Question

Calculate the boiling point of a solution containing 0.61 g of benzoic acid (122 \( g mol^{-1} \)) in 5 g of \( CS_2 \) in which it dimerises to 88%. \( T_b^\circ(CS_2) = 46.2^\circ C, K_b = 2.3 K kg mol^{-1} \).

Hint:

Always check if the solute associates or dissociates. Dimerization (n=2) is common for organic acids in non-polar solvents like $CS_2$ or Benzene.

Answer 1

Step 1: Understanding the Concept:
Solutes that associate in solution show a lower boiling point elevation than expected. We use the van't Hoff factor (\( i \)).
Step 2: Detailed Explanation:
Moles of benzoic acid \( = 0.61 / 122 = 0.005 \) mol.
Molality \( m = 0.005 / (5 \times 10^{-3}) = 1.0 \) m.
\( \alpha = 0.88 \). \( i = 1 - (1 - 0.5) \times 0.88 = 1 - 0.44 = 0.56 \).
\( \Delta T_b = 0.56 \times 2.3 \times 1.0 = 1.288^\circ C \).
Boiling point of solution \( = 46.2 + 1.288 = 47.488^\circ C \).
Step 3: Final Answer:
The boiling point is \( 47.488^\circ C \).