Question

The freezing point of a solution containing 0.2 g of acetic acid in 20.0g of benzene is lowered by 0.45${}^o$C. Calculate the degree of association of acetic acid in benzene. (K${}_f$ for benzene $=$ 5.12 K mol${}^{-1}$ kg)

• A. 94.5%
• B. 95.5%
• C. 96.5%
• D. 97.5%

Answer 1

Answer: (A)

94.5%

The expression for the depression in the freezing point and the molar mass of the solute is as shown below.
$\Delta T_f=\frac{K_f\times W_2}{M_2\times W_1}$
Substitute values in the above expression.
$0.45=\frac{5.12\times 0.2}{M_2\times 20}$
Thus the molecular weight of acetic acid is 0.1137 kg/mol or 113.7 g/mol.
When acetic acid is not associated, its molecular weight is 60 g/mol.
Thus, its percentage of association is $\frac{100\times 113.7}{2\times 60}=94.5$%