Ethanoic acid (normal molar mass 60 g mol^-1) is dissolved in benzene, where it associates as 2 A leftharpoons A_2. The molar mass determined from freezing point depression is 96 g mol^-1. The degree of association is

JEE Main levelSingle correct optionDifficulty 2 of 3so098

Ethanoic acid (normal molar mass $60 g mo l^{- 1}$ ) is dissolved in benzene, where it associates as $2 A ⇌ A_{2}$ . The molar mass determined from freezing point depression is $96 g mo l^{- 1}$ . The degree of association is

i = M_{n or ma l} / M_{o b ser v e d} = 60/96 = 0.625

. For dimerisation,

i = 1 - β /2

, so

β = 2 (1 - 0.625) = 0.75

, i.e.

75%