Which of the following aqueous solutions is expected to have the highest boiling point?

• A. 0.5 molal CaCl2
• B. 0.5 molal NaCl
• C. 0.5 molal C6H12O6 (glucose)
• D. 0.5 molal KBr

Answer: (A)

Verified step by step guidance

1. Recall that the boiling point elevation of a solution depends on the number of solute particles dissolved in the solvent, not just the concentration of the solute. This is a colligative property described by the formula: \(\Delta T_b = i \cdot K_b \cdot m\), where \(\Delta T_b\) is the boiling point elevation, \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(K_b\) is the boiling point elevation constant of the solvent, and \(m\) is the molality of the solution.
2. Identify the van't Hoff factor (\(i\)) for each solute: - For \(\mathrm{CaCl_2}\), it dissociates into 3 ions (\(\mathrm{Ca^{2+}}\) and 2 \(\mathrm{Cl^-}\)), so \(i = 3\). - For \(\mathrm{NaCl}\), it dissociates into 2 ions (\(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\)), so \(i = 2\). - For glucose (\(\mathrm{C_6H_{12}O_6}\)), it does not dissociate, so \(i = 1\). - For \(\mathrm{KBr}\), it dissociates into 2 ions (\(\mathrm{K^+}\) and \(\mathrm{Br^-}\)), so \(i = 2\).
3. Since all solutions have the same molality (\$0.5\( molal), the difference in boiling point elevation depends on the van't Hoff factor \)i\(. Calculate the effective concentration of particles for each solution by multiplying \)i\( by the molality \)m$.
4. Compare the values of \(i \cdot m\) for each solution to determine which has the highest boiling point elevation. The solution with the highest \(i \cdot m\) will have the highest boiling point.
5. Conclude that the solution with \$0.5\( molal \)\mathrm{CaCl_2}\(, having the highest van't Hoff factor (\)i=3$), produces the greatest number of dissolved particles and thus the highest boiling point elevation.