Colligative Properties


Vapor pressure lowering | Boiling point elevation | Freezing point depression | Osmotic pressure | Conceptual questions |

The colligative properties of a solution depend on the relative numbers (concentration) of solute and solvent particles, they do not depend on the nature of the particles. Colligative properties change in proportion to the concentration of the solute particles. We distinguish between four colligative properties: All of the colligative properties fit the relationship
property = solute concentration x constant
Property Symbol Solute Concentration Proportionality Constant
Vapor pressure DP mole fraction Po (vapor pressure of pure solvent)
Boiling Point DTb molal Kb (bpoilng point constant)
Freezing Point DTf molal Kf (freezing point constant)
Osmotic Pressure P molar RT

The determination of colligative properties allows us to

Vapor Pressure Lowering:
The vapor pressure of a solvent in a solution is always lower than the vapor pressure of the pure solvent. The vapor pressure lowering is directly proportional to the molefraction of the solute.
Psolvent = Xsolvent Posolvent
 or     DP = Xsolute Posolvent

where Psolvent  is the vapor pressure of the solvent on the solution, Xsolvent  and Xsolute is the mole fraction  of the solvent or solute, respectively. (Xsolvent + Xsolute = 1). Posolvent is the vapor pressure of the pure solvent, and DP is the lowering in vapor pressure.
Exercise: At 25oC the vapor pressure of pure benzene is 93.9 torr. When a non-volatile solvent is dissolved in benzene, the vapor pressure of benzene is lowered to 91.5 torr. What is the concentration of the solute and the solvent, expressed in mole fraction? (answer: vapor pressure lowering DP= 2.4 torr; X solute = 0.026)

Boiling Point Elevation:
The boiling points of solutions are all higher than that of the pure solvent. Difference between the boiling points of the pure solvent and the solution is proportional to the concentration of the solute particles:
 

DTb = Tb (solution) - Tb (solvent) = Kb x m
where DTb  is the boiling point elevation, Kb is the boiling point elevation constant, and m is the molality (mol/kg solvent) of the solute.

Exercise: A solution is prepared when 1.20 g of a compound is dissolved in 20.0 g of benzene. The boiling point of the solution is 80.94 oC.


Freezing Point Depression:
The freezing points of solutions are all lower than that of the pure solvent. The freezing point depression is directly proportional to the molality of the solute.

DTf = Tf(solvent) - Tf (solution) = Kf x m
where DTf  is the freezing point depression,Tf (solution)  is the freezing point of the solution, Tf(solvent) is the freezing point of the solvent,Kf  is the freezing point depression constant, and m is the molality.
Exercise: Benzophenone has a freezing point of 49.00oC. A 0.450 molal solution of urea in this solvent has a freezing point of 44.59oC. Find the freezing point depression constant for the solvent.(answ.: 9.80oC/m)
 

Osmotic pressure
Osmosis is the diffusion of a fluid through a semipermeable membrane. When a semipermeable membrane (animal bladders, skins of fruits and vegetables) separates a solution from a solvent, then only solvent molecules are able to pass through the membrane. The osmotic pressure of a solution is the pressure difference needed to stop the flow of solvent across a semipermeable membrane. The osmotic pressure of a solution is proportional to the molar concentration of the solute particles in solution.

    P = nRT/V = MM RT

where is the osmotic pressure, R is the ideal gas constant (0.0821 L atm / mol K), T is the temperature in Kelvin, n is the number of moles of solute present, and V is the volume of the solution (n/V is then the molar concentration of the solute), and MM is the molar mass of the solute.
Exercise:Hemoglobin is a large molecule that carries oxygen in human blood. A water solution that contains 0.263 g of hemoglobin (Hb) in 10.0 mL of solution has an osmotic pressure of 7.51 torr at 25oC. What is the molar mass of the hemoglobin? (answ.: 6.51 x 104 g/mol)

Conceptual Questions