1. Dissociation is separation of ions that occurs when an ionic compound dissolves.

2. Colligative Properties of Solutions
Properties that depend on the concentration of solute particles but not on their identity
Vapor-Pressure Lowering
Freezing-Point Depression
Boiling-Point Elevation
Osmotic Pressure

3. Vapor-Pressure Lowering
A nonvolatile substance : has little tendency to become a gas under existing conditions.

4. Freezing-point depression(∆tf ) : the difference between the freezing points of the pure solvent and that of a solution
∆tf = i Kf m
where: i = # of dissociated moles of solute particles per mole of compound (van't Hoff factor )
Kf = freezing-point constant
m = molality of the solution

5. Boiling-point elevation (∆tb ) : the difference between the boiling points of the pure solvent and that of the solution
∆tb = i Kb m
where: i = # of dissociated moles of solute particles per mole of compound (van't Hoff factor )
Kb = boiling-point constant
m = molality of the solution

7. Molal Freezing-Point and Boiling-Point Constants

8. Sample Problem
What is the freezing-point depression of water in a solution of 17.1 g of sucrose, C12H22O11, in 200. g of water?
Given: solute mass and chemical formula = 17.1 g C12H22O11
solvent mass and identity = 200 g water
Solution:
Calculate the molality of the solution

9. 2. Solve for change is freezing point. ∆tf = iKfm
Solution:
2. Solve for change is freezing point.
∆tf = iKfm
Since sugar does not dissociate , i = 1
∆tf = (1)(−1.86°C/m)(0.250 m) = − °C

10. Sample Problem
What is the boiling-point elevation of a solution made from 20.1 g of NaCl and g of water? The molar mass of NaCl is g.
Given: solute mass = 20.1 g
solute molar mass = g
solvent mass and identity = g of water
Solution:
Calculate the number of moles of the solute.
20.1 g NaCl ÷ g/mol = mol
2. Calculate the molality of the solution.
0.344 mol = m
0.400 kg
3. Calculate the difference in boiling points.
Since salt dissociates into two difference ions , i = 2
∆tb = (2)( m)(0.51°C/m) = °C