CHAPTER 6

Properties of Solutions
 Definitions
• A solution is a homogeneous mixture
• A solute is dissolved in a solvent.
• A solvent is the liquid in which the solute is dissolved
solute
and present in greater amount

an aqueous solution has water as solvent

Solvent
• A saturated solution is one where the
concentration is at a maximum - no more solute
is able to dissolve.
 SATURATED SOLUTION
UNSATURATED SATURATED SUPERSATURATED
SOLUTION SOLUTION SOLUTION
more solute no more solute becomes unstable,
dissolves dissolves crystals form

concentration
 Solubility
• is the mass of solute that forms a saturated
solution with a given mass of solvent at a
specified temperature

• The units of Solubility

• (g solute/100 ml solvent)
• A dynamic equilibrium exists between the solute
in the two phases
 Intermolecular Forces & Solubility
• Substances with similar types of intermolecular
forces (IMFs) dissolve in each other
• “Like Dissolves Like”
1- polar molecules are soluble in polar solvents
C2H5OH in H2O
non-polar molecules are soluble in non-polar solvents
CCl4 in C6H6
2- ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
Ways of Expressing Concentration
Four ways of expressing concentration
–Molarity(M) = moles solute / Liter solution
–Mass percent = (mass solute / mass of solution) x 100
–Molality* (m) = moles solute / Kg solvent
–Mole Fraction( A) = moles solute / total moles solution

* Note that Molality is the only concentration unit in which

denominator contains only solvent information rather than solution
 Mass Percent
the mass of the solute in grams per 100 grams of the
solution

Mass of solute
Mass % of solute = _____________ x 100
Mass of solution
A 5 % solution of sodium chloride means that means ; 5 g of
NaCl is present in 100g of the solution, or 95 g solvent
 Mole Fraction (X)

moles of A
XA = ____________________
total moles in solution

In some applications, one needs the mole fraction of

solvent, not solute, make sure you find the quantity
you need
 Molarity
•Molarity defined as the number of moles of solute dissolved per
litre (dm3) of the solution. It is denoted by the symbol M.
Measurements in Molarity can change with the change in
temperature because solutions expand or contract accordingly.
number of moles of solute
Molarity of Solution = Volume of solution in liters
The Molarity of the solution can also be expressed in terms of
mass and molar mass
Mass of solute
Molarity of Solution =
Molar mass of the solute x Volume of solution in liters

In terms of weight, molarity of the substance can be expressed as:

Wq W mol/L
Molarity = =
M g mol-1 X V litre MXV
Ex: If 0.435 g of KMnO4 is dissolved in enough water to give
250. mL of solution, what is the molarity of KMnO4?

0.435 g KMnO4 = 1 mol KMnO4 = 0.00275 mol KMnO4

158.0 g KMnO4

Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M

0.250 L solution
 Molality (m)

Molality is the number of moles of solute dissolved in 1 kg (1000 g) of solvent.

the units of molality are mol/kg.

EXAMPLE

Calculate the molality of a sulfuric acid solution containing 24.4 g of sulfuric acid in 198 g
of water. The molar mass of sulfuric acid is 98.09 g.
The effect of temperature on
solubility
 Solid Solubility and Temperature
In most but certainly not all cases,
the solubility of a solid substance
increases with temperature.
 Gas Solubility and Temperature
The solubility of gases in water usually decreases
with increasing temperature
Colligative properties of
nonelectrolyte solutions
 Colligative properties (or collective properties) are properties that depend
only on the number of solute particles in solution and not on the nature of
the solute particles.

The colligative properties are:

• Vapor pressure lowering.
• Boiling point elevation.
• Melting point depression.
• Osmotic pressure.
 Vapor-Pressure Lowering
If a solute is nonvolatile (that is, it does not have a measurable vapor pressure), the vapor
pressure of its solution is always less than that of the pure solvent.
The extent to which a nonvolatile solute lowers the vapor pressure is proportional to its
concentration. This relationship is expressed by Raoult’s law.

the decrease in vapor pressure, ∆P, is directly proportional to the solute concentration
(measured in mole fraction).
 EXAMPL
E
Calculate the vapor pressure of a solution made by dissolving 218 g of glucose (molar mass =
180.2 g/mol) in 460 mL of water at 30°C. What is the vapor-pressure lowering? The vapor
pressure of pure water at 30°C = 31.82 mmHg. Assume the density of the solution is 1.00 g/mL.

First we calculate the number of moles of glucose and water in the solution:

The mole fraction of water, X1,

Therefore, the vapor pressure of the glucose solution is

Finally, the vapor-pressure lowering is (31.82 - 30.4) = 1.4 mmHg.

If both components of a solution are volatile (that is, have measurable vapor pressure), the vapor
pressure of the solution is the sum of the individual partial pressures.

Raoult’s law holds equally well in this case:

The total pressure is given by Dalton’s law of partial pressure

 Boiling-Point Elevation

The boiling point of a solution is the temperature at which its vapor pressure equals the
external atmospheric pressure

Nonvolatile solute-solvent interactions also cause solutions to have higher boiling points
than the pure solvent.

At the normal boiling point of the pure liquid, the vapor pressure of the solution will be less than
1 atm. Therefore, a higher temperature is required to attain a vapor pressure of 1 atm. Thus, the
boiling point of the solution is higher than that of the pure liquid.
The change in boiling point is proportional to the molality (m) of the
solution:

Tb = Kb m
where:
Kb is the molal boiling point elevation constant, a property of the solvent.
Tb is added to the normal boiling point of the solvent.

The increase in boiling point relative to that of the pure solvent, Tb is a +ve quantity
obtained by subtracting the boiling point of the pure solvent from the boiling point of the
solution.

Tb = Tsoln – Tpure solvent (Tsoln > Tpure solvent)

 Freezing-Point Depression
Nonvolatile solute-solvent interactions also cause solutions to have lower freezing points
than the pure solvent.

The change in freezing point can be found similarly:

Tf = Kf m

where:
Kf is the molal freezing point depression constant of the solvent.
Tf is subtracted from the normal freezing point of the solvent.
The decrease in freezing point, Tf is a +ve quantity obtained by subtracting the
freezing point of the solution from the freezing point of the pure solvent.

Tf = Tpure solvent – Tsoln (Tpure solvent > Tsoln)

 Boiling Point Elevation
and
Tb = Kb m
Freezing Point Depression
Tf = Kf m

Molal boiling-point elevation and freezing-point depression constants of several common liquids
EXAMPL
E
Ethylene glycol, CH2(OH)CH2(OH), is a common automobile antifreeze. It is water soluble and
fairly nonvolatile (b.p. 197°C). Calculate the freezing point of a solution containing 651 g of this
substance in 2505 g of water. Would you keep this substance in your car radiator during the
summer? The molar mass of ethylene glycol is 62.01 g.

Because pure water freezes at 0°C, the solution will freeze at (0 - 7.79) = -7.79°C.
We can calculate boiling-point elevation in the same way as follows:

Because the solution will boil at (100 + 2.2) = 102.2°C, it would be preferable to leave the
antifreeze in your car radiator in summer to prevent the solution from boiling.
Example
Automotive antifreeze consists of ethylene glycol, CH2(OH)CH2(OH), a nonvolatile
nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of
ethylene glycol in water.

Solution
The molality of the solution is calculated as follows:

We can now use Equations to calculate the changes in the boiling and freezing points:

Hence, the boiling and freezing points of the solution are

Comment: Notice that the solution is a liquid over a larger temperature range than the pure solvent.
 Osmotic Pressure

The pressure required to stop or to prevent osmosis by pure solvent, known as

osmotic pressure, , of the solution.
The osmotic pressure obeys a law similar in form to the ideal gas law:

where: M is the molarity of solution, R is the gas constant (0.0821 L atm/K mol), and T is the absolute
temperature. The osmotic pressure, π, is expressed in atm.
 EXAMPL
E
The average osmotic pressure of seawater, measured in the kind of osmotic pressure
apparatus, is about 30.0 atm at 25°C. Calculate the molar concentration of an aqueous
solution of sucrose (C12H22O11) that is isotonic with seawater.
 Using Colligative Properties to Determine
Molar Mass

The colligative properties of nonelectrolyte solutions provide a means of determining the

molar mass of a solute. Theoretically, any of the four colligative properties is suitable for
this purpose. In practice, however, only freezing-point depression and osmotic pressure are
used because they show the most pronounced changes.

freezing-point depression → molality → number of moles → molar mass

osmotic pressure → molarity → number of moles → molar mass

 EXAMPL
E
A 7.85-g sample of a compound with the empirical formula C5H4 is dissolved in 301 g of
benzene. The freezing point of the solution is 1.05°C below that of pure benzene. What are
the molar mass and molecular formula of this compound?

Now we can determine the ratio

Therefore, the molecular formula is

(C5H4)2 or C10H8 (naphthalene).
 EXAMPL
E
A solution is prepared by dissolving 35.0 g of hemoglobin (Hb) in enough water to make up
1 L in volume. If the osmotic pressure of the solution is found to be 10.0 mmHg at 25°C,
calculate the molar mass of hemoglobin.

The volume of the solution is 1 L, so it must contain 5.38 x 10-4 mol of Hb.