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CHAPTER: 0 2 SOLUTIONS
INTRODUCTION:
In normal life we rarely come across pure substances. Most of these are mixtures containing
two or more pure substances. Their utility or importance in life depends on their composition.
Example :
The properties of alloys are quite different from those of the respective metals.
Example: Brass (mixture of copper and zinc), German silver (mixture of copper, zinc
and nickel) , bronze (mixture of copper and tin), Gunmetal ( 86% copper , 9.5% tin,
2.5% lead and 2% zinc. ).
Concentration of 1 part per million (ppm) of fluoride ions in water prevents tooth
decay , while 1.5 ppm causes the tooth to become mottled (mark with spots or smears
of colour) and high concentrations of fluoride ions can be poisonous (sodium fluoride is
used in rat poison).
Intravenous injections are always dissolved in water containing salts at particular ionic
concentrations that match with blood plasma concentrations and so on.
In this Unit, we will consider mostly liquid solutions and their formation, followed
by studying the properties of the solutions, like vapour pressure and colligative properties.
Let us begin with components of solution, types of solutions and then with
various alternatives in which concentrations of a solute can be expressed in liquid solution.
SOLUTION:
A solution is a homogeneous mixture of two or more components. The components are
present in different proportions within certain limits.
NOTE: By homogenous mixture we mean that its composition and properties are uniform
throughout the mixture.
SOLUTE:
The component present in a smaller proportion in the solution is called a solute.
One or more solute components may be present in smaller proportion in the solution.
SOLVENT:
The component present in a larger proportion in the solution is called a solvent.
Solvent determines the physical state of the solution in which it exists. i.e The physical
state of the solution is same as that of the solvent.
Example: 1) In the solution of common salt (brine solution), common salt is the solute and
water is the solvent.
- In the aqueous solution of ethanol, ethanol is the solute and water is the solvent.
TYPES OF SOLUTIONS:
In this Unit we shall consider only binary solutions (i.e., consisting of two components). Here
each component may be solid, liquid or in gaseous state.
II. BASED ON THE PHYSICAL STATE OF SOLUTE AND SOLVENT:
There are three types of solutions based on the physical states of the solute and the
solvent.
1. SOLID SOLUTIONS:
Solutions containing solid / liquid /gaseous solute dissolved in solid solvent are
called solid solutions. Example: Brass, Bronze, amalgams…. etc.
2. LIQUID SOLUTIONS:
Solutions containing solid / liquid /gaseous solute dissolved in liquid solvent are
called liquid solutions.
Example: Glucose in water, Carbon dioxide in water (soda water), Ethanol in water,
Acetone in water, etc.
3. GASEOUS SOLUTIONS:
Solutions containing solid / liquid /gaseous solute dissolved in gaseous solvent are
called gaseous solutions. Example: Atmospheric air, oxygen cylinder etc.
SOME COMMON EXAMPLES OF SOLID, LIQUID AND GASEOUS SOLUTIONS:
TYPE OF SOLUTION SOLUTE SOLVENT COMMON EXAMPLES
SOLID SOLUTIONS
Solid Solid Copper dissolved in gold
Liquid Solid Amalgam of mercury with sodium
Gas Solid Solution of hydrogen in palladium
LIQUID SOLUTIONS
Solid Liquid Glucose dissolved in water
Liquid Liquid Ethanol dissolved in water
Gas Liquid Oxygen dissolved in water
GASEOUS SOLUTIONS
Solid Gas Camphor in nitrogen gas
Liquid Gas Chloroform mixed with nitrogen gas
Gas Gas Mixture of oxygen and nitrogen gases
III. BASED ON THE AMOUNT (CONCENTRATION) OF SOLUTE:
I. Based on the number of components:
1. Binary solution: A solution containing two components is called a **binary solution.
- Ternary solution:** A solution containing three components is called a **ternary solution.
- Quaternary solution:** A solution containing four components is called a quarternary
solution.
and so on…..
3 ) MASS BY VOLUME PERCENTAGE (w/v%): It is the number of grams of the solute
present in 100 ml of the solution.
Mass of the solute x 100
Mass-Volume percentage =
Total volume of the solution
Note: This unit is commonly used in medicine and pharmacy.
4) PARTS PER MILLION (ppm): It is the number of parts by mass of the solute present
per million parts by mass of the solution.
6 Mass of the solute x 10 Parts per million (ppm) = Mass of the solution
Note:
- When the solute is present in trace quantities in the solution, concentration of such a
solution is expressed in terms of ppm.
- As in the case of percentage, concentration in parts per million can also be expressed as
mass to mass, volume to volume and mass to volume. A litre of sea water (which weighs 1030
g) contains about 6 × 10
- 3 g of dissolved oxygen (O 2 ). Such a small concentration is also
expressed as 5.8 g per 106 g (5.8 ppm) of sea water. The concentration of pollutants in water
or atmosphere is often expressed in terms of μg mL
- 1 or ppm.
2) MOLE FRACTION ( x ): It is the ratio of number of moles of one component to that of
the total number of moles of all the components present in the solution.
Let us consider a solution containing 2 components namely ‘A’ and ‘B’. If nA is the number of
moles of a solute ‘ A’ dissolved in nB moles of a solvent ‘ B’ , then mole fraction of ‘A’ and ‘B’
are given as
A B A B A B A B A B
n n
and Note: 1
n n n n
x x x x
Mole fraction unit is very useful in relating some physical properties of solutions, say vapour
pressure with the concentration of the solution and quite useful in describing the calculations
involving gas mixtures.
3) MOLARITY (M): It is the number of moles of a solute present in 1 dm
3 (1 litre) of the
solution.
The unit of molarity is mol dm
- 3 .
3 Mass /dm No. of moles of solute mass of solute x 1000 Molarity= = = Molecular mass Volume of solvent in litre Mol. mass of solute x Vol. in litres
NOTE : If 1 mole of a solute is dissolved in 1 dm
3 of the solvent then the concentration of such
a solutionis 1M.
EXAMPLE: 0.25 mol L
- 1 or 0.25 M solution of NaOH means that 0.25 mol of NaOH has been
dissolved in one litre (or one cubic decimetre).
4) MOLALITY (m): It is the number of moles of a solute dissolved in 1 kg of solvent.
The Unit of molality is mol kg
- 1 .
Mass /kg of the solvent No. of moles of solute Mass of solute x 1000 Molality = = = Molecular mass Mass of solvent in kg Mol. mass of solute x mass of solvent in gms
NOTE : If 1 mole of a solute is dissolved in 1 kg of the solvent then the concentration of such
a solutionis 1m.
EXAMPLE: 1.00 mol kg
- 1 or 1.00 m solution of KCl means that 1 mol i.e 74.5 g of KCl is
dissolved in 1 kg of water.
Note:
1. Each method of expressing concentration of the solutions has its own merits and
demerits. Mass %, ppm, mole fraction and molality are independent of temperature,
whereas molarity and normality are function of temperature. This is because volume
depends on temperature and the mass does not.
2. With increase in temperature volume increases, therefore molarity and normality
decreases.
DISSOLUTION:
When a solid solute is added to the solvent, some solute dissolves and its
concentration increases in solution. This process is known as dissolution.
CRYSTALLIZATION:
Some solute particles in solution collide with the solid solute particles and get
separated out of solution. This process is known as crystallisation.
SATURATED SOLUTION :
It is a state at which the two processes i.e dissolution and crystallisation occur at the
same rate. Under such conditions, number of solute particles going into solution will
be equal to the solute particles separating out and a state of dynamic equilibrium is
reached.
5 ) NORMALITY (N): It is the number of gram equivalents of a solute present in 1 dm
- 3 of a
solution.
The Unit of normality is N.
It is expressed as,
No. gram equivalents of the solute mass of solute in grams Normality = = Volume of solution in litres Eq. mass of solute x Volume of solution in litre
NOTE : If 1 gram equivalent of a solute is dissolved in 1 litre of a solvent then the concentration
of such a solutionis 1N.
EXAMPLE: If 53 g of Na 2 CO 3 is dissolved in 1 litre of water, the concentration of such a solution is
1N, where as if 106 g of Na 2 CO 3 is dissolved in 1 litre of water, the concentration of such a solution
is 1M.
3. EFFECT OF PRESSURE:
Pressure does not have any significant effect on solubility of solids in liquids. It is so
because solids and liquids are highly incompressible and practically remain unaffected by
changes in pressure.
II. SOLUBILITY OF A GAS IN A LIQUID:
**1. EFFECT OF NATURE OF SOLUTE AND SOLVENT: Like dissolves like.
- EFFECT OF TEMPERATURE: Solubility of gases in liquids decreases with rise in**
temperature.
When gases are dissolved in liquids, the gas molecules are present in liquid phase
and the process of dissolution can be considered similar to condensation and heat is
evolved in this process. And this dissolution process involves dynamic equilibrium and
thus must follow Le Chatelier’s Principle. As dissolution is an exothermic process, the
solubility of gases in liquids should decrease with increase of temperature.
3. EFFECT OF PRESSURE: Solubility of gases in liquids increases with increase in
pressure.
Quantitative relation between pressure and solubility of a gas in a solvent
is given by Henry’s law.
Many gases dissolve in water. Oxygen dissolves only to a small extent in water. It is this
dissolved oxygen which sustains all aquatic life. On the other hand, hydrogen chloride gas
(HCl) is highly soluble in water. Solubility of gases in liquids is greatly affected by pressure
and temperature. The solubility of gases increases with increase in pressure over the
solution. For solution of gases in a solvent, consider a system as shown in following diagram.
W (^1)
W 1 W 2 W (^3)
Effect of pressure on the solubility of a gas
Piston
Gas molecules
Undissolved
Gas molecules
Dissolved
The lower part is solution and the upper part is gaseous system at pressure p and temperature
T. Assume this system to be in a state of dynamic equilibrium, i.e., under these conditions rate
of gaseous particles entering and leaving the solution phase is the same. Now increase the
pressure over the solution phase by compressing the gas to a smaller volume. This will
increase the number of gaseous particles per unit volume over the solution and also the rate
at which the gaseous particles are striking the surface of solution to enter it. The solubility of
the gas will increase until a new equilibrium is reached resulting in an increase in the pressure
of a gas above the solution and thus its solubility increases.
Henry was the first person to give a quantitative relation between pressure and
solubility of a gas in a solvent which is known as Henry’s law.
HENRY’S LAW:
It states that “At constant temperature the partial pressure of the gas in vapour phase
(P) is directly proportional to the mole fraction of the gas ( x )in the solution”
Mathematically,
P = KH x
Where,
KH = Henry’s law constant.
P = partial pressure of the gas in vapour phase.
x = mole fraction of the gas in solution
Graphical interpretation;
500
1000
0.01 0.
Partial pressure of HCl /torr
Mole fraction of HCl in its solution in cyclohexane
.
.
.
.
.
0
DIFFERENT GASES HAVE DIFFERENT KH VALUES AT THE SAME TEMPERATURE AS
SHOWN IN THE FOLLOWING TABLE.
Gas Temperature/K KH / k bar Gas Temperature/K^ KH/ k bar
He 293 144.97 Argon 298 40.
H 2 293 69.16 CO 2 298 1.
N 2 293 76.48 Formaldehyde 298 1.83×10-^5
N 2 303 88.84 Methane 298 0.
O 2 293 34.86 Vinyl chloride 298 0.
O 2 303 46.
Characteristics of Henry’s constant (KH):
1. This suggests that KH is a function of the nature of the gas. 2. Higher the value of KH at a given pressure, the lower is the solubility of the gas in
the liquid.
EXAMPLE: KH values for both N 2 and O 2 increase with increase of temperature indicating
that the solubility of gases increases with decrease of temperature. It is due to this reason that
aquatic species are more comfortable in cold waters rather than in warm waters.
1 mol of solvent
1 mol of solute
Solute molecules
Pure solvent
Solvent molecules
Decrease in the vapour pressure of solvent molecules in presence of solute
If P
o is the vapour pressure of the pure solvent and P is the vapour pressure of the solution,
the lowering of vapour pressure (∆P) is given by ∆P = P
o
- P
LOWERING OF VAPOUR PRESSURE (∆P):
The difference between the vapour pressure of the pure solvent and that of the solution
is called lowering of vapour pressure.
II. VAPOUR PRESSURE OF LIQUID IN LIQUID SOLUTIONS:
Let us consider a binary solution of two volatile liquids and denote the two components as 1
and 2. When taken in a closed vessel, both the components would evaporate and eventually
equilibrium would be established between vapour phase and the liquid phase. Let the total
vapour pressure at this stage be Ptotal and P 1 and P 2 be the partial vapour pressures of the
two components 1 and 2 respectively. These partial pressures are related to the mole fractions
1
x and
2
x of the two components 1 and 2 respectively. The French chemist, Francois Marte
Raoult (1886) gave the quantitative relationship between them. The relationship is known as
the Raoult’s law
RAOULT’S LAW:
It states that for a solution of volatile liquids, the partial vapour pressure of each
component in the solution is directly proportional to its mole fraction.
Thus, for component 1
1 1
0 1 1 1
P
P P
x
x
Where,
0
P 1 is the vapour pressure of pure component 1 at the same temperature,
Similarly, for component 2
0
2 2 2
P P x
Where,
0
P 2 represents the vapour pressure of the pure component 2.
According to Dalton’s law of partial pressures, the total pressure (Ptotal) over the solution phase
in the container will be the sum of the partial pressures of the components of the solution and
is given as:
total 1 2
P P P
Substituting the values of P 1 and P 2 , we get
0 0 total 1 1 2 2
0 0 2 1 2 2
0 0 0 1 2 1 2
P P P
(1 )P P
P (P P )
x x
x x
x
FOLLOWING CONCLUSIONS ARE DRAWN FROM THE ABOVE EQUATION.
i. Total vapour pressure over the solution can be related to the mole fraction of any one
component.
ii. Total vapour pressure over the solution varies linearly with the mole fraction of component
iii. Depending on the vapour pressures of the pure components 1 and 2, total vapour
pressure over the solution decreases or increases with the increase of the mole fraction of
component 1.
A plot of p 1 or p 2 versus the mole fractions x 1 and x 2 for a solution gives a linear plot as shown
in the following graph. These lines (I and II) pass through the points and respectively when x 1
and x 2 equal unity. Similarly the plot (line III) of Ptotal versus x 2 is also linear. The minimum
value of Ptotal is
0
P 1 and the maximum value is
0
P 2 , assuming that component 1 is less volatile
than component 2, i.e.,
0 0
P 1 P 2.
PTotal^ = p
1 + p^2
p 1
o P 1
vapour pressure (in atm)
mole fraction X 2
Plot of V.P and mole fraction of an ideal solution at constant temperature
o P 2
x 1 = 1 x 1 = 0
x 2 = 0 x 2 = 1
I
II
III
p^2
The composition of vapour phase in equilibrium with the solution is determined by the partial
pressures of the components. If y 1 and y 2 are the mole fractions of the components 1 and 2
respectively in the vapour phase then, using Dalton’s law of partial pressures:
p 1 = y 1 Ptotal
p 2 = y 2 Ptotal
In general pi = yi Ptotal
RAOULT’S LAW AS A SPECIAL CASE OF HENRY’S LAW:
An ideal solution is one where the intermolecular interactions between the components of the
solution are of the same magnitude as that of the intermolecular interactions found in the pure
components.
i.e., experimentally found vapour pressure of a solution is equal to the value calculated by
Raoult's law.
0 0 total 1 2 1 1 2 2
P P P P x P x
EXAMPLE:
i) n-hexane and n-heptane ii) Bromoethane and chloroethane
iii) benzene and toluene iv) Bromobenzene and Chlorobenzene.
(A perfectly ideal solution is rare but some solutions are nearly ideal in behaviour.)
CHARACTERSTIC FEATURES OF IDEAL SOLUTIONS:
- Ideal solution obeys 'Raoult's law of liquid mixtures'. For an ideal solution total vapour
pressure is equal to the sum of the vapour pressure of the components as given by Raoult's
law. P = P 1 + P 2 = P
o 1 x 1 + P
o 2 x 2
- ∆V (^) mixing = 0 ; i.e. total volume of the components mixed is equal to the volume of the mixture
obtained.
- ∆H (^) mixing = 0 ; i.e. no enthalpy change takes place when the two components are mixed.
NON-IDEAL SOLUTIONS:
The solutions which do not obey Raoult's law of solutions over the entire range of
concentrations and temperature are called non-ideal solutions.
A non-ideal solution is one where the intermolecular interactions between the components are
different from the intermolecular interactions between the molecules of the pure components.
i.e., experimentally found vapour pressure of the solution is either more or less than the
value of vapour pressure calculated by Raoult's law.
P ≠ P 1 + P 2 P ≠ P
o 1 x 1 + P
o 2 x 2
EXAMPLES:
Ethanol + water 2) carbon tetrachloride + toluene
benzene + acetone 4) water + n-propyl alcohol
CHARACTERSTIC FEATURES OF NON IDEAL SOLUTIONS:
- Non ideal solution does not obey 'Raoult's law of liquid mixtures'. For an ideal solution total
vapour pressure is not equal to the sum of the vapour pressure of the components as given
by Raoult's law.
P ≠ P 1 + P 2 P ≠ P
o 1 x 1 + P
o 2 x 2
- ∆V (^) mixing ≠ 0 ; i.e. total volume of the components mixed is not equal to the volume of the
mixture obtained.
- ∆H (^) mixing ≠ 0 ; i.e. change in enthalpy takes place when the two components are mixed.
Heat is either absorbed or liberated.
DIFFERENCES BETWEEN IDEAL AND NON-IDEAL SOLUTIONS:
SI.
No.
Ideal solutions Non ideal solutions
The intermolecular interactions
between the components of the
solution are similar to those found in
the pure components.
The intermolecular interactions between
the components of the solution are different
from those found in the pure components.
Obeys Raoult's law of solution. Does not obey Raoult's law of solution.
Shows either positive or negative deviation.
There is no volume change during
mixing (∆V = 0)
Volume of the solution is either more or less
than the total volume of the components
mixed. (∆V ≠ 0)
Enthalpy of mixing is zero (∆H = 0) i.e.,
no heat is evolved or absorbed when a
solution is prepared.
Enthalpy of mixing is not zero (∆H ≠0) i.e.
heat is either evolved or absorbed during
mixing of the components.
THERE ARE TWO TYPES OF NON-IDEAL SOLUTIONS:
Type 1: Solutions showing positive deviation from Raoult's law.
Examples:
(i) Methanol + water (ii) Ethanol + water (iii) Ethanol + acetone (iv) Benzene + acetone
(v) CS 2 +Acetone
- If there is repulsion between the molecules of the two components of the solution then
the system shows positive deviation from Raoult's law.
In this case P 1 > x 1 P
o 1 and P 2 >^ x 2 P
o 2 Hence P >^ x 1 P
o 1 +^ x 2 P
o 2
- ∆V mixing is positive. i.e. the volume of the solution is more than the sum of the volumes of
the two individual liquids mixed.
- ∆H mixing is positive i.e. the process is endothermic. Hence, during the preparation of the
solution heat is absorbed and temperature decreases.
Type 2: Solutions showing negative deviation from Raoult's law
Examples:
(i) HCl + water (ii) HNO 3 + water (ii) chloroform + acetone (iv) chloroform + benzene,
(v) acetone + aniline
- If there is attraction between the molecules of the two components of the solution, then
the system shows negative deviation from Raoult's law.
In this case P 1 < x 1 P
o 1 and P 2 <^ x 2 P
o 2 Hence P <^ x 1 P
o 1 +^ x 2 P
o 2
2) ∆Vmixing is negative ; i.e. the volume of the solution is less than the sum of the volumes of
the two individual liquids mixed.
3) ∆Hmixing is negative; i.e. the process is exothermic. Hence during the preparation of the
solution heat is evolved and temperature rises.
DIFFERENCES BETWEEN NON-IDEAL SOLUTIONS SHOWING POSITIVE AND
NEGATIVE DEVIATIONS:
1. Mixture of ethanol and acetone behave in this manner. In pure ethanol, molecules are
hydrogen bonded. On adding acetone, its molecules get in between the host molecules
and break some of the hydrogen bonds between them (ethanol). Due to weakening of
interactions, the solution shows positive deviation from Raoult’s law.
2. Mixture of carbon disulphide and acetone, the dipolar interactions between solute-solvent
molecules are weaker than the respective interactions among the solute-solute and
solvent-solvent molecules. This solution also shows positive deviation.
NON-IDEAL SOLUTIONS SHOWING NEGATIVE DEVIATION:
In case of negative deviations from Raoult’s law, A-B interactions are stronger than
those between A-A or B-B i.e. the intermolecular attractive forces between A-A and B-B are
weaker than those between A-B and leads to decrease in vapour pressure resulting in
negative deviations.
Example:
1. Mixture of phenol and aniline. In this case the intermolecular hydrogen bonding between
phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective
intermolecular hydrogen bonding between similar molecules.
2. Mixture of chloroform and acetone forms a solution with negative deviation from Raoult’s
law. This is because chloroform molecule is able to form hydrogen bond with acetone
molecule as shown below.
H 3 C
C O
Cl
H
H 3 C
C Cl
Cl
........
This decreases the escaping tendency of molecules for each component and consequently
the vapour pressure decreases resulting in negative deviation from Raoult’s law.
AZEOTROPES:
Binary liquid mixtures forming constant boiling mixtures having the same composition
in liquid phase and vapour phase and boil at a constant temperature are called as
azeotropes.
Note: it is not possible to separate the components azeotropes by fractional distillation.
There are two types of azeotropes namely,
1. MINIMUM BOILING AZEOTROPE: The solutions which show a large positive
deviation from Raoult’s law form minimum boiling azeotrope at a specific
composition.
Example: Ethanol-water mixture
(Rectified spirit obtained by fermentation of sugars on fractional distillation gives a solution
containing approximately 95.5%by volume of ethanol and 4.5% by volume of water. Once this
composition, known as azeotrope composition, has been achieved, the liquid and vapour have
the same composition, and no further separation occurs.)
2. MAXIMUM BOILING AZEOTROPE : The solutions that show large negative deviation
from Raoult’s law form maximum boiling azeotrope at a specific composition.
Example: Nitric acid – water mixture
(This azeotrope has the approximate composition, 68% nitric acid and 32% water by mass,
with a boiling point of 393.5 K.)
COLLIGATIVE PROPERTIES:
Colligative properties are the properties of a dilute solution containing a non volatile
solute whose values depend only upon the number of the solute particles present in
the solution but not on their nature, size or chemical composition.
Some of the Colligative properties of dilute solutions are:
1. Relative lowering of vapour pressure
o
o
P - P
P
2. Elevation of boiling point ^ T b .
3. Depression of freezing point T f .
4. Osmotic pressure ^^ π .
RELATIVE LOWERING OF VAPOUR PRESSURE
o
o
P - P
P
It is the ratio of lowering of vapour pressure to that of the vapour pressure of the pure
solvent.
We know that the vapour pressure of a solvent in solution is less than that of the pure solvent
and Raoult’s established that the lowering of vapour pressure depends only on the
concentration of the solute particles and it is independent of their identity. The following
equation establishes a relation between vapour pressure of the solvent in solution, mole
fraction and vapour pressure of the solvent.
According to Raoult’s law of vapour pressure of liquid mixtures,
0
P 1 P 1 x 1
We know that lowering of vapour pressure,
0 P 1 P 1 -P 1 .....(1)
Substituting the value of P 1 in equation (1) we get,
0 0 1 1 1 1
0 1 1 1
P P -P
P P (1 - )
x
x
We know that x 1 +x 2 = 1 hence 1-x 1 =x 2
0 1 1 2
P P x
Substituting the value of ∆P 1 in the above equation we get,
0 0 1 1 1 2
P - P P x
Liquid solvent Solution
o vapour pressure (in atm)^ Tb
Temperature (in K)
Tb
Graph showing Tb elevation in boiling
point of solution
Tb
1 Atm
The above graph shows the variation of vapour pressure of the pure solvent and solution as
a function of temperature.
Example: The vapour pressure of an aqueous solution of sucrose is less than 1.013 bar at
373.15 K. In order to make this solution boil, its vapour pressure must be increased to 1.
bar by raising the temperature above the boiling temperature of the pure solvent (water). Thus,
the boiling point of a solution is always higher than that of the boiling point of the pure solvent
in which the solution is prepared as shown in the graph.
Similar to lowering of vapour pressure, the elevation of boiling point also
depends on the number of solute molecules rather than their nature. A solution of 1 mol of
sucrose in 1000 g of water boils at 373.52 K at one atmospheric pressure.
The difference between boiling point of the solution and the pure solvent is called as
elevation on boiling point.
Let
o Tb be the boiling point of pure solvent and Tb be the boiling point of
solution. The increase in the boiling point
o T^ b^ ^ Tb^ Tb is known as elevation of boiling
point.
Experiments have shown that for dilute solutions the elevation of boiling point (Δ T b) is directly
proportional to the molal concentration of the solute in a solution.
Thus
T (^) b m
Tb K b m
2
2 2
(^1 1 )
we know that
w
M w
m
w w M
substituting the value of ' m ' in the above equation we get,
b 2 b 1 2
K 1000
T
w
w M
Where,
m = Molality of the solution.
K b = Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant).
w 2 = mass of solute in grams.
M 2 = molar mass of the solute.
w 1 = mass of solvent in grams.
Ebullioscopic Constant ( K b):
It is the elevation in boiling point when the molar concentration of the solution is one
molal.
Note:
Unit of K b is K kg mol
- 1 .
Values of K b are different for different solvents.
Thus, in order to determine M 2 , molar mass of the solute, known mass of solute in a known
mass of the solvent is taken and Δ T b is determined experimentally for a known solvent whose
K b value is known.
DEPRESSION OF FREEZING POINT (∆Tf):
The lowering of vapour pressure of a solution causes a lowering of the freezing point compared
to that of the pure solvent as shown in the graph.
Liquid solvent
Solution
Frozen solvent
Tf
o Tf
vapour pressure (in atm)
Temperature (in K)
Tf
Graph showing Tf depression of the freezing point of a solution
FREEZING POINT:
It is the temperature at which the vapour pressure of the substance in its liquid phase
is equal to its vapour pressure in the solid phase.
Note:
A solution will freeze when its vapour pressure equals the vapour pressure of the pure
solid solvent as is clear from graph.
At the freezing point of a substance, the solid phase is in dynamic equilibrium with the
liquid phase.
