SOLUTIONS

A solution is a
Mixture of two or more
substances in a single phase.

One constituent is usually

regarded as the solvent and the
others as solute.
 PARTS OF A SOLUTION

• SOLUTE – The part of a solution Solute Solvent Example

solid solid Salt in flour, Alum
that is being dissolved (usually the solid liquid Salt in water
lesser amount) Solid Gas Iodine in air, carbon in
air
gas solid Hydrogen in palladium
• SOLVENT – The part of a solution gas liquid Oxygen in water
that dissolves the solute (usually gas gas Air
Liquid Solid Moisture in paper
the greater amount)
liquid liquid Ethanol in water
Liquid Gas Moisture in air
• Solute + solvent = solution
Solutions can be classified as saturated or unsaturated.

A saturated solution contains the maximum quantity of solute that

dissolves at that temperature.

An unsaturated solution contains less than the maximum amount

of solute that can dissolve at a particular temperature

Supersaturated solutions contain more solute than is possible to

be dissolved
Ideal Solution Non Ideal solution
A solution which obeys Raoult’s A solution which does not obeys Raoult’s
law in all conditions law in all conditions

A-A interactions are same as that A-A interactions are not same as that of
of A-B and B-B interactions A-B and B-B interactions

Change in heat is zero. Change in heat is not zero

No change in volume i.e volume The volume changes the volume of

of solution is additive of volume solution may be smaller or greater than
of solute and solution the volumes of solute and solvent

Example: Toluene and Benzene Example: Water and HCl

 Raoult’s law
The partial pressure (Pi ) of each
component in a solution is directly
proportional to the vapor pressure of
the corresponding pure substance (Pi
*) and that the proportionality
constant is the mole fraction (xi ) of
the component in the liquid.

Pi = Xi . Pi* i=1,2
• PROPAN-1-OL AND PROPAN-2-OL

Hydrogen bond

• BROMOMETHANE AND IODOMETHANE

Dipole-dipole attraction

• HEXANE AND HEPTANE

VDW forces
 POSITIVE DEVIATION FROM RAOULT’S LAW
* Vapour pressure of a liquid mixture is greater than that predicted by
Raoult’s law,
i.e.PA > pao x A and PB > pbo x B

* Intermolecular intermolecular intermolecular

Attraction between < attraction between + attraction between
A and B A and A B and B
(In mixture) (in pure A) (in pure B)
 Example of non-ideal solution showing positive
deviation
Mixture of tetrachloromethane and ethanol
Intermolecular Intermolecular Intermolecular
Attraction between < Attraction between + Attraction between
CCl4 and C2H5OH CCl4 molecules C2H5OH molecules

Dipole-dipole Hydrogen
VDW forces
attraction bond
** Weakening of intermolecular attraction in mixture results in
1. Volume expansion,
2. Absorption of heat (i.e. Temp. Drop) when mixing.
 Phase diagram for positive deviation

Weaker intermolecular attraction Easier for molecules to escape

Higher vapour pressure
Weaker intermolecular attraction Lower boiling temperature
 Negative deviation from Raoult’s law

*Vapour pressure of a liquid mixture is smaller than that predicted by

raoult’s law,
I.E. PA < pao x A and PB < pbo x B

*Intermolecular intermolecular intermolecular

Attraction between > attraction between +attraction between
A and B A and A B and B
(In mixture) (in pure A) (in pure B)
 Example of non-ideal solution showing negative deviation
Mixture of trichloromethane and ethoxyethane

Intermolecular Intermolecular Intermolecular

Attraction between > Attraction between + Attraction between
CHCl3 and C2H5OC2H5 CHCl3 molecules C2H5OC2H5 molecules

Hydrogen bond Dipole -Dipole Dipole-dipole

attraction attraction

** Strengthening of intermolecular attraction in mixture results in

1. Volume contraction,
2. Evolution of heat (i.E. Temp. Rise) when mixing.
 Phase diagram for negative deviation

More difficult for molecules to escape

Stronger intermolecular attraction
Lower vapour pressure
Stronger intermolecular attraction Higher boiling temperature
 CONCENTRATION
CONCENTRATION OF
OF SOLUTE
SOLUTE

The amount of solute in a solution is given by its concentration.

Concentration of solutions can be expressed as;
• Molarity M
• Molality m
• Normality N
• Mole fraction X
• Percent %
• Weight percent w/w%
• Volume percent v/v%
• Weight in volume percent w/v%
Expression Symbol Definition
Molarity M Moles (gram molecular weight) of solute in 1 liter of
solution
Normality N Gram equivalent weight of solute in 1 liter of
resolution
Molality m Moles of solute in 1000 gm of solvent
Mole Fraction X Ration of the moles of constituents of solution to the
total moles of all constituents (solute and solvent)
Mole Percent M% Moles of one constituent in 100 moles of the solution
Percent by Weight %w/w Grams of solute in 100 grams of solution
Percent by volume %v/v Ml of solute in 100 ml of solution
Percent weight in volume %w/v Grams of solute in 100 ml of solution
 moles solute
Molarity (M) =
liters of solution

1.0 L of water was

used to make 1.0 L of
solution. Notice the
water left over.
PROBLEM:
PROBLEM: Dissolve
Dissolve 5.00
5.00 gg of
of NiCl
NiCl22 in
in enough
enough
water
water to
to make
make 250
250 ml
ml of
of solution.
solution. Calculate
Calculate the
the
molarity.
molarity.
Step 1: Calculate moles of NiCl2
1 mol
5.00 g • = 0.0385 mol
129.6 g

Step 2: Calculate Molarity

0.0385 mol
= 0.154 M
0.250 L

[NiCl2] = 0.154 M
 USING
USING MOLARITY
MOLARITY
What mass of oxalic acid, H2C2O4, is required to make
250. mL of a 0.0500 M solution?
moles = M x V
Step 1: change ml to L.
250 ml x 1L/1000ml = 0.250 L
Step 2: calculate moles.
Moles = (0.0500 mol/L) (0.250 L) = 0.0125 moles
Step 3: convert moles to grams.
(0.0125 mol)(90.00 g/mol) = 1.13 g
 LEARNING CHECK

How many grams of NaOH are required to

prepare 400 ml of 3.0 M NaOH solution?

1) 12 g
2) 48 g
3) 300 g
 Dissolve
Dissolve62.1
62.1gg(1.00
(1.00mol)
mol)ofofethylene
ethyleneglycol
glycolin
in250.
250.ggof
ofHH22O.
O.
Calculate
Calculatem m&&% %ofofethylene
ethyleneglycol
glycol(by
(bymass).
mass).

CALCULATE
CALCULATE MOLALITY
MOLALITY

1.00 mol glycol

conc (molality) =  4.00 molal
0.250 kg H2O
Calculate weight %

62.1 g
%glycol = x 100% = 19.9%
62.1 g + 250. g
 LEARNING CHECK

A SOLUTION CONTAINS 15 G NA2CO3 AND 235 G

OF H2O? WHAT IS THE MASS % OF THE
SOLUTION?

1) 15% NA2CO3
2) 6.4% NA2CO3
3) 6.0% NA2CO3
COLLIGATIVE PROPERTIES
HOW DO YOU GET FROM THIS…
…to this?
ADD AN IONIC COMPOUND!
 COLLIGATIVE PROPERTIES
On adding a solute to a solvent, the properties of the solvent are
modified.
• Vapor pressure decreases (Lowering of Vapour Pressure)
• Melting point decreases (Depression of Freezing Point)
• Boiling point increases (Elevation of Boiling Point)
• Osmosis is possible (osmotic pressure)
These changes are called COLLIGATIVE PROPERTIES.
They depend only on the number of solute particles relative to solvent
particles, not on the kind of solute particles.
 Lowering of Vapour Pressure

• The particles of solute are surrounded by and attracted to

particles of solvent.
• Now the solvent particles have less kinetic energy and tend
less to escape into the space above the liquid.
• So the vapor pressure is less.
 IONIC VS MOLECULAR SOLUTES

• Ionic solutes produce two or more ion particles in solution.

• They affect the colligative properties proportionately more than
molecular solutes (that do not ionize).

• The effect is proportional to the number of particles of the solute

in the solution.
• 1 Mole of sugar = 1 mole of Na +
According Raoults Law;

Pi = X1 . Po eq. 1
We know that
X 1 + X2 = 1 eq. 2
X1 = 1 – X 2 eq. 3
Where X1, X2 are mole fraction of solvent and solute respectively
replacing value of X1from eq. 3 into eq. 1 we get
Pi = Po. (1 – X2)= Po - PoX2
Pi + PoX2 = Po
PoX2 = Po - Pi
X2 = Po – Pi = ΔP/ Po = X2 = n2/n1 + n2 eq. 4
 Vapor-pressure lowering
Higher vapor pressure
Lower vapor pressure

Solution containing nonvolatile solute Solute particle

Pure solvent Solvent particle

Equilibrium is established In a solution, solute particles reduce

between the liquid and the number of solvent particles able to
escape the liquid. Equilibrium is
vapor in a pure solvent.
established at a lower vapor pressure.
Boiling Point Elevation

1. The normal boiling point is the

temperature at which the vapor pressure of
the liquid becomes equal to the external
Pressure of 760 mm of Hg.
2. While a solution will boil at higher
Temperature than the pure solvent.
3. The boiling point raises because the
vapor pressure decreases.

ΔTb= T – To , the ratio of ΔTb/ Δp at 100 oC is approximately constant and can be

written as;
ΔTb/ Δp =k΄ eq.1
ΔTb = k΄Δp eq.2
As po is constant the relative lowering of vapor pressure is given by;
Δp/po
which is equal to the mole fraction of the solute hence we can write;
Δp/po = X2 eq.3
ΔTb = k΄Δp = k΄X2 eq.4
ΔTb = k΄Δp = k΄X2 eq.5
In dilute solutions, X2 is approximately equal to m/(1000/M1).
Hence equation 5 can be written as;
ΔTb = k΄m/(1000/M1) eq.6
ΔTb = k΄ M1 m eq.7
1000
ΔTb = kbm eq.8
ΔTb is known as boiling point elevation and kb is known as molal elevation
Ebullioscopic constant
The ratio of the elevation of the boiling point of a solvent caused by dissolving
a solute to the molality of the solution, taken at extremely low concentrations.
Also known as molal elevation of the boiling point.

Applying clapeyron equation we can rewrite the eq. 8 as

ΔTb = Tb Vv – Vl eq.9
ΔP ΔHv
Where Vv and Vl are the molar volume of the gas and liquid, respectively, Tb is
the boiling point of solvent and ΔHv is the molar heat of vaporization.
we can also write;
ΔTb = Tb Vv eq.10
ΔP ΔHv
As; Vv = RTb/Po by replacing in eq.10 we get
 ΔTb = RTb2 eq.11
ΔP PoΔHv

ΔTb = RTb2 ΔP eq.12

ΔHv Po
We know that ΔP = X2, by putting in eq.12 we get;
Po
ΔTb = RTb2 X2 eq.13
ΔHv
We can also write
ΔTb = RTb2 M1 m eq.14
1000 ΔHv
 Substance Boiling Point Kb Freezing Point Kf
o
C o
C
Acetic Acid 118.0 3.56 16.7 3.9
Acetone 56.0 1.71 -94.82 2.40
Benzene 80.1 2.53 5.5 5.12
Camphor 208.3 5.95 178.4 37.7
Chloroform 61.20 3.54 -63.5 4.96
Ethyl Alcohol 78.4 1.22 -114.49 3.0
Ethyl Ether 34.6 2.02 -116.3 1.79
Phenol 181.4 3.56 42.0 7.27
Water 100.0 0.51 0.00 1.86
 Measurement of osmotic pressure (π)
van’t Hoff and Morse equation:
van’t Hoff established that the that the osmotic pressure is analogous to the
vapour pressure exerted by the gases provided both the liquid or gas have same
volume.
As we know; PV= nRT eq.1
As van’t Hoff established that P = π, we can rewrite the eq. 1 as;
πV= nRT eq.2
π = nRT n/V = Molarity
V
π = RTM eq.3 (van’t Hoff equation)
Morse established when concentration is expressed in terms of molality gives
accurate result as that of using molarity, thus the eq.3 can be written as;
π = RTm eq.4 (Morse equation)
 Thermodynamics of osmotic pressure and vapour pressure lowering
Facts:
1. ΔP = po – p
2. π = P – Po
3. At equilibrium there is no escaping tendency and the free energy equals zero
i.e. ΔG = 0
• In first case ΔG = RT ln p/po eq.5
• In second case ΔG = -Vl (P-Po) = -Vl π eq .6

ΔG = -Vl π = RT ln p/po or Vl π = - RT ln p/po eq. 7

π = RT ln po/p eq.8 (imp)
Vl
Assuming the solution obeys Raoult’s law, then
p = po X1, p/ po = X1 = 1 – X2
Putting p/po = 1-X2 in eq.7 we obtain;

πVl = - RT ln (1-X2) eq. 9

ln (1-X2) can be expanded into a series;
ln (1-X2) = -X2
When the solution is dilute X2 is very small, thus all the terms in the expansion
may be neglected, hence
ln (1-X2) -X2
πVl = RT X2 eq.10
X2 = n2/n1
π = n2 RT
n1Vl
n1Vl = V
 Molecular Weight Determination

We know that mole fraction of solvent and solute are given by;
1. By lowering of vapour pressure
n1= w1/M1 and n2= w2/M2 respectively,

Po – Pi = X2 = n2 = w2/M2 = w2/M2
Po n1+n2 w1/M1 +w2/M2 w1/M1

ΔP = w2/M2
Po w1/M1
M2= w2 M1 Po
w1 ΔP
1. Elevation of boiling point