Answer :

Given: ΔT_{f} = 2.12K

K_{f} = 5.12 K kg mol^{-1}

Weight of benzoic acid (solute) = 2.5g

Molar mass of benzoic acid, (M_{solute}) = 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:__

ΔT_{f} = i × K_{f} × 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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