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 sol^{n }.

Given:

W_{A} = 2 g

W_{B}= 25 g,

K_{f} = 4.9 K kg mol^{–1}

T_{f} = 1.62 K

Molar mass of the solute can be found out using the equation:

=

M_{B }= 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:

2C_{6}H_{5}COOH ⇔ (C_{6}H_{5}COOH)_{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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