1. Chapter 14: Solutions Chem 103: Chapter 14
Copyright 2006, David R. Anderson

2. More Concentration Units
Molar concentration
Mole fraction and mole percent
Weight fraction and weight percent
Molal concentration
The Solution Process
“Like dissolves like”
Heats of solution
Factors affecting solubility
Colligative Properties
Vapor pressure: Raoult's law
Boiling point elevation and freezing point depression
Molar masses from colligative properties
Osmotic pressure
Solutions of electrolytes

3. mole fraction of compound A in a mixture, XA = (no units)
More Concentration Units
Molar concentration
Mole fraction and mole percent
moles of solute
liter of solution
molarity, M =
mole fraction of compound A in a mixture,
XA = (no units)
mole percent of A = XA · 100
1.0 mol of NaCl is dissolved in 4.0 mol of water:
XNaCl = mol % NaCl =
Xwater = mol % H2O = nA
ntotal

4. The solubility of NaCl at 0ºC is 35.7 g/100 g of water. What is the
More Concentration Units
Weight fraction and weight percent
Molal concentration
mass of A
total mass
wA = % (w/w) = wA · 100
Vinegar is 5.00 % (w/w) acetic acid in water. How many grams of acetic acid are in one cup (237 mL) of vinegar? (dvinegar = g/cm3)
moles of solute
kg of solvent
molality, m =
The solubility of NaCl at 0ºC is 35.7 g/100 g of water. What is the
molality of this solution? (fw NaCl 58.44)

5. More Concentration Units
Molal concentration
Vinegar is 5.00 % (w/w) acetic acid in water. The density of vinegar is g/mL. What are X, m, and M? (mw HC2H3O )

6. More Concentration Units
Molal concentration
Antifreeze is typically a 50/50 (v/v) mixture of ethylene glycol (C2H6O2) in water. What are m and M? (mw C2H6O )
d(C2H6O2) = g/mL
d(H2O) = g/mL
d(50/50) = g/mL

7. CH3OH(l) + H2O(l)  miscible
The Solution Process
“Like dissolves like”
CH3OH(l) + H2O(l)  miscible
H-bonds in pure substances are
replaced by H-bonds in solution
H-bonding
H-bonding
CCl4(l) + C6H6(l)  miscible
London forces in pure substances are
replaced by London forces in solution
London
London
but CCl4(l) + H2O(l)  immiscible
or C6H6(l) + H2O(l)
Weak dipole-induced dipole
interactions in solution can’t replace
strong H-bonding in pure water

8. CH3OH(l) + H2O(l)  miscible
The Solution Process
“Like dissolves like”
CH3OH(l) + H2O(l)  miscible
Strong H-bonds in pure substances are
replaced by strong H-bonds in solution.
H-bonding
H-bonding
H2O(g) + CH3OH(g)
H2O(g)
Energy 
CH3OH(l)
Mixture is lower energy than separate components, therefore the solution process occurs.
H2O(l)
mixture

9. CHCl3(l) + C6H12(l)  miscible
The Solution Process
“Like dissolves like”
CHCl3(l) + C6H12(l)  miscible
Weak London/dipole forces in pure substances are replaced by weak dipole-induced dipole forces in solution.
London
London
CHCl3(g) + C6H12(g)
Energy 
CHCl3(g)
C6H12(l)
Again, mixture is lower energy than separate components, so the solution process occurs.
CHCl3(l)
mixture

10. but CHCl3(l) + H2O(l)  immiscible or C6H12(l) + H2O(l)
The Solution Process
“Like dissolves like”
but CHCl3(l) + H2O(l)  immiscible
or C6H12(l) + H2O(l)
Weak dipole-induced dipole
interactions in solution can’t replace
strong H-bonding in pure water.
H2O(g) + C6H12(g)
H2O(g)
C6H12(l)
Energy 
Mixture is higher energy than separate components, so the solution process does not occur.
mixture
H2O(l)

11. Ionic solids: ion-dipole attractions in solution
The Solution Process
“Like dissolves like”
Ionic solids: ion-dipole attractions in solution

12. solute + solvent  solution DHsoln
The Solution Process
Heats of solution
solute + solvent  solution DHsoln
CaCl2(s) Ca2+(aq) + 2Cl–(aq) DHsoln = -83 kJ/mol
NH4NO3(s) NH4+(aq) + NO3–(aq) DHsoln = 26 kJ/mol
Spontaneity of dissolution is promoted by:
decrease in potential energy
increase in disorder
H2O
H2O

13. Pressure: Henry’s law - dissolving gases in liquids
The Solution Process
Factors affecting solubility
Pressure: Henry’s law - dissolving gases in liquids
Mgas = kH·pgas
higher pgas
higher Mgas
Temperature
gases: solubility decreases with increasing temperature
solids: solubility generally increases with increasing temperature

14. A. Vapor pressure: Raoult’s law
Colligative Properties
Vapor pressure: Raoult's law
Colligative properties: depend on the number (concentration) of particles in solution, not on their identity.
A. Vapor pressure: Raoult’s law
Solution of a nonvolatile (pvap  0) solute in a solvent:
pA = XA·pAº
Raoult’s law
pAº
solvent A
solvent A + nonvolatile solute B

15. Colligative Properties
Vapor pressure: Raoult's law
What is the vapor pressure, at 100ºC, of a 50/50 (v/v) solution of ethylene glycol in water? Why is this mixture used as a coolant in your automobile?
From previous example: 500 mL C2H6O2  558 g, 8.99 mol
500 mL H2O  500 g, 27.8 mol

16. DTb = Kb·msolute molal boiling point elevation Kb = 0.51ºC/m
Colligative Properties
Boiling point elevation and freezing point depression
Nonvolatile solutes: raise the boiling point and lower the freezing point
for H2O:
DTb = Kb·msolute molal boiling point elevation Kb = 0.51ºC/m
DTf = Kf·msolute molal freezing point depression Kf = 1.86ºC/m
We saw that a 50/50 (v/v) solution of ethylene glycol in water had a molality of 18.0 m. What are its freezing and boiling points?

17. (For benzene, Kf = 5.12ºC/m, Tf = 5.53ºC.)
Colligative Properties
Molar masses from colligative properties
A solution made by dissolving 3.46 g of an unknown compound in 85.0 g of benzene had a freezing point of 4.13ºC. What is the molecular weight of the compound?
(For benzene, Kf = 5.12ºC/m, Tf = 5.53ºC.)

18. In dilute solution, m  M  P = MRT (similar to ideal gas equation:
Colligative Properties
Osmotic pressure
osmotic
pressure, P
equilibrium
solvent
solution
solvent
solution
semipermeable membrane
-allows solvent molecules
to pass, but not solute
tries to equalize
concentration,
creates pressure
Find: P = mRT
In dilute solution, m  M  P = MRT
(similar to ideal gas equation:
P = (n/V)RT or PV = nRT)

19. Useful for measuring very high molecular weights.
Colligative Properties
Osmotic pressure
Useful for measuring very high molecular weights.
e.g., 0.96 g polyvinyl alcohol (PVA, mw 10,000 g/mol) dissolved in 100 g
of water at 25ºC.
0.96 g x = 9.6 x 10–5 mol  9.6 x 10–4 m, 9.6 x 10–4 M
By freezing point depression:
DTf = Kf·m = (1.86ºC/m)(9.6 x 10–4 m) = ºC can’t measure!
By osmotic pressure:
P = MRT = (9.6 x 10–4 mol/L)( L·atm/mol·K)(298 K)
= atm
= 18 torr
= 9.6 inches of water easy to measure
1 mol
10,000 g

20. Colligative Properties
Osmotic pressure
A 1.40-g sample of polyethylene was dissolved in benzene to give exactly mL of solution. The osmotic pressure of the solution was 1.86 mm Hg at 25ºC. What is the molecular weight of the polyethylene?

21. NaCl(s) Na+(aq) + Cl–(aq) 1.0 m  2.0 m particles
Colligative Properties
Solutions of electrolytes
H2O
NaCl(s) Na+(aq) + Cl–(aq)
1.0 m  m particles
DTf(theor) = (1.86 ºC/m)(2.0 m) = 3.72 ºC
But, this is only true for very dilute solutions. In more concentrated solutions,
the effect diminishes at higher concentrations of ions:
For NaCl
m DTf (calc*) DTf (meas) DTf(meas)/DTf(calc)
van’t Hoff factor, i
*calculated assuming no ionization
 DTf = Kf · m · i

22. Colligative Properties
Solutions of electrolytes
What is the freezing point of a solution containing 150. g of water and 35.0 g of CaCl2? Assume i = 2.7 for CaCl2 (fw CaCl )