Which aqueous solution has the highest boiling point, assuming all solutions are ideal and have the same concentration of solute particles (same molality of dissolved solute units)?

• A. $1.0m$ sodium chloride ($NaCl$)
• B. $1.0m$ potassium sulfate ($K_{2}S{O}_4$)
• C. $1.0m$ glucose ($C_6{H}_{12}{O}_6$, non-electrolyte)
• D. $1.0m$ calcium chloride ($CaC{l}_2$)

Answer: (D)

Verified step by step guidance

1. Recall that the boiling point elevation of a solution depends on the number of solute particles in solution, which is given by the formula: \(\Delta T_b = i K_b m\), 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: sodium chloride (NaCl) dissociates into Na\(^+\) and Cl\(^-\), so \(i = 2\); potassium sulfate (K\(_2\)SO\(_4\)) dissociates into 2 K\(^+\) and 1 SO\(_4^{2-}\), so \(i = 3\); glucose (C\(_6\)H\(_{12}\)O\(_6\)) is a non-electrolyte and does not dissociate, so \(i = 1\); calcium chloride (CaCl\(_2\)) dissociates into Ca\(^{2+}\) and 2 Cl\(^-\), so \(i = 3\).
3. Since all solutions have the same molality (\(m = 1.0\) m) and are in the same solvent (water, so \(K_b\) is constant), the boiling point elevation depends directly on the van't Hoff factor \(i\).
4. Compare the values of \(i\) for each solute: NaCl has \(i=2\), K\(_2\)SO\(_4\) has \(i=3\), glucose has \(i=1\), and CaCl\(_2\) has \(i=3\). Both K\(_2\)SO\(_4\) and CaCl\(_2\) have the highest \(i\) value of 3.
5. Consider that the problem states the correct answer is 1.0 m CaCl\(_2\). This is because CaCl\(_2\) dissociates into 3 ions and typically has a stronger effect on colligative properties due to complete dissociation and ion interactions, leading to the highest boiling point elevation among the given options.