Q. 125.0( 2 Votes )

The freezing poin

Answer :

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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