = 1 + α (since ideal dissociation produces 2 particles)
1.923 = 1 + α ⇒ α = 0.923 or 92.3%

Q.2. If the apparent degree of ionization of KCl (KCl = 74.5 g mol^-1) in water at 290 K is 0.86. Calculate the mass of KCl which must be made up to 1 dm³ of aqueous solution so as to produce the same osmotic pressure as the 4.0% solution of glucose at that temperature.

Solution: Degree of ionization a = 0.86 ⇒ i = 1 + a = 1.86
For 4.0% glucose solution: mass of glucose = 40 g in 1 L
Molar mass of glucose = 180 g mol^-1
Moles of glucose = 40 / 180 = 0.22222 mol
Osmotic pressure of glucose solution π_glucose = n/V × R T = 0.22222 × R T
For KCl solution to have same π:
π_KCl = i (n_KCl/V) R T = π_glucose
Thus i × (m_KCl/M_KCl) = 0.22222
m_KCl = (0.22222 × M_KCl) / i
m_KCl = (0.22222 × 74.5) / 1.86
m_KCl = (16.5559) / 1.86 = 8.9 g (approximately)

Summary (optional): When solute particles associate or dissociate, the number of particles in solution changes; colligative properties must be corrected using the van't Hoff factor i. For dissociation i = 1 + (n - 1)α and for association i = 1 - [(n - 1)/n]α. Use ΔT_b = i K_b m, ΔT_f = i K_f m and π = i C R T for computations.