Skip to content
Question

Q.Calculate the boiling point of a solution containing 0.61 g of benzoic acid (Molar mass =122 g mol−1= 122\ g\ mol^{-1}) in 5 g of CS2CS_2 in which it dimerises to the extent of 88%. The boiling point and KbK_b of CS2CS_2 are 46.2 ∘C46.2\ ^{\circ}C and 2.3 K kg mol−12.3\ K\ kg\ mol^{-1} respectively.

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The boiling point is found by first calculating the effective molality after accounting for 88% dimerisation of benzoic acid in CS₂, then applying the boiling point elevation formula. The final boiling point is 47.0 ∘C47.0\ ^{\circ}C.

Why This Approach Works

This is a colligative properties problem with a twist — the solute undergoes dimerisation. Benzoic acid molecules associate in pairs in non-polar solvents like carbon disulphide (CS₂). When molecules dimerise, the number of independent particles in solution decreases, which directly lowers the observed boiling point elevation compared to an ideal (non-associating) solution.

The key relationship is:

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

where ii is the van't Hoff factor (actual number of particles per formula unit dissolved), KbK_b is the ebullioscopic constant, and mm is the molality if no association occurred. The dimerisation reduces ii below 1.


Step-by-Step Solution

1. Calculate the molality assuming no dimerisation

First, find the moles of benzoic acid:

Moles of benzoic acid=0.61 g122 g mol−1=0.005 mol\text{Moles of benzoic acid} = \frac{0.61\ \text{g}}{122\ \text{g mol}^{-1}} = 0.005\ \text{mol}

Mass of solvent (CS₂) = 5 g=0.005 kg5\ \text{g} = 0.005\ \text{kg}.

So the theoretical molality (if all molecules remained as monomers) is:

mtheoretical=0.005 mol0.005 kg=1.0 mol kg−1m_{\text{theoretical}} = \frac{0.005\ \text{mol}}{0.005\ \text{kg}} = 1.0\ \text{mol kg}^{-1}

2. Determine the van't Hoff factor from the extent of dimerisation

Let the degree of dimerisation be α=0.88\alpha = 0.88 (88%). Consider the equilibrium:

2 monomers⇌1 dimer2\ \text{monomers} \rightleftharpoons 1\ \text{dimer}

If we start with 1 mole of monomer (as formula units), then after dimerisation:

  • Moles of monomer remaining = 1−α1 - \alpha
  • Moles of dimer formed = α/2\alpha/2
  • Total moles of particles = (1−α)+α/2=1−α/2(1 - \alpha) + \alpha/2 = 1 - \alpha/2

The van't Hoff factor ii is the ratio of actual particles to formula units dissolved:

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

Watch out

A common mistake is to use i=1+αi = 1 + \alpha (which is for dissociation) or to forget that each dimerisation event removes one particle (two monomers become one dimer), so the reduction is α/2\alpha/2, not α\alpha.

3. Calculate the effective molality

The effective molality that determines ΔTb\Delta T_b is: …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.