1. Colligative Properties Honors Chemistry Unit 8 Chapter 15

2. Colligative Properties These are the effects that a solute has on a solvent. When water has something dissolved in it, its physical properties change. It will no longer boil at 100 o C and it will no longer freeze at 0 o C like pure water.

3. Three Main Effects Lowers the vapor pressure of a solvent. –Lower vapor pressure means that fewer water molecules can escape from the liquid phase into the gas phase at given temperature. Remember, a lower vapor pressure means a higher boiling point! 2) Raises the boiling point of a solvent. 3) Lowers the freezing point of a solvent.

4. Freezing Point Depression

5. Boiling Point Elevation

6. Applications –salting icy roads –making ice cream –antifreeze cars (-64°C to 136°C) fish & insects

7. Colligative Properties -These depend only on the number of dissolved particles -Not on what kind of particle General Rule: The more solute particles that are present in a solvent, the greater the effect.

8. # of Particles –Nonelectrolytes (covalent) remain intact when dissolved 1 particle –Electrolytes (ionic) dissociate into ions when dissolved 2 or more particles –Electrolytes have a stronger affect in lowering the freezing point and elevating the boiling point because it puts more particles into the solution.

9. Remember the rule, the more particles, the greater the effect! The Dissociation Factor (d.f.) for an electrolyte is the number of ions a compound dissociates into. NaCl gives Na + ions and Cl - ions, which is 2 particles, therefore d.f. = 2 What is “ d.f. “ for Al(NO 3 ) 3 ? Al(NO 3 ) 3  Al 3+ + 3 NO 3 - = 4 particles

10. Non-electrolytes A compound that does not conduct electricity when dissolved in water. Examples are glucose or any other sugars and alcohols, such as ethanol (CH 3 CH 2 OH) The d.f. = 1 for any non-electrolyte. Why? Covalent molecules do not break apart when they become solvated by water molecules.

11. Calculations  t :change in temperature (° C ) K :constant based on the solvent (° C·kg/mol ) Use K b for boiling point elevation and K f for freezing point depression. Each solvent has it’s own unique factors! m :molality ( m ) = moles solute/Kg solvent d.f. :# of particles  t = m · d.f. · K

12. Example 1: If 48 moles of ethylene glycol, C 2 H 4 (OH) 2, is dissolved in 5.0kg of water. What is the boiling point of the solution? ∆Tbp = m  d.f.  Kb d.f. = 1 ( it is a non-eletrolyte) Kb = 0.52 o C/m (a constant for water) Molality = m = moles solute/kg solvent ∆Tbp = (48 moles) (1) (0.52 o C/molal) (5.0 Kg) (5.0 Kg) ∆Tbp = 5.0 o C New BP = 100 + 5 = 105 o C

13. Example 2: 55.5 grams of CaCl 2 are dissolved in 2.0kg of water. What is the freezing point of this solution? ∆Tfp = m  d.f.  Kf d.f. = 3 (CaCl 2 produces 3 particles) CaCl 2  Ca 2+ + 2 Cl - Kf = 1.86 o C/m (a constant for water) ∆Tfp = (0.25molal) (3) (1.86 o C/molal) Must calculate molality Molality = mass/molar mass Kg Solvent = (55.5 gram/111g/mole) 2.0 kg = 0.25m = 1.4 o C New FP = -1.4 o C