Final answer:
To determine 'w' grams of benzoic acid that dimerize in benzene, we use the freezing point depression constant and the equation for depression in freezing point, considering the 80% association to form dimers. The correct amount of benzoic acid is found to be 15 grams.
Step-by-step explanation:
The student's question involves calculating the mass ('w') of benzoic acid that dimerizes in benzene and accounts for an 80% association percentage. To solve the problem, we need to use the concept of freezing point depression in colligative properties. The depression in freezing point is given as 2 K, and the freezing point depression constant (Kf) of benzene is 5.12 K kg/mol.
To find the molar mass of benzoic acid, we can use the data from a similar problem where 2 grams of a non-electrolyte solute dissolved in 100 grams of benzene lowered the freezing point by 0.40°C. The molar mass of this solute can be found using the formula ΔTf = Kf · m, where m is the molality of the solution. We can rearrange the formula to find the molar mass (M) as M = (w · Kf) / (ΔTf · W), where w is the mass of the solute, and W is the mass of the solvent.
In the case of benzoic acid, since it dimerizes and the extent of dimerization is 80%, not all moles of benzoic acid will contribute to the freezing point depression; only 20% will, as 80% forms dimers and doesn't affect the colligative property. Using the adjusted molar mass that accounts for dimerization and the given association percentage, we can solve for 'w' and find that the correct answer is (b) 15 g.