Question

61 g benzoic acid (M = 122 g mol$^{-1}$) dissolved in 500 g benzene. Vapour pressure of pure benzene = 66 torr. Assume complete dimerisation. Calculate vapour pressure of solution.

Hint:

Complete dimerisation: $i = 0.5$. Effective moles halved. Always find mole fraction of solute after accounting for association.

Answer 1

Step 1: Concept
Relative lowering of vapour pressure is a colligative property: $\dfrac{p^\circ - p_s}{p^\circ} = x_{solute}$ (modified by van't Hoff factor for association/dissociation).

Step 2: Calculation
Moles of benzoic acid $= \dfrac{61}{122} = 0.5$ mol.
With complete dimerisation, 2 molecules associate into 1, so $i = 0.5$ (degree of association $\alpha = 1$):
$i = 1 - \left(1 - \dfrac{1}{2}\right)(1) = 1 - 0.5 = 0.5$.
Effective moles of solute $= 0.5 \times 0.5 = 0.25$ mol.
Moles of benzene (solvent, M = 78 g mol$^{-1}$) $= \dfrac{500}{78} = 6.41$ mol.

Step 3: Calculation of mole fraction and vapour pressure
$x_{solute} = \dfrac{0.25}{0.25 + 6.41} = \dfrac{0.25}{6.66} = 0.0375$
Using Raoult's law: $\dfrac{p^\circ - p_s}{p^\circ} = x_{solute}$
$p^\circ - p_s = 0.0375 \times 66 = 2.48$ torr
$p_s = 66 - 2.48 = 63.52$ torr.

Final Answer: Vapour pressure of solution $\approx 63.52$ torr.