\(2.0\, \ g\) of benzoic acid dissolved in \(25\, \ g\) of benzene shows a depression in freezing point equal to \(1.62\, \ K\). Molal depression constant, \(K_f\) of benzene is \(4.9\, \ Kkg\, \ mol^{-1}\). What is the percentage association of the acid?

99.2%

The correct answer is option 3. 99.2%.

To determine the percentage association of benzoic acid in benzene, we first need to calculate the experimental molar mass of benzoic acid and then compare it with its theoretical molar mass. Benzoic acid is known to dimerize in benzene, which will affect the observed molar mass.

Given:

Mass of benzoic acid \((w_1) = 2.0\, \text{g}\)

Mass of benzene \((w_2) = 25\, \text{g} = 0.025\, \text{kg}\)

Depression in freezing point \((\Delta T_f) = 1.62\, \text{K}\)

Molal depression constant \((K_f) = 4.9\, \text{K kg mol}^{-1}\)

The molality \(m\) is given by the formula:

\(m = \frac{\Delta T_f}{K_f} = \frac{1.62\, \text{K}}{4.9\, \text{K kg mol}^{-1}} = 0.33\, \text{mol kg}^{-1}\)

The number of moles of benzoic acid \((n)\) can be calculated using:

\(n = m \times w_2 = 0.33\, \text{mol kg}^{-1} \times 0.025\, \text{kg} = 0.00825\, \text{mol}\)

Now, using the definition of molar mass:

\(\text{Molar mass (experimental)} = \frac{\text{Mass of solute}}{\text{Number of moles of solute}} = \frac{2.0\, \text{g}}{0.00825\, \text{mol}} = 242.42\, \text{g mol}^{-1}\)

Theoretical molar mass of benzoic acid \((C_6H_5COOH)\) is approximately \(122\, \text{g mol}^{-1}\).

Assume that \(x\) molecules of benzoic acid associate to form a dimer, so \(x = 2\).

The degree of association (\(\alpha\)) is related to the molar mass by:

\(\text{Observed molar mass} = \frac{\text{Theoretical molar mass}}{1 - \alpha + \frac{\alpha}{2}} = \frac{122\, \text{g mol}^{-1}}{1 - \frac{\alpha}{2}}\)

Now, set this equal to the experimental molar mass:

\(242.42 = \frac{122}{1 - \frac{\alpha}{2}}\)

Solving for \(\alpha\)

\(1 - \frac{\alpha}{2} = \frac{122}{242.42} \approx 0.503\)

\(\frac{\alpha}{2} = 1 - 0.503 = 0.497\)

\(\alpha = 0.994\)

\(\text{Percentage association} = \alpha \times 100\% = 0.994 \times 100\% = 99.4\%\)

Given that the closest option is 99.2%, we conclude that the correct answer is option 3. 99.2%