Question

Among the following 0.1 m aqueous solutions, which one will exhibit the lowest boiling point elevation, assuming complete ionization of the compounds in solution?

• A. Aluminium sulphate
• B. Potassium sulphate
• C. Sodium chloride
• D. Aluminium chloride

Hint:

When dealing with colligative properties like boiling point elevation or freezing point depression, remember that the more ions a solute dissociates into, the greater the effect on the property.

Answer 1

The Correct Option is C

Boiling point elevation (\(\Delta T_b\)) is directly proportional to the molality of the solution and the number of ions formed by dissociation. The formula for boiling point elevation is: \[ \Delta T_b = i \times K_b \times m \] Where: - \(i\) is the van 't Hoff factor (number of ions formed per formula unit), - \(K_b\) is the ebullioscopic constant (depends on the solvent), - \(m\) is the molality of the solution. Since all the solutions have the same molality (0.1 m), the key factor in determining the boiling point elevation is the van 't Hoff factor \(i\). The ions produced by each compound are: - Aluminium sulphate (\( \text{Al}_2(\text{SO}_4)_3 \)) dissociates into 5 ions. - Potassium sulphate (\( \text{K}_2\text{SO}_4 \)) dissociates into 3 ions. - Sodium chloride (\( \text{NaCl} \)) dissociates into 2 ions. - Aluminium chloride (\( \text{AlCl}_3 \)) dissociates into 4 ions. The boiling point elevation will be lowest for the solution with the lowest \(i\), which is Sodium chloride (\( \text{NaCl} \)) with \(i = 2\). Thus, the answer is Sodium chloride.