Question 243 marksShort AnswerSection C

Calculate the boiling point of a solution containing 0.61 g of benzoic acid (Molar mass=122 g mol−1\text{Molar mass} = 122 \text{ g mol}^{-1}) in 5 g of CS₂ in which it dimerises to the extent of 88%. The boiling point and Kb of CS₂ are 46.2 °C and 2.3 K kg mol−12.3 \text{ K kg mol}^{-1} respectively.

Solutions
Elevation of boiling point with association (dimerisation)

Official answer

Step 1 — molality of benzoic acid


  • Moles=0.61122=0.005 mol\text{Moles} = \frac{0.61}{122} = 0.005 \text{ mol}
  • Mass of solvent=5 g=0.005 kg\text{Mass of solvent} = 5 \text{ g} = 0.005 \text{ kg}
  • molality m=0.0050.005=\text{molality } m = \frac{0.005}{0.005} = 1.0 mol kg−11.0 \text{ mol kg}^{-1}

Step 2 — van't Hoff factor for dimerisation


For 2 molecules → 1 dimer, with degree of association α = 0.88:


  • i=1−α2=1−0.882=1−0.44=i = 1 - \frac{\alpha}{2} = 1 - \frac{0.88}{2} = 1 - 0.44 = 0.56

Step 3 — elevation of boiling point


ΔTb=i⋅Kb⋅m=0.56×2.3×1.0=\Delta T_b = i \cdot K_b \cdot m = 0.56 \times 2.3 \times 1.0 = 1.288 K≈1.29 K1.288 \text{ K} \approx 1.29 \text{ K}


Step 4 — boiling point of solution


Tb=Tb∘(solvent)+ΔTb=46.2+1.288=T_b = T_b^{\circ}(\text{solvent}) + \Delta T_b = 46.2 + 1.288 = 47.49 ∘C47.49\,^{\circ}\text{C} (≈47.5 ∘C\approx 47.5\,^{\circ}\text{C})

Answer keywords:colligative propertydimerisation of benzoic acidvan't Hoff factor i = 1 − α/2i = 0.56molality 1.0 mol kg⁻¹elevation of boiling point ΔTbΔTb = i Kb mboiling point 47.49 °C

Marking scheme

  1. 1.1 mark: molality of benzoic acid = 1.0 mol kg−11.0 \text{ mol kg}^{-1} (moles=0.005\text{moles} = 0.005, solvent=0.005 kg\text{solvent} = 0.005 \text{ kg}).
  2. 2.1 mark: van't Hoff factor for dimerisation i=1−α2=0.56i = 1 - \frac{\alpha}{2} = 0.56.
  3. 3.1 mark: ΔTb=i⋅Kb⋅m=1.29 K\Delta T_b = i \cdot K_b \cdot m = 1.29 \text{ K} and Tb=46.2+1.29=47.49 ∘CT_b = 46.2 + 1.29 = 47.49\,^{\circ}\text{C} (accept ≈ 47.5 °C).

Hint

molality=0.61/1225/1000=1.0\text{molality} = \frac{0.61/122}{5/1000} = 1.0; for dimerisation i=1−α2=0.56i = 1 - \frac{\alpha}{2} = 0.56; ΔTb=i⋅Kb⋅m=1.29 K\Delta T_b = i \cdot K_b \cdot m = 1.29 \text{ K}; add to 46.2 °C.

Quick oral answer

The benzoic acid is 1.0 molal; because 88% dimerises, i=1−0.882=0.56i = 1 - \frac{0.88}{2} = 0.56, so ΔTb=0.56×2.3×1.0=1.29 K\Delta T_b = 0.56 \times 2.3 \times 1.0 = 1.29 \text{ K}, making the boiling point 46.2+1.29=47.49 ∘C46.2 + 1.29 = 47.49\,^{\circ}\text{C}.

Common mistakes

  1. 1.Using the dissociation formula i=1+(n−1)αi = 1 + (n-1)\alpha instead of the association formula i=1−α2i = 1 - \frac{\alpha}{2}, giving a wrong i.
  2. 2.Forgetting to include the van't Hoff factor and computing ΔTb as just Kb⋅m=2.3 KK_b \cdot m = 2.3 \text{ K}.
  3. 3.Reporting only ΔTb (1.29 K) and forgetting to add it to the solvent's boiling point 46.2 °C.

Analysis and explanation

Concept — association lowers particle count


Colligative properties depend on the number of solute particles. Benzoic acid dimerises in a non-polar solvent like CS₂ through hydrogen bonding, so two molecules act as one particle. This reduces the effective number of particles, making i less than 1 and the boiling-point elevation smaller than expected.


Deriving the van't Hoff factor


  • Start with 1 mole; a fraction α associates. Particles left = (1 − α) monomers + α/2 dimers.
  • Total particles = 1−α+α2=1−α21 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}, which is the value of i.
  • With α=0.88,i=0.56\alpha = 0.88, i = 0.56 — nearly half the particles have paired up.

Exam trap


  • Use i=1−α2i = 1 - \frac{\alpha}{2} for dimerisation; the general form is i=1+(1n−1)α with n=2i = 1 + \left(\frac{1}{n} - 1\right)\alpha \text{ with } n = 2. Do not use the dissociation form i=1+(n−1)αi = 1 + (n-1)\alpha.
  • Add ΔTb to the solvent's boiling point (46.2 °C), and keep ΔTb in kelvin (numerically same size as °C change).

Real-world link


Carboxylic acids famously form such dimers; this molecular pairing is why their measured molar masses in benzene or CS₂ come out roughly double the true value.

Interesting facts

Benzoic acid was one of the first molecules shown to have an 'abnormal' molar mass — nearly double the real value — because it dimerises in non-polar solvents.

The dimer is held together by two O–H···O hydrogen bonds, forming a stable eight-membered ring that survives even in the vapour phase.

Association reduces colligative effects, whereas dissociation (as in salts) increases them — the same formula family (i=1+(1n−1)αi = 1 + \left(\frac{1}{n} - 1\right)\alpha) describes both.

Question details

Q24 of 37 questions

Board
CBSE
Class
Class 12
Subject
Chemistry
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Solutions
Topic
Elevation of boiling point with association (dimerisation)
Bloom's level
apply
Syllabus topic
Unit: Solutions — Colligative properties, van't Hoff factor and abnormal molar mass (association)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is the van't Hoff factor less than 1 here?

Benzoic acid molecules associate into dimers in CS₂, so two molecules behave as a single particle. This lowers the total number of solute particles below the number dissolved, making i less than 1. With 88% association, i=1−0.882=0.56i = 1 - \frac{0.88}{2} = 0.56, reflecting that nearly half the particles have paired up.

Why use i = 1 − α/2 instead of i = 1 + (n−1)α?

The form i=1+(n−1)αi = 1 + (n-1)\alpha is for dissociation, where one particle splits into n. For association, n particles combine into one, so the correct expression is i=1+(1n−1)αi = 1 + \left(\frac{1}{n} - 1\right)\alpha. For dimerisation n = 2, giving i=1−α2i = 1 - \frac{\alpha}{2}. Using the dissociation formula here would wrongly increase the particle count.