Question

0.3 g of acetic acid (Molar mass = 60 g mol\(^{-1}\)) dissolved in 30 g of benzene shows a depression in freezing point equal to 0.45°C. Calculate the percentage association of acid if it forms a dimer in the solution.
(Given: \( K_f \) for benzene = 5.12 K kg mol\(^{-1}\))

Hint:

Remember, depression in freezing point is directly related to the number of particles in solution. If association occurs, the number of particles is reduced, affecting the freezing point.

Answer 1

The depression in freezing point (\(\Delta T_f\)) is given by the formula: \[ \Delta T_f = \frac{K_f \cdot m}{w} \] where \(m\) is the molality of the solution, and \(w\) is the molar mass of the solute. From the given data: \[ \Delta T_f = 0.45^\circ C, \, K_f = 5.12 \, \text{K kg mol}^{-1}, \, \text{and mass of acetic acid} = 0.3 \, \text{g}. \] Using the formula and calculating the molality, we can solve for the degree of association. Assuming the dimerization of acetic acid, the apparent molar mass increases, which will affect the depression in freezing point. Solving for the percentage association, we find that the acid forms a dimer with a degree of association of approximately 60%.