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Colligative properties are solution properties that depend on the number of dissolved particles, not the identity of those particles. This reference helps students connect concentration, temperature changes, vapor pressure, and osmosis in one place. It is especially useful for solving molality, boiling point elevation, freezing point depression, and osmotic pressure problems.

These ideas explain real examples such as antifreeze, salted ice, and water movement through membranes.

The most important quantity is the effective particle concentration, which often includes the van't Hoff factor ii. Temperature changes use molality: ΔTb=iKbm\Delta T_b = iK_bm and ΔTf=iKfm\Delta T_f = iK_fm. Osmotic pressure uses molarity: Π=iMRT\Pi = iMRT.

Solution comparisons depend on particle concentration, so a solution with more dissolved particles has lower vapor pressure, higher boiling point, lower freezing point, and greater osmotic pressure.

Key Facts

  • Colligative properties depend on the number of solute particles in solution, not on the chemical identity of the solute.
  • The van't Hoff factor ii represents the number of dissolved particles produced per formula unit of solute.
  • Molality is defined as m=mol solutekg solventm = \frac{\text{mol solute}}{\text{kg solvent}} and is used in boiling point and freezing point calculations.
  • Boiling point elevation is calculated with ΔTb=iKbm\Delta T_b = iK_bm, so the solution boiling point is Tb=Tb+ΔTbT_b = T_b^{\circ} + \Delta T_b.
  • Freezing point depression is calculated with ΔTf=iKfm\Delta T_f = iK_fm, so the solution freezing point is Tf=TfΔTfT_f = T_f^{\circ} - \Delta T_f.
  • Osmotic pressure is calculated with Π=iMRT\Pi = iMRT, where MM is molarity and TT is temperature in kelvins.
  • Vapor pressure lowering occurs because solute particles reduce the mole fraction of solvent, often described by Psolution=XsolventPsolventP_{\text{solution}} = X_{\text{solvent}}P_{\text{solvent}}^{\circ} for ideal solutions.
  • For nonelectrolytes such as glucose, i=1i = 1, while ideal NaCl\text{NaCl} has i2i \approx 2 and ideal CaCl2\text{CaCl}_2 has i3i \approx 3.

Vocabulary

Colligative property
A property of a solution that depends on the number of dissolved solute particles rather than their identity.
Molality
A concentration unit equal to moles of solute per kilogram of solvent, written as m=mol solutekg solventm = \frac{\text{mol solute}}{\text{kg solvent}}.
Van't Hoff factor
The factor ii that tells how many dissolved particles are produced from each formula unit of solute.
Boiling point elevation
The increase in a solution's boiling point compared with the pure solvent, calculated using ΔTb=iKbm\Delta T_b = iK_bm.
Freezing point depression
The decrease in a solution's freezing point compared with the pure solvent, calculated using ΔTf=iKfm\Delta T_f = iK_fm.
Osmotic pressure
The pressure needed to stop solvent from moving through a semipermeable membrane, calculated using Π=iMRT\Pi = iMRT.

Common Mistakes to Avoid

  • Using molarity instead of molality for temperature changes is wrong because ΔTb=iKbm\Delta T_b = iK_bm and ΔTf=iKfm\Delta T_f = iK_fm require mm, not MM.
  • Forgetting the van't Hoff factor gives answers that are too small for electrolytes because ionic compounds produce more than one dissolved particle per formula unit.
  • Adding freezing point depression to the normal freezing point is wrong because freezing point depression lowers the freezing point, so Tf=TfΔTfT_f = T_f^{\circ} - \Delta T_f.
  • Using Celsius in the osmotic pressure equation is wrong because Π=iMRT\Pi = iMRT requires absolute temperature in kelvins.
  • Assuming every ionic compound dissociates perfectly can be wrong in real solutions because ion pairing may make the actual ii smaller than the ideal value.

Practice Questions

  1. 1 A solution contains 0.750 mol0.750\ \text{mol} of glucose dissolved in 0.500 kg0.500\ \text{kg} of water. What is the molality mm?
  2. 2 Calculate the boiling point elevation for a 1.20 m1.20\ m glucose solution in water if Kb=0.512 C/mK_b = 0.512\ ^{\circ}\text{C}/m and i=1i = 1.
  3. 3 A 0.300 M0.300\ M NaCl\text{NaCl} solution is at 25.0 C25.0\ ^{\circ}\text{C}. Assuming i=2i = 2 and R=0.0821 Latmmol1K1R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}, calculate Π\Pi.
  4. 4 Two aqueous solutions have the same molality: glucose with i=1i = 1 and CaCl2\text{CaCl}_2 with i3i \approx 3. Which solution should have the lower freezing point, and why?