Calculate the boiling point of a solution containing 0.61 g of benzoic acid (Molar mass = 122 g mol\(^{-1}\)) in 5 g of \(CS_2\) in which it dimerises to the extent of 88%. The boiling point and \(K_b\) of \(CS_2\) are 46.2 \(^\circ C\) and 2.3 K kg mol\(^{-1}\) respectively.
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Solution and Explanation
When a solute associates (dimerises) in solution, the number of particles falls, so the van't Hoff factor i is less than 1 and the boiling point elevation is smaller than for a non-associating solute.
Step 1:
Calculate the van't Hoff factor. For dimerisation 2 molecules give 1 particle, so \(i = 1 - \frac{\alpha}{2}\) where \(\alpha = 0.88\). Thus \(i = 1 - \frac{0.88}{2} = 1 - 0.44 = 0.56\).
Step 2:
Calculate molality. Moles of benzoic acid = \(\frac{0.61}{122} = 0.005\) mol. Mass of solvent \(CS_2 = 5\) g \(= 0.005\) kg. Molality \(m = \frac{0.005}{0.005} = 1.0\) mol kg\(^{-1}\).
Step 3:
Boiling point elevation: \(\Delta T_b = i\,K_b\,m = 0.56 \times 2.3 \times 1.0 = 1.288\ K \approx 1.29\ ^\circ C\).
Answer: Boiling point of solution \(= 46.2 + 1.29 = 47.49\ ^\circ C \approx 47.5\ ^\circ C\).
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