Equal molar amounts (1 mol each) of glucose and NaCl are dissolved in separate 1 kg samples of water. Which solution will have the greater freezing point depression, and why?
AThe glucose solution, because glucose has a higher molar mass
BThey will be equal, because both solutions contain the same number of moles of solute
CThe NaCl solution, because NaCl dissociates into approximately two ions per formula unit, doubling the effective particle count
DThe NaCl solution, because ionic solutes always lower the freezing point more than molecular solutes regardless of concentration
Colligative properties depend on the number of dissolved particles, not the chemical identity or mass of the solute. NaCl dissociates into Na⁺ and Cl⁻ (i ≈ 2), effectively producing ~2 mol of particles per mol of NaCl dissolved. Glucose does not dissociate (i = 1). ΔTf = iKfm, so the NaCl solution has roughly twice the freezing point depression. Option D is wrong because it's particle count, not ionic character per se, that matters.
Question 2 True / False
Two solutions at the same molality should have the same boiling point elevation if they are at the same temperature and pressure.
TTrue
FFalse
Answer: False
Boiling point elevation depends on the total concentration of dissolved particles (ΔTb = iKbm), so the van't Hoff factor i matters. A 1.0 m NaCl solution (i ≈ 2) has roughly twice the boiling point elevation of a 1.0 m glucose solution (i = 1). Equal molality does not guarantee equal colligative effects unless the solutes have the same degree of dissociation.
Question 3 Short Answer
Why does salting an icy road cause the ice to melt at temperatures below 0°C? Invoke the relevant colligative property in your answer.
Think about your answer, then reveal below.
Model answer: Salt (NaCl) dissolves in a thin film of liquid water on the ice surface and dissociates into Na⁺ and Cl⁻, increasing the total dissolved particle concentration. This lowers the freezing point of the solution below 0°C via freezing point depression (ΔTf = iKfm). As long as the ambient temperature is above the new (depressed) freezing point of the brine, the ice melts into liquid solution rather than remaining solid.
This question connects the abstract formula to a concrete physical process. The key insight is that adding dissolved particles disrupts the equilibrium between liquid and solid water — the solution's chemical potential is lower than that of pure ice at 0°C, so ice melts to reach a new equilibrium. This is the same thermodynamic logic behind all colligative properties.