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A 1 molal solution of a carboxylic acid in benzene shows a boiling point elevation of 1.518 K. Given that the ebullioscopic constant Kb for benzene is 2.53 K kg/mol, find the degree of association of the acid due to dimerization in benzene.
- 60%
- 70%
- 75%
- 80%
Correct answer: 80%
Solution
DeltaTb = i * Kb * m. 1.518 = i * 2.53 * 1 => i = 1.518/2.53 = 0.6. For dimerization 2A -> A2: if alpha is the fraction associated, initially 1 mol A. At equilibrium: moles of A = 1 - alpha, moles of A2 = alpha/2. Total moles = 1 - alpha + alpha/2 = 1 - alpha/2. So i = (1 - alpha/2)/1 = 1 - alpha/2. Setting i = 0.6: 1 - alpha/2 = 0.6 => alpha/2 = 0.4 => alpha = 0.8 = 80%.
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