Which 1 m solution would have the highest vapor pressure at a given temperature?

• A. 1 m aluminum sulfate (Al2(SO4)3) in water
• B. 1 m sodium chloride (NaCl) in water
• C. 1 m magnesium chloride (MgCl2) in water
• D. 1 m glucose (C6H12O6) in water

Answer: (D)

Verified step by step guidance

1. Understand that vapor pressure lowering is a colligative property, which depends on the number of solute particles in solution, not their identity.
2. Recall that the vapor pressure of a solvent decreases when a non-volatile solute is dissolved because the solute particles reduce the number of solvent molecules at the surface.
3. Calculate the van't Hoff factor (\(i\)) for each solute, which represents the number of particles the solute dissociates into in solution: - For glucose (C6H12O6), \(i = 1\) (no dissociation), - For NaCl, \(i = 2\) (Na\(^+\) and Cl\(^-\)), - For MgCl\(_2\), \(i = 3\) (Mg\(^{2+}\) and 2 Cl\(^-\)), - For Al\(_2\)(SO\(_4\))\(_3\), \(i = 5\) (2 Al\(^{3+}\) and 3 SO\(_4^{2-}\)).
4. Recognize that the effective concentration of particles in solution is \(i \times m\), where \(m\) is the molality. Since all solutions are 1 m, the total particle concentration differs due to \(i\).
5. Conclude that the solution with the smallest \(i\) (glucose, \(i=1\)) produces the fewest solute particles, thus causing the least vapor pressure lowering and therefore the highest vapor pressure at the given temperature.