# a 2g of benzoic acid dissolve in 25g of benzene shows a depression in freezing point of 1.62 k.molal depression constant for benzene is 4.9 KKg/mol. what is the % of association of acid.if it forms a dimer in soln .

Given:

WA = 2 g

WB= 25 g,

Kf = 4.9 K kg mol–1

Tf = 1.62 K

Molar mass of the solute can be found out using the equation:  = MB = 4.9 K kg mol–1 × 2g × 1000 g kg –1/1.62 K × 2g = 241.98 g mol–1

Therefore, molar mass of benzoic acid in benzene is= 241.98 g mol–1

Considering the following equilibrium for the acid:

2C6H5COOH ⇔ (C6H5COOH)2

If x represents the degree of association of the solute then we would have (1 – x ) mol of benzoic acid left in unassociated form and x/2 as associated moles of benzoic acid at equilibrium. Correspondingly x/2 as associated moles of benzoic acid at equilibrium.

Therefore, total number of moles at equilibrium will be:

= 1−x+x/2

1-x/2 = i

Total number of moles of particles at equilibrium equals van’t Hoff factor i.

We know that,

i = Normal molar mass / Abnormal molar mass

= 122 g mol-1/241.98 g mol

1-x/2 = i = 122 g mol-1/241.98 g mol

or x/2 = 1 – 122 /241.98 = 1 – 0.0504 = 0.496

or x = 2 × 0 .496 = 0.992

Therefore, degree of association of benzoic acid in benzene is 99.2 %.

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