Chapter 12
Physical Properties of Solutions
12.1 Types of Solutions
12.2 Concentration Units
12.4 The Effect of Temperature on Solubility
12.5 The Effect of Pressure on the Solubility of Gases
12.6 Colligative Properties of Nonelectrolyte Solutions
12.1 Types of Solutions
Solution: homogeneous mixture of two or more substances
6 types of solutions are (i) Gas + Gas
Gas solution (Air)
(ii) Liquid + Liquid
Liquid solution (Soda water)
(iii) Gas + Solid
Solid solution (H2 gas in Palladium)
(iv) Liquid + Liquid
Liquid solution (Ethanol in Water)
(v) Solid + Liquid
Liquid solution (NaCl in water)
(vi) Solid + Solid
Solid solution (Cu / Zn Brass)
Saturated solution
Unsaturated solution
Supersaturated solution
Contains the maximum
Contains less solute than it
Contains more solute than
amount of a solute that will
has the capacity to dissolve
is present in a saturated
dissolve in a given solvent
solution
at a specific temperature
Crystallization..the process in which dissolved solute comes out of solution and forms crystal.
12.3 Concentration Units
Concentration…
The amount of solute present in a given quantity of solvent or solution.
(i) Percent by Mass of solute

Percent by Mass of solvent
massofsolute
x100%
(massofsolute  massofsolvent )

massofsolvent
x100%
(massofsolute  massofsolvent )
mass of solution = (mass of solute + mass of solvent)
(ii) Mole Fraction (X A)
X
A

Mole Fraction (XB)
molesofA
sumofmolesofallcomponents
X
B

molesofB
sumofmolesofallcomponents
(iii) Molarity (M)
M 
molesofsolute
litersofso lution
(iv) Molality (m)
m
molesofsolute
massofsolvent (kg)
12.4 The Effect of Temperature on Solubility
(a) Solid Solubility and Temperature
(b) Gas Solubility and Temperature
(a) Solid Solubility and Temperature
solubility
In most cases solubility of solid
substance increases with temperature
(g solute/100g H2O)
Temperature(‘C)
Fractional Crystallization…separation of a mixture of substances into pure components on the basis
of their differing solubilities
(b) Gas Solubility and Temperature
Solubility of gases in water
usually decreases with
increasing temperature
Solubility
(g solute/100g
H2O)
Temperature (‘C)
12.5 The Effect of Pressure on the Solubility of Gases
Henry’s Law
The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution:
cP
i.e c = kP (c = molar concentration (mol/L)
P = pressure in atm(gas over the solution)
k = constant (mol/L.atm)
(Ex: soft drink when the cap of the bottle is removed)
Before cap is sealed (pressurized with a mixture of air & CO2 )
Example 12.6 .The solubility of nitrogen gas at 25’C and 1 atm is
6.8x10 -4 mol/L.What is the concentration of nitrogen dissolved in water under
atmospheric conditions?The partial pressure of nitrogen gas in the atmosphere is
0.78 atm.
c (solubility) = 6.8 x 10 -4 mol/L
p (partial pressure) = 0.78 atm
Apply Henry’s Law equation c = kp and find constant k = 6.8 x10 -4 mol/L.atm
Then find the solubility of nitrogen gas in water where
c = (6.8 x 10 -4 mol/L.atm)(0.78 atm)
= 5.3 x 10 -4 M (ANSWER)
12.6 Colligative properties of Nonelectrolyte Solutions
Colligative properties..properties that depend only on the number of solute particles
in solution and not on the nature of the solute particles.
Types of colligative properties are..
(i)
Vapor-Pressure Lowering
(ii)
Boling-Point Elevation
(iii)
Freezing-Point Depression
(iv)
Osmotic Pressure
(i)
Vapor-Pressure Lowering
Case (i) The vapor pressure of solution with nonvolatile solute (no measureable
vapor pressure) is always less than that of pure solvent.
Raoult’s Law
The partial pressure of a solvent over a solution (P1) is equal to the vapor pressure
of the pure solvent ( P10) times the mole fraction of the solvent in te solution X1.
P1 = X1 . P10
Ref: mole fraction…X1 = 1- X2 (X2 = mole fraction of the solute)
P1 = (1 – X2) P10
P1 = P10 – X2 P10
P10 – P1 = ∆ P = X2P10
THEREFORE DECREASE IN THE VAPOR PRESSURE ∆ P, IS DIRECTLY
PROPORTIONAL TO THE SOLUTE CONCENTRATION (MEASURED IN
MOLE FRACTION)
Example 12.7 Explain
Question..Why is the vapor pressure of a solution less than that of the pure
solvent?
Vaporization increases the disorder of a system because molecules in a vapor
have less order than those in a liquid. Solution is more disordered than a pure
solvent due to that reason difference in disorder between a solution and a vapor is
less than that between a pure solvent and a vapor. Therefore solvent molecules
have less tendency to leave a solution than to leave the pure solvent to become
vapor. That is the reason the vapor pressure of a solution is less than that of the
solvent.
Case (ii) If both components of solution are volatile (measurable vapor pressure),
the vapor pressure of the solution is the sum of the individual partial pressures.
Raoult’s Law can be expressed as…
PA = XA PA0
PB = XB PB0
PT = PA + PB
THEREFORE
PT = XAPA0 + XBPB0
Ideal solution..Any solution that obeys Raoult’s law and
the heat of solution ∆ Hsoln = 0
(ii)
& (iii) Boiling – Point Elevation & Freezing –Point Depression
Boiling Point…of a solution is the temperature at which its vapor pressure equals
the external atmospheric pressure.
∆Tb = Tb – Tb0 (+ quantity)
∆Tb = Kbm
/
/
∆Tf = Tf0 – Tf (+ quantity)
∆Tf = Kfm
Kb & Kf = molal boiling –point elevation constant (or) molal freezing-point
depression constant (‘C /m) m = molality (mol/kg of solvent)
(iii)
Osmotic pressure (∏)
Osmosis…phenomenon which allows free passage of solvent molecules to pass
through a porous membrane from a dilute solution to a more concentrated one.
Semipermeable membrane allows the passage of solvent molecules but blocks the
passage of solute molecules.
Osmotic pressure…( ∏ ) the pressure required to stop osmosis.
∏ = MRT ( M = molarity of the solution mol/L)