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Q.0.3 g of acetic acid (M = 60 g mol−1mol^{-1}) dissolved in 30 g of benzene shows a depression in freezing point equal to 0.45∘C0.45^\circ C. Calculate the percentage association of acid if it forms a dimer in the solution. (Given: KfK_f for benzene = 5.125.12 K kg mol−1mol^{-1})

CBSECBSE Class XII Board 2023Subjective· 3mImportance★★★★★
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The problem uses the observed freezing point depression to find the experimental van’t Hoff factor ii, then relates ii to the degree of dimerisation α\alpha for acetic acid in benzene. The calculated percentage association is 94.6%.


Concept and Intuition

Acetic acid in a non-polar solvent like benzene does not remain as isolated molecules. Instead, two molecules of acetic acid form a dimer through hydrogen bonding:

2 CH3COOH⇌(CH3COOH)22\,\text{CH}_3\text{COOH} \rightleftharpoons (\text{CH}_3\text{COOH})_2

This dimerisation reduces the total number of particles in solution. Since colligative properties (like freezing point depression) depend only on the number of solute particles, the observed depression will be smaller than expected for a non-associating solute.

The van’t Hoff factor ii is defined as:

i=observed colligative effectexpected colligative effect for no associationi = \frac{\text{observed colligative effect}}{\text{expected colligative effect for no association}}

For association, i<1i < 1. Once we find ii from experimental data, we can relate it to the degree of association α\alpha — the fraction of acetic acid molecules that have dimerised.


Step-by-step solution

1. Calculate the expected (theoretical) molality

Mass of acetic acid = 0.30.3 g

Molar mass of acetic acid = 6060 g mol−1^{-1}

Mass of benzene = 3030 g = 0.0300.030 kg

Moles of acetic acid:

n=0.360=0.005 moln = \frac{0.3}{60} = 0.005 \text{ mol}

Expected molality (if no association):

mtheoretical=0.0050.030=0.1667 mol kg−1m_{\text{theoretical}} = \frac{0.005}{0.030} = 0.1667 \text{ mol kg}^{-1}

2. Calculate the observed molality from freezing point depression

The formula for depression in freezing point is:

ΔTf=i⋅Kf⋅mtheoretical\Delta T_f = i \cdot K_f \cdot m_{\text{theoretical}}

Given ΔTf=0.45∘\Delta T_f = 0.45^\circC and Kf=5.12K_f = 5.12 K kg mol−1^{-1}:

0.45=i×5.12×0.16670.45 = i \times 5.12 \times 0.1667

i=0.455.12×0.1667i = \frac{0.45}{5.12 \times 0.1667}

First compute 5.12×0.16675.12 \times 0.1667:

5.12×0.1667≈0.85355.12 \times 0.1667 \approx 0.8535

Then:

i=0.450.8535≈0.5272i = \frac{0.45}{0.8535} \approx 0.5272

Watch out

A common mistake is to forget that KfK_f is in K kg mol−1^{-1} and ΔTf\Delta T_f in ∘^\circC — but since the size of 1 K and 1∘^\circC are identical, no conversion is needed here. However, always check units: KfK_f uses kg of solvent, so mass must be in kg.

3. Relate ii to the degree of dimerisation

Let α\alpha be the fraction of acetic acid molecules that have dimerised.

Initially, we have nn moles of monomer.

After dimerisation:

  • Moles that dimerise = nαn\alpha …

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