Question

2.0 g of benzoic acid dissolved in 25.0g of benzene shows a depression in freezing point equal to 1.62K. Molal depression constant (K${}_f$) of benzene is 4.9 K. kg. mol${}^{-1}$. What is the percentage association of the acid?

• A. 99.2%
• B. 98.2%
• C. 97.2%
• D. 96.2%

Answer 1

The correct option is A 99.2%

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.

$1.62=\frac{4.9\times 2.0}{M_2\times 25}$

Thus the molecular weight of benzoic acid is 0.242 kg/mol or 242 g/mol.

When benzoic acid is not associated, its molecular weight is 122 g/mol.

Thus, its percentage of association is $\frac{100\times 242}{2\times 122}=99.2$%

Hence, the correct option is $\text{A}$