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Step-by-Step Solution
Step 1: Identify the Known Data
• The solute A undergoes dimerization in water.
• The boiling point of the solution is 100.52 °C.
• The boiling point of pure water is 100 °C.
• The molality (M) of the solution is 2 mol/kg.
• The ebullioscopic constant (Kb) for water is 0.52 K·kg·mol–1.
Step 2: Calculate the Elevation in Boiling Point
The elevation in boiling point, \Delta T_b , is given by:
\Delta T_b = \text{(Boiling point of solution)} - \text{(Boiling point of pure solvent)}
Substituting the values:
\Delta T_b = 100.52 - 100 = 0.52 \,^\circ\text{C}
Step 3: Apply the Colligative Property Formula
For a solution, the elevation in boiling point also follows:
\Delta T_b = K_b \, i \, M
where:
• K_b is the ebullioscopic constant (0.52 K·kg·mol–1).
• i is the van't Hoff factor.
• M is the molal concentration (2 mol/kg).
Substitute the known values:
0.52 = (0.52) \times i \times 2
Step 4: Solve for the van't Hoff Factor (i)
From the above equation:
0.52 = 0.52 \times i \times 2
Simplifying:
0.52 = 1.04 \times i
i = \frac{0.52}{1.04} = \frac{1}{2} = 0.5
Step 5: Relate Van't Hoff Factor to Degree of Dimerization
When a solute dimerizes, the van't Hoff factor i is given by:
i = 1 + \biggl(\frac{1}{n} - 1\biggr)\beta
Here,
• n = 2 (because the solute forms a dimer),
• \beta is the degree (or fraction) of dimerization.
Substitute i = \tfrac{1}{2} and n = 2 :
\frac{1}{2} = 1 + \biggl(\frac{1}{2} - 1\biggr)\beta
\frac{1}{2} = 1 - \frac{1}{2}\beta
Step 6: Solve for the Degree of Dimerization (β)
\frac{1}{2} - 1 = -\frac{1}{2} = - \frac{1}{2}\beta
So, -\frac{1}{2} = -\frac{1}{2}\beta
\beta = 1.0
Step 7: Convert Degree of Dimerization to Percentage
If the degree of dimerization \beta = 1 , then it means 100% of the solute molecules are associated into dimers. Hence, the percentage association is 100%.
Final Answer: 100%
