Which of the following aqueous solutions would have the lowest vapor pressure?

• A. 0.5 mol KBr in 1.0 kg H2O
• B. 0.5 mol NaCl in 1.0 kg H2O
• C. 1.0 mol C6H12O6 (glucose) in 1.0 kg H2O
• D. 1.0 mol Na2SO4 in 1.0 kg H2O

Answer: (D)

Verified step by step guidance

1. Recall that the vapor pressure of a solution is lowered compared to pure solvent due to the presence of solute particles, a phenomenon known as vapor pressure lowering, which is a colligative property depending on the number of solute particles in solution.
2. Calculate the van't Hoff factor (\(i\)) for each solute, which represents the number of particles the solute dissociates into in solution: for KBr and NaCl, \(i = 2\) (since they dissociate into two ions), for glucose, \(i = 1\) (non-electrolyte, no dissociation), and for Na\(_2\)SO\(_4\), \(i = 3\) (dissociates into 2 Na\(^+\) and 1 SO\(_4^{2-}\) ions).
3. Determine the molality (\(m\)) of each solution, which is moles of solute per kilogram of solvent. Here, all solutions have 1.0 kg of water, so molality equals the moles of solute given.
4. Calculate the effective concentration of particles in solution by multiplying the molality by the van't Hoff factor: \(i \times m\). This value correlates directly with the extent of vapor pressure lowering.
5. Compare the values of \(i \times m\) for all solutions; the solution with the highest value will have the greatest vapor pressure lowering and thus the lowest vapor pressure.