Question

A solute X is found to exist as a dimer in water. A 4 molal solution of X shows a boiling point of 101.04$^\circ\text{C}$. What is the percentage association of X? (K_b for water = 0.52 K/m).

• A. 75
• B. 100
• C. 80
• D. 40

Hint:

When calculating the degree of association, make sure to apply the correct van't Hoff factor and understand that the solute's association directly influences the van't Hoff factor in colligative properties like boiling point elevation.

Answer 1

The Correct Option is B

The elevation in boiling point (\(\Delta T_b\)) is related to the molality (m) of the solution by the formula: \[ \Delta T_b = K_b \times m \times i \] Where: - \(K_b\) is the ebullioscopic constant of the solvent, - \(m\) is the molality of the solution, - \(i\) is the van't Hoff factor, which represents the number of particles the solute dissociates into. For dimerization, \(i = 2(1 - \alpha) + \alpha\), where \(\alpha\) is the degree of association. Given: \[ \Delta T_b = 101.04^\circ \text{C} - 100^\circ \text{C} = 1.04^\circ \text{C}, \quad K_b = 0.52 \text{ K/m}, \quad m = 4 \text{ molal} \] Substituting into the equation: \[ 1.04 = 0.52 \times 4 \times (2(1 - \alpha) + \alpha) \] Simplifying: \[ 1.04 = 2.08 \times (2 - \alpha) \] \[ \frac{1.04}{2.08} = 2 - \alpha \] \[ 0.5 = 2 - \alpha \] \[ \alpha = 1.5 \] Since the degree of dissociation (\(\alpha\)) can range from 0 to 1, the actual dimerization factor (\(\alpha = 1\)) corresponds to 100% association.
Thus, the percentage association of X is 100%.