Given information:
- One molal solution of a carboxylic acid in benzene
- Boiling point elevation = 1.518 K
- Kb for benzene = 2.53 K kg mol-1

To determine: Degree of association for dimerization of the acid in benzene

Formula:
ΔTb = Kb ∙ molality ∙ degree of association

where,
ΔTb = elevation of boiling point
Kb = ebullioscopic constant for solvent
molality = moles of solute per kg of solvent
degree of association = extent to which the solute associates or dissociates in solution

Solution:
Let's first calculate the molality of the solution.

Molality (m) = 1 mol of solute / 1000 g of solvent

We need to convert the mass of benzene (solvent) to kg.

Mass of benzene = 1000 g
Mass of solute = 1 mol
Molar mass of solute = unknown

Let's assume the molar mass of the solute to be M.

1 mol of solute = M g of solute
Total mass of solution = 1000 g (benzene) + M g (solute)

Molality (m) = 1 mol / (1000 g + M g)

Now, let's use the formula to calculate the degree of association.

ΔTb = Kb ∙ molality ∙ degree of association
1.518 = 2.53 ∙ (1 mol / (1000 g + M g)) ∙ degree of association

Simplifying the equation,

Degree of association = 1.518 ∙ (1000 g + M g) / (2.53 ∙ M)

We can further simplify the equation by assuming the molar mass of the solute to be twice its empirical formula weight (dimerization assumption).

Molar mass of solute = 2 ∙ Empirical formula weight

Degree of association = 1.518 ∙ (1000 g + 2 ∙ EFw g) / (2.53 ∙ 2 ∙ EFw)
Degree of association = (1.518 ∙ 1000) / (5.06 ∙ EFw + 1000)

Let's assume the empirical formula weight of the acid to be x.

Empirical formula weight (EFw) = x
Molar mass of solute = 2 ∙ x

Substituting the values,

Degree of association = (1.518 ∙ 1000) / (5.06 ∙ x + 1000)

We need to find the value of x that gives a degree of association of 80%.

Degree of association = 0.8 = (1.518 ∙ 1000) / (5.06 ∙ x + 1000)

Solving for x,

x = 90 g/mol

Therefore, the empirical formula of the acid is CH2O2 and the degree of association for dimerization of the acid in benzene is 80%.