1.1 What is Chemistry?

16.3. Colligative Properties http://www.ck12.org

molesC 12 H 10 = 5 .00 g C 12 H 10 × 154 1 mol C.22 g C^1212 HH^1010 = 0 .0324 mol C 12 H 10

molesC 10 H 8 = 7 .50 g C 10 H 8 × 128 1 mol C.18 g C^1010 HH^88 = 0 .0585 mol C 10 H 8

mass of benzene= 200 .0 g benzene×1000 g1 kg = 0 .2000 kg benzene

molality=^0 .0324 mol C 0 .2000 kg benzene^12 H^10 +^0 .0585 mol C^10 H^8 = 0. 455 m

∆Tf= 5. 12 ◦C/m× 0. 455 m× 1 = 2. 33 ◦C

The freezing point will be lowered by 2.33 °C. Subtracting this from the normal freezing point of benzene (5.5 °C),

this solution will freeze at 3.2 °C.

Values of kfand kbfor some other common solvents are listed in theTable16.3.

TABLE16.3: Molal Boiling-Point Elevation and Freezing-Point Depression Constants of Several
Common Liquids

Solvent Normal Freezing

Point (°C)

kf(°C/m) Normal Boiling

Point (°C)

kb(°C/m)

Water 0 1.86 100 0.52

Benzene 5.5 5.12 80.1 2.53

Ethanol -117.3 1.99 78.4 1.22

Acetic Acid 16.6 3.90 117.9 2.93

Cyclohexane 6.6 20.0 80.7 2.79

Notice that kfis generally larger than kbfor a given substance.

Lesson Summary

• The degree to which particles of a given liquid tend to escape into the gas phase is measured by the liquid’s
  vapor pressure. The vapor pressure of a solution is a colligative property, which means that it is affected only
  by the concentration of solute particles and not their identity.


• The vapor pressure of a solution is lower than the vapor pressure of the pure solvent (P <P°).


• Raoult’s Law states that the vapor pressure of a solution is equal to the product of the vapor pressure of the
  pure solvent and the mole fraction of the solvent (P =χsolventP°).


• The mole fraction of a component in a mixture (χ) can be calculated by dividing the number of particles (or
  moles) of the component by the total number of particles (or moles) in the complete mixture.


• Adding a solute increases the boiling point of a pure solvent. This change can be calculated using the equation
  ∆Tb=kb×m×i.


• Similarly, adding a solute decreases the freezing point of a pure solvent. This change can be calculated using
  the equation∆Tf=kf×m×i.

Review Problems

1. Would you expect water or ethanol (C 2 H 5 OH) to have a higher vapor pressure at a given temperature?


2. Based onFigure16.11, which substance do you suppose has the strongest intermolecular attractions? Which
   has the weakest intermolecular attractions? Which substance would evaporate most quickly?


3. What would be the vapor pressure of a 1.0 molal NaCl solution?