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: van't Hoff factor for complete dimerisation.
Moles of benzoic acid $= \dfrac{61}{122} = 0.5$ mol. Complete dimerisation: every 2 molecules form 1 dimer, so $i = 0.5$. Effective moles of solute $= 0.5 \times 0.5 = 0.25$ mol.
Step 2: Moles of solvent.
Moles of benzene ($M = 78$ g mol$^{-1}$) $= \dfrac{500}{78} \approx 6.41$ mol.
Step 3: Mole fraction of solute.
$x_{\text{solute}} = \dfrac{0.25}{0.25 + 6.41} = \dfrac{0.25}{6.66} \approx 0.0376$.
Step 4: Apply Raoult's law.
$\dfrac{p^\circ - p_s}{p^\circ} = x_{\text{solute}}$, so $p^\circ - p_s = 0.0376 \times 66 \approx 2.48$ torr. $p_s = 66 - 2.48 \approx 63.52$ torr. \[ \boxed{p_s \approx 63.52 \text{ torr}} \]