Transcript Back to School Night

•Dissociation is separation of ions that occurs when an ionic
compound dissolves.
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
Vapor-Pressure Lowering
• A nonvolatile substance : has little tendency to become a gas
under existing conditions.
• 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
Kf
m
= # of dissociated moles of solute particles per
mole of compound (van't Hoff factor )
= freezing-point constant
= molality of the solution
•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
Kb
m
= # of dissociated moles of solute particles
per mole of compound (van't Hoff factor )
= boiling-point constant
= molality of the solution
Molal Freezing-Point and Boiling-Point Constants
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:
1. Calculate the molality of the solution
17.1 g C12H22O11 
1 mol solute
 0.0500 mol C12H22O11
342.34 g C12H22O11
0.0500 mol C12H22O11
1000 g water

200. g water
1 kg water

0.250 mol C12H22O11
kg water
 .250 m
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) = −
0.465 °C
Sample Problem
What is the boiling-point elevation of a solution made from 20.1 g of
NaCl and 400.0 g of water? The molar mass of NaCl is 58.44 g.
Given: solute mass = 20.1 g
solute molar mass = 58.44 g
solvent mass and identity = 400.0 g of water
Solution:
1. Calculate the number of moles of the solute.
20.1 g NaCl ÷ 58.44 g/mol = 0.344 mol
2.
Calculate the molality of the solution.
0.344 mol
= 0.860 m
0.400 kg
3. Calculate the difference in boiling points.
Since salt dissociates into two difference ions , i = 2
∆tb = (2)( 0.860 m)(0.51°C/m) = 0.877 °C