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Solutions - Abnormal molecular mass

Grade 12ICSEChemistry

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

🔑Concepts

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Abnormal molecular mass is observed when the solute undergoes association or dissociation in the solvent, leading to a change in the total number of particles.

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Dissociation: When a solute (usually electrolytes like NaClNaCl or KClKCl) breaks into ions, the number of particles increases. This results in a higher observed colligative property and a lower observed molecular mass than the theoretical value (i>1i > 1).

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Association: When solute molecules (like carboxylic acids in benzene) join to form dimers or polymers, the number of particles decreases. This results in a lower observed colligative property and a higher observed molecular mass than the theoretical value (i<1i < 1).

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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 calculated (theoretical) value.

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Colligative properties depend on the number of solute particles. To account for association or dissociation, the standard equations must be multiplied by the factor ii.

📐Formulae

i=Normal Molar MassObserved Molar Massi = \frac{\text{Normal Molar Mass}}{\text{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}}

ΔTb=i⋅Kb⋅m\Delta T_b = i \cdot K_b \cdot m

ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m

π=i⋅C⋅R⋅T\pi = i \cdot C \cdot R \cdot T

α=i−1n−1 (For Dissociation)\alpha = \frac{i - 1}{n - 1} \text{ (For Dissociation)}

α=i−11n−1 (For Association)\alpha = \frac{i - 1}{\frac{1}{n} - 1} \text{ (For Association)}

💡Examples

Problem 1:

A 0.5%0.5\% aqueous solution of KClKCl was found to freeze at −0.24∘C-0.24^{\circ}C. Calculate the Van't Hoff factor and the degree of dissociation of KClKCl at this concentration. (KfK_f for H2O=1.86 K kg mol−1H_2O = 1.86\, K\, kg\, mol^{-1}, Molar mass of KCl=74.5 g mol−1KCl = 74.5\, g\, mol^{-1})

Solution:

  1. Molality (mm) = 0.5×100074.5×99.5≈0.0674 m\frac{0.5 \times 1000}{74.5 \times 99.5} \approx 0.0674\, m.
  2. Calculated ΔTf=Kf×m=1.86×0.0674=0.1254∘C\Delta T_f = K_f \times m = 1.86 \times 0.0674 = 0.1254^{\circ}C.
  3. Observed ΔTf=0.24∘C\Delta T_f = 0.24^{\circ}C.
  4. i=Observed ΔTfCalculated ΔTf=0.240.1254≈1.91i = \frac{\text{Observed } \Delta T_f}{\text{Calculated } \Delta T_f} = \frac{0.24}{0.1254} \approx 1.91.
  5. For KCl→K++Cl−KCl \rightarrow K^+ + Cl^-, n=2n=2.
  6. α=i−1n−1=1.91−12−1=0.91\alpha = \frac{i - 1}{n - 1} = \frac{1.91 - 1}{2 - 1} = 0.91.

Explanation:

Since KClKCl is an electrolyte, it dissociates. The experimental freezing point depression is nearly double the calculated value because the number of particles almost doubles, resulting in i≈1.91i \approx 1.91 and 91%91\% dissociation.

Problem 2:

Acetic acid (CH3COOHCH_3COOH) undergoes dimerization in benzene. If the observed molar mass is 110 g mol−1110\, g\, mol^{-1}, calculate the Van't Hoff factor (ii). (Normal molar mass of CH3COOH=60 g mol−1CH_3COOH = 60\, g\, mol^{-1})

Solution:

i=Normal Molar MassObserved Molar Massi = \frac{\text{Normal Molar Mass}}{\text{Observed Molar Mass}} i=60110≈0.545i = \frac{60}{110} \approx 0.545

Explanation:

In non-polar solvents like benzene, acetic acid forms hydrogen-bonded dimers (2CH3COOH→(CH3COOH)22CH_3COOH \rightarrow (CH_3COOH)_2). This reduces the particle count, making i<1i < 1 and the observed molecular mass higher than the actual molar mass.