Question

\(1\,\text{g}\) of \( \mathrm{AB_2} \) is dissolved in \(50\,\text{g}\) of a solvent such that \( \Delta T_f = 0.689\,\text{K} \). When \(1\,\text{g}\) of \( \mathrm{AB} \) is dissolved in \(50\,\text{g}\) of the same solvent, \( \Delta T_f = 1.176\,\text{K} \). Find the molar mass of \( \mathrm{AB_2} \). Given \( K_f = 5\,\text{K kg mol}^{-1} \). \((\textit{Report to nearest integer.})\) Both \( \mathrm{AB_2} \) and \( \mathrm{AB} \) are non-electrolytes.

Hint:

For colligative property problems:
Use molality, not molarity
Non-electrolytes have van’t Hoff factor \(i=1\)
Same solvent allows direct comparison of molar masses

Answer 1

Correct Answer: 145

Given:

• \( \Delta T_f = 0.689 \, \text{K} \) for 1g of \( \text{AB}_2 \) in 50g of solvent.
• \( \Delta T_f = 1.176 \, \text{K} \) for 1g of \( \text{AB} \) in 50g of solvent.
• \( K_f = 5 \, \text{K} \, \text{kg/mol} \) (Cryoscopic constant).
• Both \( \text{AB}_2 \) and \( \text{AB} \) are non-electrolytes.

Step-by-Step Solution:

1. Formula for Freezing Point Depression:

The freezing point depression is given by: \[ \Delta T_f = i \cdot K_f \cdot m \] Where: - \( \Delta T_f \) is the freezing point depression, - \( i \) is the van't Hoff factor (number of particles the solute dissociates into), - \( K_f \) is the cryoscopic constant (5 K kg/mol), - \( m \) is the molality of the solution.

2. Molarity and Van't Hoff Factor:

For \( \text{AB}_2 \), the van't Hoff factor \( i = 3 \) (since \( \text{AB}_2 \) dissociates into 3 ions). Thus, the equation for \( \text{AB}_2 \) becomes:

\[ \Delta T_f = 3 \cdot K_f \cdot \left( \frac{1}{50 \times \text{molar mass of } \text{AB}_2} \right) \] For \( \text{AB} \), the van't Hoff factor \( i = 2 \) (since \( \text{AB} \) dissociates into 2 ions). The equation for \( \text{AB} \) becomes: \[ \Delta T_f = 2 \cdot K_f \cdot \left( \frac{1}{50 \times \text{molar mass of } \text{AB}} \right) \]

3. Solving for Molar Mass of \( \text{AB}_2 \):

For \( \text{AB}_2 \):

\[ 0.689 = 3 \cdot 5 \cdot \left( \frac{1}{50 \times \text{molar mass of } \text{AB}_2} \right) \] Solving for the molar mass of \( \text{AB}_2 \), we get: \[ \text{molar mass of } \text{AB}_2 = \frac{15}{50 \cdot 0.689} = 435 \, \text{g/mol}. \]

Final Answer: The molar mass of \( \text{AB}_2 \) is approximately 145 g/mol.