GENERAL PROPERTIES OF SOLUTIONS
1. A solution is a homogeneous mixture of two or
more components.
2. It has variable composition.
3. The dissolved solute is molecular or ionic in
size.
4. A solution may be either colored or colorless
nut is generally transparent.
5. The solute remains uniformly distributed
throughout the solution and will not settle out
through time.
6. The solute can be separated from the solvent
by physical methods.
Colligative Properties of Solutions
Depends on the concentration of the solute
particles and not on the identity of the
solute.
Dissolved particles alter and interfere with
the dynamic process of a solution.
NOTE: DT=Tf-Ti or in this case DT=Tsolution-Tsolvent
Boiling point elevation
Freezing point depression
Osmosis
Vapor pressure lowering
Colligative Properties
DT = k m
1. Calculate the molality of a solution consisting of
5.34 g of aluminum sulfate in 200.0 mL of pure
water at 24oC?
2. The boiling point of a solution containing 5.55 g
of an unknown nonvolatile substance dissolved in
15.0 g of water is 103.3 oC. Calculate the molar
mass of the compound.
3. Calculate the freezing point of a 500.0 mL
sample of aluminum sulfate solution (kf = 1.86 oC
kg / mol) containing 16.0 g of solute.
MOLALITY
• Molality = moles of solute per kg of solvent
• m = nsolute / kg solvent
• If the concentration of a solution is given in terms of
molality, it is referred to as a molal solution.
Q. Calculate the molality of a solution consisting of 25 g of KCl in
250.0 mL of pure water at 20oC?
First calculate the mass in kilograms of solvent using the density of
solvent:
250.0 mL of H2O (1 g/ 1 mL) = 250.0 g of H2O (1 kg / 1000 g) = 0.2500 kg of H2O
Next calculate the moles of solute using the molar mass:
25 g KCl (1 mol / 54.5 g) = 0.46 moles of solute
Lastly calculate the molality:
m = n / kg = 0.46 mol / 0.2500 kg = 1.8 m (molal) solution
Freezing Point Depression
DTf = - kf m
Q. Estimate the freezing point of a 2.00 L sample of seawater (kf = 1.86 oC kg / mol), which
has the following composition:
0.458 mol of Na+
0.052 mol of Mg2+
0.010 mol Ca2+
0.010 mol K+
0.533 mol Cl0.002 mol HCO30.001 mol Br0.001 mol neutral species.
Since colligative properties are dependent on the NUMBER of particles and not
the character of the particles, you must first add up all the moles of solute in
the solution.
Total moles = 1.067 moles of solute
Now calculate the molality of the solution:
m = moles of solute / kg of solvent = 1.067 mol / 2.00 kg
= 0.5335 mol/kg
Last calculate the temperature change:
DTf = - kf m = -(1.86 oC kg/mol) (0.5335 mol/kg) = 0.992 oC
The freezing point of seawater is Tsolvent - DT = 0 oC - 0.992 oC
= - 0.992 oC
Boiling Point Elevation
DTb = kb m
Q. The boiling point of a solution containing 40.0 g of an unknown substance dissolved in
100.0 g of water is 105.3 oC . Calculate the molar mass of the compound.
Since the solvent is water, the change in temperature (DT) would be 105.3 - 100.0
oC = 5.3 oC. You can also find the k in the table in your textbook, k = 0.512 oC
b
b
kg/mol.
From this data, you can calculate the molality:
m = DTb / kb = 5.3 oC / 0.512 oC kg/mol = 10.4 mol/kg
Molality is also defined as the moles of solute per kg of solvent:
m = n /(kg solvent), can be rearranged to be n = m (kg of solvent)
n = 10.4 mol /kg (0.1000 kg) = 1.04 mol of solute
The molar mass can be calculated by using the equation, MW = m/n
MW = 40.0 g / 1.04 mol = 38.5 g/mol
PRACTICE PROBLEMS # CP
Short essay: (answers on next slide)
1. Excluding any possible chemical reactions, which would be more
effective as an antifreeze; a solution containing 25 m methyl
alcohol (CH3OH) or 25 m KCl?
Application:
2. What is the freezing point of an aqueous sugar (C12H22O11)
solution that boils at 110oC?
-36.3oC
3.
When 256 g of a nonvolatile, nonelectrolyte unknown were
dissolved in 499 g of water, the freezing point was found to be –
2.79oC. The molar mass of the unknown solute is?
a) 357
b) 62.0
c) 768
d) 342
D
Short essay answers:
1. Excluding any secondary chemical reactions, which would be
more effective as an antifreeze; a solution containing 25 m methyl
alcohol (CH3OH) or 25 m KCl?
KCl because colligative properties depend on the number of
solute particles and not on the nature of the particles therefore
you must calculate the amount of solute particles in each case.
Proof:
If you assume you have 1 kg of solution, a 25 molal (moles
of solute per kg of solvent) solution of methyl alcohol
would contain 25 moles of solute. In one 1 kg of a 25 molal
KCl solution there would be 25 moles of K+ ions and 25
moles of Cl- ions (because KCl completely dissociates in
water) therefore KCl has twice as many particles as methyl
alcohol and thus be more effective.
GROUP STUDY PROBLEMS # CP
Short Essay (write answers on back)
1. Which would be more effective as an antifreeze; a solution
containing 25 m methyl alcohol (CH3OH) or 25 m ethyl alcohol
(C2H5OH)?
Application:
___2. An aqueous solution freezes at -7.98 oC. What is its boiling
temperature?
___ 3. When 42.8 g of a nonvolatile, nonelectrolyte unknown were
dissolved in 423 g of water, the freezing point was found to be –
2.29oC. The molar mass of the unknown solute is?