Calculate the boiling point of a solution containing 0.61 g of benzoic acid ( $\text{Molar mass} = 122 \text{ g mol}^{-1}$ ) 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 $2.3 \text{ K kg mol}^{-1}$ respectively.

Q: Calculate the boiling point of a solution containing 0.61 g of benzoic acid ($\text{Molar mass} = 122 \text{ g mol}^{-1}$) 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 $2.3 \text{ K kg mol}^{-1}$ respectively.

**Step 1 — molality of benzoic acid** - $\text{Moles} = \frac{0.61}{122} = 0.005 \text{ mol}$ - $\text{Mass of solvent} = 5 \text{ g} = 0.005 \text{ kg}$ - $\text{molality } m = \frac{0.005}{0.005} = $**$1.0 \text{ mol kg}^{-1}$** **Step 2 — van't Hoff factor for dimerisation** For 2 molecules → 1 dimer, with degree of association α = 0.88: - $i = 1 - \frac{\alpha}{2} = 1 - \frac{0.88}{2} = 1 - 0.44 = $**0.56** **Step 3 — elevation of boiling point** $\Delta T_b = i \cdot K_b \cdot m = 0.56 \times 2.3 \times 1.0 = $**$1.288 \text{ K} \approx 1.29 \text{ K}$** **Step 4 — boiling point of solution** $T_b = T_b^{\circ}(\text{solvent}) + \Delta T_b = 46.2 + 1.288 = $**$47.49\,^{\circ}\text{C}$** ($\approx 47.5\,^{\circ}\text{C}$)

Solutions

Elevation of boiling point with association (dimerisation)

Official Answer

Step 1 — molality of benzoic acid

$\text{Moles} = \frac{0.61}{122} = 0.005 \text{ mol}$
$\text{Mass of solvent} = 5 \text{ g} = 0.005 \text{ kg}$
$\text{molality } m = \frac{0.005}{0.005} =$ $1.0 \text{ mol kg}^{-1}$

Step 2 — van't Hoff factor for dimerisation

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

$i = 1 - \frac{\alpha}{2} = 1 - \frac{0.88}{2} = 1 - 0.44 =$ 0.56

Step 3 — elevation of boiling point

$\Delta T_b = i \cdot K_b \cdot m = 0.56 \times 2.3 \times 1.0 =$ $1.288 \text{ K} \approx 1.29 \text{ K}$

Step 4 — boiling point of solution

$T_b = T_b^{\circ}(\text{solvent}) + \Delta T_b = 46.2 + 1.288 =$ $47.49\,^{\circ}\text{C}$ ( $\approx 47.5\,^{\circ}\text{C}$ )

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

11 mark: molality of benzoic acid = $1.0 \text{ mol kg}^{-1}$ ( $\text{moles} = 0.005$ , $\text{solvent} = 0.005 \text{ kg}$ ).
21 mark: van't Hoff factor for dimerisation $i = 1 - \frac{\alpha}{2} = 0.56$ .
31 mark: $\Delta T_b = i \cdot K_b \cdot m = 1.29 \text{ K}$ and $T_b = 46.2 + 1.29 = 47.49\,^{\circ}\text{C}$ (accept ≈ 47.5 °C).

Hint

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

Quick Oral Answer

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

Analysis & 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 = $1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$ , which is the value of i.
With $\alpha = 0.88, i = 0.56$ — nearly half the particles have paired up.

Exam trap

Use $i = 1 - \frac{\alpha}{2}$ for dimerisation; the general form is $i = 1 + \left(\frac{1}{n} - 1\right)\alpha \text{ with } n = 2$ . Do not use the dissociation form $i = 1 + (n-1)\alpha$ .
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.

Common Mistakes

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

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 ( $i = 1 + \left(\frac{1}{n} - 1\right)\alpha$ ) describes both.

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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, $i = 1 - \frac{0.88}{2} = 0.56$ , reflecting that nearly half the particles have paired up.

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

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