# The freezing poin

Given: ΔTf = 2.12K

Kf = 5.12 K kg mol-1

Weight of benzoic acid (solute) = 2.5g

Molar mass of benzoic acid, (Msolute) = 12×6+1×5+12+16+16+1 =122 g/mol

Weight of benzene (solvent) = 25g

First we will find out the van’t Hoff factor by applying the formula:

ΔTf = i × Kf × m

Where m = 2.12K = i × 5.12 K kg mol-1 × i = 0.505

Thus, the van’t Hoff factor of benzoic acid is 0.505

Now to calculate percentage association of benzoic acid, we apply the formula given below:

i= 1 - where α is the degree of association

By putting the values in the formula, we get

α = 0.99

Thus, the percentage association of benzoic acid is 99%

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