Solutions of nonelectrolytes-Physical Pharmacy

JAYAPAL REDDY GANGADI
Concentration Expressions,
Equivalent Weights,
Ideal And Real Solutions,
Colligative Properties,
Molecular Weight Determination
Chapter:III
a.Solutions of Nonelectrolytes:

Definitions
 A solution is a homogeneous mixture.
 A solute is dissolved in a solvent.
 Substance whose aqueous solutions does not
conduct electricity are called as non-
electrolytes
Eg sugar, glycerol
 Dilute solutions-
 the concentration of solute is relatively low.
 Conc. Solutions-
 the conc. of solute is relatively high.

Types of Solutions
 Solutions are homogeneous mixtures of two
or more pure substances.
 In a solution, the solute is dispersed
uniformly throughout the solvent.

Concentration Expressions
 Molarity(M): moles solute / Liter solution
◦Mass percent: (mass solute / mass of solution) * 100
◦Molality (m) - moles solute / Kg solvent
◦Mole Fraction(A) - moles solute / total moles solution

% Concentration
 % (w/w) =
 % (w/v) =
 % (v/v) =
100x
solutionmass
solutemass
100x
solutionvolume
solutemass
100x
solutionvolume
solutevolume
100
/
x
moleeqvivalent
weightmolecular
Eqvivalent weignt =
Equivalent weight =
Atomic weight
number of equivalents per
atomic weight (valence)

 Example:
An aqueous solution of ferrous sulfate was prepared by adding 41.5 g of
FeSO4 to enough water to make 1000 ml of solution at 18°C. The density of
the solution is 1.0375 and the molecular weight of FeSO4 is 151.9. Calculate:
a) The molarity.
b) The normality.
c) The molality.
d) The mole fraction and mole percent for both solute and solvent.
e) The percentage by weight of FeSO4.
Answer:
a) Molarity = moles of FeSO4/Liters of solution
= (41.5/151.9)/ 1 L = 0.2732 M.
b) Normality = (41.5/ 75.95)/ 1L= 0.5464 N
c) Molality = Grams of solution = volume x density
= 1000 * 1.0375= 1037.5 g ;
Grams of solvent = 1037.5- 41.5= 996 g
=moles of FeSO4/ kg of solvent
Molality = 0.2732/ 0.996= 0.2743 m Equivalent weight= 151.9/2= 75.95 g/Eq

Example…….cont.:
d) Mole fraction and mole percent:
Moles of water = n1= 996/18.02= 55.27 moles.
Moles of FeSO4 = n2 = 0.2732 moles.
Mole fraction of water = X1 = 55.27/ (55.27 + 0.2732) = 0.9951
Mole percent of water = 0.9951 * 100%= 99.51%.
Mole fraction of FeSO4= X2= 0.2732/ (55.27 + 0.2732) = 0.0049
Mole percent of FeSO4= 0.49%
e) Percentage by weight of FeSO4= (g of FeSO4/ g of solution) *100
= (41.5/ 1037.5) *100 = 4 %

Solution properties
 Solution properties may be classified as extensive properties ,
depending on the quantity of the matter in the system (e.g.
mass and volume) and intensive properties, which are
independent of the amount of substances in the system (e.g.
temperature, pressure, density, surface tension, and viscosity of
pure liquids).
 Solution properties can also be classified as
additive, constitutive and colligative.
 Additive properties are derived from the sum of the properties
of the individual atoms or functional groups within the molecule
e.g. mass
 Constitutive properties are dependent on the structural
arrangement of the atoms within the molecule e.g. optical
rotation.
 Colligative properties will be discussed soon later.

Ideal and Real solutions
 Ideal solution is defined as a solution in which
there is No change in the properties of the
components other than dilution when they
are mixed to form the solution.
 Molecules exhibit complete freedom of
motion and randomness of distribution in the
solution.
Characteristics
 The volume change due to mixing of solute
with solvent should be zero

Ideal and Real solutions
 The final volume of the solution is
simply the. sum of the volumes of its
constituents
Vmixing = Observed vol of the solution- expected volume of the solution
 There is no contraction or expansion
when they are mixed.
 The enthalpy H change due to mixing
of solute in a solvent must be zero
(H=0)

 If the solute and solvent molecules having
same type of intermolecular attraction H
=0
 No heat is absorbed or evolved during mixing
of the components.
 The number of moles of solute does not
change after the dissolution process.
 Solute molecules should neither dissociate
nor associate themselves in the solution.
 The solution must be very dilute.
 If the attractive forces between the
constituents are as the same order the
solution can be considered to be an ideal .
 Ideal solutions is “one in which each
component of solution obeys Raoult’s law of
all temperatures and concentrations.

A) Ideal solutions and Raoult’s Law(solution of liquid in liquid ;
Both are volatile)
In solutions that have a volatile solute and a volatile solvent, the
partial vapor pressure of each component is less than the vapor
pressure of the pure component at the same temperature.

Raoult’s Law
 At any particular constant temperature,
the partial pressure of one component of a mixture =
the mole fraction of that component x (multiplied by) its
vapour pressure in the pure state at this temp.
OR
 The partial pressure of the component varies linearly
from zero to the full vapour pressure as its mole
fraction varies from zero to unity.
 The lowering of the vapour pressure when a non-volatile
solute is dissolved in a volatile solvent (A) can be
described by Raoult’s Law:
Describes vapor pressure lowering mathematically.

PA = P°A XA
PA = vapour pressure of solvent A above the solution
XA = mole fraction of the solvent A in the solution
P°A = vapour pressure of pure solvent A
PB = XB P°B
PB= vapour pressure of solvent B above the solution
P°B= Vapour pressure of pure ‘B’
only the solvent (A) contributes to
the vapour pressure of the solution

XA= 1
Vapor pressure-
composition curve for
an ideal binary system
benzene and Ethylene
chloride at 50 0C.
The three lines
represent the partial
pressure of ethylene
chloride, the partial
pressure of benzene
and the total
pressure of the
solution as the
function of the mole
fraction of the
constituents.

Now
PT=PA+ PB
PT= XAP°A + P°BXB
LIMITATIONS OF RAOULT’S LAW
 Applicable to dilute solutions because intermolecular attraction
are weak.
 To solutions containing non-volatile solutes.
 Does not apply if the solute dissociates
 Does not apply if the solute is associated
 Apply to a mixtures in which the components are similar
chemically
 Components of mixture do not interact
Eg :NaOH in water does not follow Raoults law
Glycerol in water follows Raoults law

NON IDEAL(REAL) SOLUTION
 A solution formed by mixing two
liquids to be non-ideal if it does not
obey Raoult’s law.
 Most liquid mixtures show varying degree
of deviation from raoults law depending on
the nature of the liquids and the
temperature.
Properties
 No volume change on mixing Vmix= zero
 No Ethanlpy change on mixing Hmix=
zero

Real (non-ideal) solutions
Many examples of solution pairs are known in which
the “cohesive” attraction of A for A exceeds the
“adhesive” attraction existing between A and B.
Similarly, the attractive forces between A and B may
be greater than those between A and A or B and B.
This may occur even though the liquids are miscible in
all proportions.
Such mixtures are real or non-ideal solutions

In case of non-ideal solution Raoults law is
modified
By replacing conc. By effective concentration
indicating activity.
PA = aA P°A
PA = vapour pressure of solvent A above the
solution
aA = activity 0f A
PB = aB P°B
aB = activity 0f B
P°B = Vapour pressure of pure ‘B’

Two Deviations are recognized from
Raoult’s Law - Nonideal Solutions
Positive Deviation
When the van der Waal’s forces between the solute and
solvent molecules in the solution are weaker than the
forces of attraction between the solute-solute molecules in
the pure solute and the solvent-solvent molecules in the
pure solvent, then the partial vapor pressure of each
component in the solution is greater than the pressure
predicted by Raoult’s Law.
Such solutions are said to exhibit a positive deviation and
to have heats of solution that are ENDOTHERMIC,
H > 0 solution
Example: Acetone - carbon disulfide solution at 35.2 0C

Negative Deviation
When the van der Waal’s forces between the solute and
solvent molecules in the solution are stronger than the
forces of attraction between the solute - solute molecules
in the pure solute and the solvent - solvent molecules in
the pure solvent, then the partial vapor pressure of each
component in the solution is less than the pressure
predicted by Raoult’s Law.
Such solutions are said to exhibit a negative deviation and to
have heats of solution that are EXOTHERMIC, H < 0
solution

Colligative Properties
 Colligative properties depend mainly on the number of
particles in a solution.
 The value of the colligative properties are approximately the
same for equal concentrations of different nonelectrolytes in
solution regardless of the species or chemical nature of the
constituents.
 The colligative properties of solutions are:
1. vapor pressure lowering,
2. boiling point elevation.
3. freezing point depression, and
4. osmotic pressure,
 In considering the colligative properties of solid-in-liquid
solutions, it is assumed that the solute is nonvolatile and
that the pressure of the vapor above the solution is provided
entirely by the solvent.

1.Lowering of Vapor Pressure, P
 When a nonvolatile solute is combined with a volatile
solvent, the vapor above the solution is provided solely
by the solvent.
 The solute reduces the escaping tendency of the
solvent, and,
 On the basis of Raoult’s law, the vapor pressure is
lowered proportional to the relative number (rather than
the weight concentration) of the solute molecules.
 Since the solute is nonvolatile, the vapor pressure of
the solvent P1 is identical to the total pressure of the
solution P.
 It is more convenient to express the vapor pressure of
the solution in terms of concentration of the solute
rather than the mole fraction of the solvent.

Lowering of Vapor Pressure, P
The sum of the mole fractions of the constituents in a solution
is unity.
X1 + X2 = 1 i.e. X1 = 1- X2
so P = P1
º (1 – X2)  P1
º – P = P1
ºX2
X1 is the solvent mole fraction and X2 is the solute mole fraction
thus
Where ΔP is the lowering of the vapor pressure and
ΔP/P1° is the relative vapor pressure lowering.
 As ΔP/P1° depends only on the number of solute molecules, it is
considered as a colligative property.
12
2
2
11
1
nn
n
X
P
P
P
PP








 For dilute solutions (n2n1) so
 When water is the solvent, 1000g = 1L
Thus
where
m is the solute molality
mX
P
P
018.02
1



5.5502.18/1000
/ 22
2
1
mMw
X
P
P



11
222
2
1
/
/
1 Mw
Mw
n
n
X
P
P




Q)Calculate the relative vapor pressure lowering at 20°C for a solution
containing 171.2 g of sucrose in 1000g of water. The molecular weight of
sucrose is 342.3 and the molecular weight of water is 18.02.
Moles of sucrose (n2) = 171.2/342.3= 0.5 mol
Moles of water (n1)= 1000/18.02= 55.5 mol
ΔP/P1°= X2= 0.5/ (0.5+55.5)= 0.0089.
Q) For the above question, providing that the vapor pressure of water at
20°C is 17.54 mm Hg. Calculate the vapor pressure when added 171.2 g
of sucrose (0.5 mol).
ΔP/P1°= (P1°- P)/ P1°= (17.54-P)/17.54=0.0089
P= 17.38 mmHg

Determination of Vapor pressure-
Isopiestic method
The solution whose vp is to be
determined and a solution containing
a standard solute KCl are placed in
separate dishes in a closed container.
The vapour of the solution with higher
pressure passes to the one with the
lower pressure until the VP of the
two solutions are the same i.e.
isopiestic equal pressures
When there is no further change in
weight, the solutions are analyzed to
determine their concentrations.
The V.P. of the test isopiestic with
KCl soluition is thus readily
obatined.
Thermoelectric Method is also
used to determine
VP of solutions

2.Elevation of the Boiling Point
 Normal B.P is the
temp. at which the VP
of liquid becomes
equal to an external
pressure of 760 mm
Hg.
 A solution will boil at
a higher temp. than
will the pure solvent
this is the colligative
property called
Boiling point
elevation.
Figure – Boiling point elevation of the
solvent due to addition of a solute
(not to scale)
The elevation of the boiling point can be written as
T – Tº = Tb.

2. Elevation of the Boiling Point
 An exact equation to calculate Tb (boiling point elevation)
where,
Hv is the latent heat of vaporization of the solvent,
Tb is the boiling point of the solvent,
R is the gas constant,
X2 is the mole fraction of the solute
 A less exact equation (more commonly used) is
where
M1 is the molecular weight of the solvent,
m is the molal concentration of the solute (mol.kg-1),
Kb is the molal elevation constant or the ebullioscopic constant (deg.kg.mol-1)
22
2
kXX
H
RT
T
v
b
b 


mKm
H
MRT
T b
v
b
b 


1000
1
2

Elevation of the Boiling Point
 Kb (ebullioscopic
constant) has a
characteristic value for
each solvent;
 Kb for water is 0.51
deg.kg/mole.
 Kb may be considered as
the boiling point elevation
for an ideal 1m solution.
 Stated another way,
Kb is the ratio of the boiling
point elevation to the
molal concentration in an
extremely dilute solution
in which the system is
approximately ideal.
Figure –The influence of concentration
on the ebulioscopic constant.

Determination of B.P Elevation
Boiling point elevation is determined by Cottrell
B.P apparatus.
In this weighed amount of solute is placed in
solvent provided with and reflux condenser and
thermometer.
The vapor is forcefully sprayed over the bulb of
thermometer to obtain an invariant equillibrium
temp. In the similar way BP of pure solvent is
also found out.

3.Depression of Freezing Point
 The normal freezing point or melting
point of a pure compound is the
temperature at which the solid and the
liquid phases are in equilibrium under
a pressure of 1atm.
 Equilibrium means that the tendency
for the solid to pass into the liquid is
the same as the tendency for the
reverse process to occur, since both
the liquid and the solid have the same
escaping tendency.
 The T value for water saturated with
air at this pressure is arbitrary
assigned a temperature of 0ºC – see
figure.
 The triple point of air-free water, at
which solid, liquid, and vapor are in
equilibrium, lies at a pressure of 4.58
mmHg and a temperature of
0.0098ºC.
Figure – Depression of the freezing point
of the solvent, water, by a solute

 The more concentrated is the solution, the farther apart are
the solvent and the solution curves and the greater is the
freezing point depression (Tf).
 Tf is proportional to the molal concentration of the solute
and can be determined as follows:
Tf = Kfm or
or
where
Tf is the freezing point depression,
Hf is molal heat of fusion,
Tf is the freezing point of the solvent and
Kf is the molal depression constant or the cryoscopic constant, which depends on
the physical and chemical properties of the solvent,
Kf for water is 1.86 deg.kg/mole.
21
21000
Mw
w
KT ff  m
H
MRT
T
f
f
f


1000
1
2
Depression of Freezing Point

 The freezing point depression of a solvent is a function only of
the number of particles in the solution, and for this reason it is
referred to as a colligative property.
 The depression of the freezing point, like the boiling point
elevation, is a direct result of the lowering of the vapor pressure
of the solvent.
 Kf may be determined experimentally by measuring Tf/m at
several molal concentrations and extrapolating to zero
concentration – i.e. Kf is the intercept.
 Electrolytes don’t follow this relationship.
 The equation is
f
f
Ktcons
m
T


tan

As seen in fig. Kf approches the
Value of 1.86 for water
solutions of Sucrose and
Glycerin as the concentration
tends to zero
and equation is valid
only in very dilute solutions.
The value of Kf for the solvent
in a solution of citric acid is
observed not to approach 1.86.
•The abnormal behavior is to be
expected when dealing with
solutions of electrolytes.
Figure – The influence of
concentration on the cryoscopic
constant for water

 Citric acid shows abnormal behavior
as seen in the previous figure –
 This abnormal behavior is to be
expected when dealing with
electrolytes.
 Freezing point depression can also be
calculated from the following equation:
where
Tf is the freezing point of the solvent,
T is the freezing point of the solution,
X2 is the mole fraction of the solute,
R is the gas constant and
Hf is the heat of fusion of the solvent.
2X
H
TRT
T
f
f
f



4.Osmotic Pressure 
 Escaping tendency can be
measured in terms of vapor
pressure or the closely related
colligative property, osmotic
pressure.
 If a solution is confined in a
membrane only permeable to
the solvent molecules (semi-
permeable), osmosis occurs.
 Osmosis is defined as the
passage of the solvent into a
solution through a semi-
permeable membrane.
Apparatus for demonstrating
osmosis.

Osmotic Pressure 
 Van’t Hoff suggested that there is an apparent analogy
between solutions and gases and that the osmotic
pressure in a dilute solution is equal to the pressure
that the solute would exert if it were a gas occupying
the same volume. The equation is
in which
 is the osmotic pressure in atm,
V is the volume of the solution in liters,
n is the number of moles of the solute,
R is the gas constant equal to 0.082L atm mole-1 deg -1, and
T is the absolute temperature.
nRTV 

 The van’t Hoff equation can be rewritten as
in which c is the concentration of the solute in moles per
liter (molarity).
 Morse has shown that when the concentration is
expressed in molality rather than molarity, the results
compare more with the experimental findings. The Morse
equation is
cRTRT
n

V

RTm

 Thermodynamic osmotic pressure equation is a more
exact expression for osmotic pressure.
 It applies to concentrated as well as dilute solutions,
provided that the vapor follows the ideal gas law
in which
 is the osmotic pressure in atm,
V1 is the volume of 1 mole of solvent (partial molar volume) in liters,
Pº is the vapor pressure of pure solvent in atm, and
p is the vapor pressure of the solution in atm,
R is the gas constant equal to 0.082L atm mole-1 deg -1, and
T is the absolute temperature.
P
P
V
RT 
 ln
1


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