1.

SOLUTIONS
Solutions are homogeneous mixtures containing two or more components. Generally, the component
that is present in larger quantity is called solvent. Solvent determines the physical state of the solution. One
or more components present in the solution other than solvent are called solutes. [Or, the substance which is
dissolved is called solute and the substance in which solute is dissolved is called solvent].
Solutions containing only two components are called binary solutions. Here each component may be
solid, liquid or in gaseous state. Based on this, solutions are of the following types:

Types of Solution Solute Solvent Examples

Gas Gas Mixture of O2 and CO2
Gaseous solutions Liquid Gas Chloroform mixed with nitrogen gas, water-vapour in air
Solid Gas Camphor in nitrogen gas, naphthalene in air
Gas Liquid Oxygen dissolved in water, soda water
Liquid solutions Liquid Liquid Alcohol dissolved in water, dilute acids and alkalies
Solid Liquid Salt in water, glucose in water
Gas Solid Hydrogen in Pd, Pt, Ni etc
Solid solutions Liquid Solid Amalgam of mercury with sodium
Solid Solid Gold ornaments, alloys of metals

Concentration of Solutions
Composition of a solution can be expressed in terms of concentration. Concentration is defined as the
number of moles of solute present per litre of the solution. The concentration of a solution can be expressed
by the following ways:
(i) Mass percentage (w/w): It is defined as the mass of the component present in 100g of the solution.
Mass of the component in the solution × 100
i.e. Mass % of a component =
Total mass of the solution

For e.g. 10% aqueous solution of glucose by mass means that 10 g of glucose is dissolved in 90 g of water
resulting in a 100 g solution.
Concentration described by mass percentage is commonly used in industrial chemical applications.
(ii) Volume percentage (v/v): It is defined as the volume of a component present in 100 mL of the
solution.
Volume of the component ×100
i.e. Volume % of a component =
Total volume of the solution

For example, 10% ethanol solution in water means that 10 mL of ethanol is dissolved in 90 mL of water such
that the total volume of the solution is 100 mL.
Concentration of solutions containing liquids is commonly expressed in this unit.
(iii) Mass by volume percentage (w/v): It is the mass of solute dissolved in 100 mL of the solution.
It is commonly used in medicine and pharmacy.
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 100
Mass/volume % of a component =
𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
(iv) Parts per million (ppm): When a solute is present in trace quantities (i.e. very small amounts), its
concentration is expressed in parts per million (ppm). It is defined as the number of parts of a
particular component in million parts of the solution.

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 Number of parts of the component × 106
i.e. Parts per million (ppm) =
Total number of parts of all the components of the solution
Concentration in parts per million can be expressed as mass to mass, volume to volume and mass to
volume.
The concentration of pollutants in water or atmosphere is expressed in terms of μg mL –1 or ppm. [μg is
microgram]
(v) Mole fraction (χ): It is defined as the ratio of the number of moles of a particular component to the
total number of moles of the solution.
Number of moles of the component
Mole fraction of a component =
Total number of moles of solution

For example, in a binary solution, if the number of moles of A and B are nA and nB respectively,
nA
then, the mole fraction of A (χA) =
nA + nB
nB
and that of the component B (χB ) =
nA + nB

nA nB
χA + χB = n + nB
+ nA + nB
=1
A
i.e. in a given solution sum of the mole fractions of all the components is unity. If there are i components,
then,
χ1 + χ2 + χ3 + .................. + χ𝑖 = 1
Mole fraction is useful in describing the calculations involving gas mixtures.
(vi) Molarity (M): It is defined as the number of moles of solute dissolved per litre of solution.
Number of moles of solute (n)
i.e. Molarity (M) =
Volume of solution in litre (V)
For example, 1 M (molar) NaOH solution means that 1 mol of NaOH (40g) is dissolved in one litre of solution.
(vii) Molality (m): It is defined as the number of moles of the solute present per kilogram (kg) of the solvent.
Number of moles of solute
i.e. Molality (m) =
Mass of solvent in kg
For example, 1 molal (m) solution of KCl means that 1 mol (74.5 g) of KCl is dissolved in 1 kg of water.
Among the different methods for expressing the concentration of solution, mass percentage, ppm (in
terms of mass), mole fraction and molality are temperature independent; while molarity, mass by volume
percentage and volume percentage are temperature dependent. This is because volume of solution depends
on temperature and the mass does not.
SOLUBILITY
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent
at a particular temperature. It depends upon the nature of solute, nature of the solvent, temperature and
pressure.
Solubility of a Solid in a Liquid
It is observed that polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents.
In general, a +solute dissolve in a solvent if the intermolecular interactions are similar in the two or the
general principle related to solubility is that “like dissolves like”.
Saturated and Unsaturated solutions
When a solid solute is added to the solvent, some solute dissolves and its concentration increases in
solution. This process is known as dissolution. Some solute particles in solution collide with the solid solute
particles and get separated out of solution. This process is known as crystallisation. After sometime, the rate
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of dissolution and crystallization becomes equal and a dynamic equilibrium is reached. At this stage the
concentration of solute in the solution remain constant and such a solution is called saturated solution.
A solution in which no more solute can be dissolved at the same temperature and pressure is called a
saturated solution. For such a solution, the concentration of the solution is equal to its solubility.
A solution in which more solute can be dissolved at the same temperature is called an unsaturated solution.
Effect of temperature on the solubility of a solid in a liquid
The solubility of a solid in a liquid mainly depends on temperature. Since the dissolution of a solid in a
liquid is an equilibrium process, it should follow Le Chatelier’s Principle. In general, in a nearly saturated
solution, if the dissolution process is endothermic (Δsol H > 0), the solubility should increase with rise in
temperature and if it is exothermic (Δsol H > 0) the solubility should decrease with temperature.
Effect of pressure on the solubility of a solid in a liquid
Since solids and liquids are highly incompressible, pressure does not have any significant effect on
solubility of solids in liquids.
Solubility of a Gas in a Liquid
Solubility of gases in liquids is greatly affected by pressure and temperature. The solubility of a gas
increases with increase of pressure.
A quantitative relation between pressure and solubility of a gas in a liquid was first given by Henry,
which is known as Henry’s law. “The law states that at a constant temperature, the solubility of a gas in a
liquid is directly proportional to the pressure of the gas”.
Or, “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas
(χ) in the solution” and is expressed as:
p = KH χ
Here KH is the Henry’s law constant. The value of KH depends on the nature of the gas and
temperature. As the value of KH increases, the solubility of the gas in the liquid decreases.
A graph of partial pressure (p) of the gas against mole fraction (χ) of the gas in solution is a straight
line as follows. The slope of the graph gives the value of KH.
Vapour pressure (p)

Mole fraction (χ)

As the temperature increases solubility of a gas in a liquid decreases. It is due to this reason that
aquatic species are more comfortable in cold water rather than in warm water.
Applications of Henry’s law
1. To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high
pressure.
2. A medical condition known as bends in scuba divers. To avoid bends, the cylinders used by scuba
divers are filled with air diluted with helium (The composition of the air in the cylinders used by scuba
divers is 32.1% oxygen, 56.2% nitrogen and 11.7% helium).
3. A medical condition known as anoxia in people living at high altitudes and in mountaineers or
climbers.

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Effect of Temperature on the solubility of a gas in a liquid
Solubility of gases in liquids decreases with rise in temperature. When dissolved, the gas molecules
are present in liquid phase. So the process of dissolution can be considered similar to condensation, which is
exothermic. Hence solubility decreases with increase of temperature.
Vapour Pressure of a liquid
In a liquid, the molecules with higher energy are escaped to vapour phase. This process is called
evaporation. As the density of the vapour increases, the molecules collide with each other and so their
energy decreases and returns to the liquid state. This process is called condensation.
After some time, the rate of evaporation becomes equal to rate of condensation and the two
processes attain equilibrium. At this condition, the pressure exerted by the vapour is called vapour pressure.
So vapour pressure is defined as the pressure exerted by the vapour in equilibrium with its own liquid. It
depends on the nature of the liquid and the temperature. As the temperature increases, the vapour pressure
also increases.
Vapour Pressure of Liquid Solutions
In liquid solutions, the solvent is always a liquid. The solute can be a gas, a liquid or a solid. Generally,
the liquid solvent is volatile. The solute may or may not be volatile. Based on the volatility of solute, the
vapour pressure of the solution is greater or less than that of the solvent.
Vapour Pressure of Liquid-Liquid Solutions – Raoult’s Law
A quantitative relationship between the vapour pressure and mole fraction of solute in a solution was
first given by a French chemist Francois Marte Raoult (F M Raoult) and it is known as 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.
Consider a binary solution of two volatile liquids 1 and 2. Let p 1 and p2 be the partial vapour pressures
of the two components 1 and 2 respectively and ptotal be the total vapour pressure. Let χ1 and χ2 be the mole
fractions of the two components 1 and 2 respectively.
Then according to Raoult’s law,
for component 1, p1 ∝ χ1
or, p1 = 𝑝10 . χ1
and for component 2, p2 ∝ χ2
or, p2 = 𝑝20 . χ2
Where p10 and p20are the vapour pressures of the pure components 1 & 2 respectively.
According to Dalton’s law of partial pressures, the total pressure (ptotal ) will be the sum of the partial
pressures of the components of the solution.
So, ptotal = p1 + p2
Substituting the values of p1 and p2, we get
ptotal = χ1. 𝑝10 + χ2 . 𝑝20
= (1 – χ2 )𝑝10 + χ2 . 𝑝20
Or, 𝐩𝐭𝐨𝐭𝐚𝐥 = 𝒑𝟎𝟏 + (𝒑𝟎𝟐 – 𝒑𝟎𝟏 ) 𝛘𝟐
Plots of p1 or p2 against the mole fractions χ1 and χ2 give
straight lines (I and II). Similarly the plot of ptotal versus χ2
(line III) is also linear.
The composition of vapour phase in equilibrium
with the solution is determined from the partial pressures
of the components. If y1 and y2 are the mole fractions of
the components 1 and 2 respectively in the vapour phase then, using Dalton’s law of partial pressures:

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 p1 = y1 ptotal and p2 = y2 ptotal
In general, pi = yi ptotal
Raoult’s Law as a special case of Henry’s Law
According to Raoult’s law, the vapour pressure of a volatile component in a solution is given by
P1 = χ1. p10 . According to Henry’s law, solubility of a gas in a liquid is given by p = KH χ.
If we compare the equations for Raoult’s law and Henry’s law, we can see that the partial pressure of
the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality
constant KH differs from p10 . Thus, Raoult’s law becomes a special case of Henry’s law in which KH becomes
equal to p10 .
Ideal and non-ideal solutions
Liquid – liquid solutions can be classified into ideal and non-ideal solutions on the basis of Raoult’s law.
1. Ideal solutions:
These are solutions which obey Raoult’s law over the entire range of concentration. For such
solutions, the vapour pressure of a component in the solution is equal to the vapour pressure of the pure
component multiplied by its mole fraction in the solution.
i.e. p1 = 𝑝10 . χ1 and p2 = 𝑝20 . χ2
For such solutions, no heat change occurs during the mixing of the pure components [i.e. ∆mixH = 0].
Also, the volume of the solution is equal to the sum of the volumes of the components. Or, there is no change
in the volume of the solution during mixing the components [i.e. ∆mixV = 0]
Thus for an ideal solution, p1 = 𝒑𝟎𝟏 . 𝝌𝟏 , p2 = 𝒑𝟎𝟐 . 𝝌𝟐 , ∆mixH = 0 and ∆mixV = 0
Ideal behaviour can be explained by considering two components A and B. In pure components, the
inter molecular attractive interactions will be of types A-A and B-B. In solution, in addition to these two
interactions, A-B type of interaction will also be present. If the A-A and B-B interactions are nearly equal to
the A-B interaction, the solution behaves ideally. i.e. solute-solute interactions and solvent-solvent
interactions are nearly equal to solute-solvent interaction.
A perfectly ideal solution is rare. But some solutions are nearly ideal in behaviour.
E.g. solutions of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene etc.
2. Non-ideal solutions:
These are solutions which do not obey Raoult’s law over the entire range of concentration. The vapour
pressure of such a solution is either higher or lower than that predicted by Raoult’s law. If it is higher, the
solution shows positive deviation from Raoult’s law and if it is lower, it shows negative deviation from
Raoult’s law.
(i) Solutions which show positive deviation from Raoult’s law:
For such solutions, p1 > 𝒑𝟎𝟏 . 𝛘𝟏 , p2 > 𝒑𝟎𝟐 . 𝛘𝟐 , ∆mixH > 0 and ∆mixV > 0
Here A-B interactions are weaker than A-A and B-B interactions. i.e., in this case solute-solvent
interactions are weaker than solute-solute and solvent-solvent interactions. So more molecules are escaped
to vapour phase and hence the vapour pressure of the solution increases.
E.g. solutions of ethanol and acetone, acetone and CS2, acetone and CCl4 etc.
(ii) Solutions which show negative deviation from Raoult’s law:
For such solutions, p1 < 𝒑𝟎𝟏 𝛘𝟏 , p2 < 𝒑𝟎𝟐 𝛘𝟐 , ∆mixH < 0 and ∆mixV < 0
Here A-B interactions are stronger than A-A and B-B interactions. i.e. solute-solvent interactions are
stronger than solute-solute interaction and solvent-solvent interaction. So number of molecules escaped to
vapour phase decreases and hence the vapour pressure of the solution decreases.
E.g. solution of phenol and aniline, chloroform and acetone etc.
The plots of vapour pressure curves for ideal solution, solutions which show positive deviation and
negative deviation from Raoult’s law are as follows:

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 Solutions showing Positive Solutions showing Negative
Ideal Solutions deviation from Raoult’s law deviation from Raoult’s law

Azeotropes [Constant Boiling Mixtures]

These are binary mixtures having the same composition in liquid and vapour phase and boil at a
constant temperature. For such solutions, it is not possible to separate the components by fractional
distillation. There are two types of azeotropes: minimum boiling azeotrope and maximum boiling azeotrope.
The solutions which show a large positive deviation from Raoult’s law form minimum boiling
azeotrope at a particular composition. E.g. 95% aqueous ethanol solution by volume at 78.20C [351.35 K].
The solutions which show large negative deviation from Raoult’s law form maximum boiling azeotrope
at a particular composition. E.g. a mixture of 68% Nitric acid and 32% water by mass forms a maximum
boiling azeotrope at 393.5 K.
Vapour Pressure of Solutions of Solids in Liquids
The vapour pressure of a liquid is the pressure exerted by the vapour in equilibrium with its own
liquid. If a non-volatile solute is added to a pure solvent, the vapour pressure of the resulting solution is
always lower than that of the pure solvent. This is because in a pure solvent, there are only solvent
molecules, which can vapourise. But when a non-volatile solute is added to the solvent, a fraction of the
surface is occupied by solute molecules. So the number of solute molecules passing to the vapour phase
decreases and hence the vapour pressure also decreases. The decrease in the vapour pressure of solvent
depends on the quantity of non-volatile solute present in the solution and not on its nature.
For such a solution the Raoult’s law can be stated as, for any solution, the partial vapour pressure of each
volatile component in the solution is directly proportional to its mole fraction in solution.
Consider a binary solution containing a solvent 1 and solute 2. Since the solute is non-volatile, only the
solvent molecules are present in vapour phase and contribute to vapour pressure.
Let p1 be the vapour pressure of the solvent, χ1 be its mole fraction, p10 be its vapour pressure in the
pure state. Then according to
Raoult’s law, p1 ∝ χ1
or, p1 = 𝑝10 χ1
A graph between the vapour pressure and the mole fraction of the solvent is linear as follows:

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 COLLIGATIVE PROPERTIES AND DETERMINATION OF MOLARMASS
The properties which depend only on the relative number of solute particles and not on their nature
are called Colligative properties. The important colligative properties are: Relative lowering of Vapour
pressure, Elevation of Boiling point, Depression of Freezing point and Osmotic Pressure.
1. Relative lowering of Vapour Pressure
When a non-volatile solute is added to a pure solvent, the vapour pressure (V.P) of the resulting
solution is lower than that of the pure solvent. The difference between the vapour pressure of pure solvent
and that of the solution is called lowering of vapour pressure (∆p).
Consider a binary solution containing a non-volatile solute 2 dissolved in a solvent 1. Let 𝑝10 be the
vapour pressure of pure solvent 1 and p1 be the vapour pressure of solution [V.P of solution = V.P of the
solvent in the solution, since the solute is non-volatile].
Then according to Raoult’s law, p1 = 𝑝10 . χ1
The lowering of vapour pressure of the solvent (∆p) = 𝑝10 – p1
= 𝑝10 – 𝑝10 χ1
Or, ∆p = 𝑝10 (1 - χ1)
But χ1 + χ2 = 1. Therefore 1 - χ1 = χ2
So ∆p = 𝑝10 . χ2
∆p
Or, = χ2 , the mole fraction of the solute.
𝑝10
∆p
Where is called relative lowering of vapour pressure.
𝑝10
It is defined as the ratio of the lowering of vapour pressure to the vapour pressure pure solvent.
n2
But χ2 =
n1 + n2
Where n1 and n2 are the number of moles of solvent and solute respectively.
For dilute solutions, n2 << n1 and hence n2 in the denominator can be neglected.
n2
So, χ2 =
n1
∆p n2
Therefore, =
𝑝10 n1
w2
∆p ⁄M
2
Or, = w1
𝑝10 ⁄M
1
∆𝐩 𝐰𝟐 𝐌𝟏
Or, = x
𝒑𝟎𝟏 𝐰𝟏 𝐌𝟐
Where w1 and w2 are the masses and M1 and M2 are the molar masses of solvent and solute respectively.
2. Elevation of Boiling Point (∆Tb)
When a liquid is heated, its vapour pressure increases. At a particular temperature, the vapour pressure
becomes equal to the external pressure (atmospheric pressure). This temperature is called boiling point of
the liquid. So boiling point is the temperature at which its vapour pressure becomes equal to the atmospheric
pressure.
When a non-volatile solute is added to a pure solvent, the boiling point of the resulting solution is always
greater than that of the pure solvent. The difference between the boiling point of solution (∆T b) and that of
the pure solvent (Tb0 ) is called elevation of boiling point (Tb).
i.e. ∆Tb = Boiling point of solution – Boiling point of pure solvent
Or, ∆Tb = Tb – Tb0

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 If we plot a graph between the vapour pressure and temperature, we get the following graphs for the
pure solvent and the solution.

For dilute solutions, the elevation of boiling point is directly proportional to molality (m).
i.e. ∆Tb α m
or, ∆Tb = Kb.m
Where Kb is a constant called Boiling Point Elevation Constant or Molal Elevation Constant or Ebullioscopic
Constant. It is defined as the elevation of boiling point for 1 molal solution (i.e. it is the increase in boiling
point of a solution containing 1 mol of a solute in 1 kg of the solvent).
The unit of Kb is K kg/mol. For water, Kb = 0.52K kg/mol.
𝑤2 𝑥 1000
But molality m =
𝑀2 𝑥 𝑤1
𝐾𝑏 𝑥 𝑤2 𝑥 1000
Therefore, ∆Tb =
𝑀2 𝑥 𝑤1
𝟏𝟎𝟎𝟎 𝐊 𝐛 .𝐰𝟐
Or, ∆Tb =
𝐌𝟐 .𝐰𝟏
Where w1 = mass of solvent, w2 = mass of solute, M2 = molar mass of solute. By using this equation, we can
calculate the molar mass of unknown solute.
3. Depression of Freezing point (∆Tf)
Freezing point is the temperature at which the solid phase and liquid phase of a substance has the
same vapour pressure.
According to Raoult’s law, when a non-volatile solute is added to a pure solvent, its vapour pressure
decreases. So the freezing point (f.p) of the solution is less than that of the pure solvent. The difference
between the freezing point (f.p) of pure solvent (Tf0 ) and that of the solution (Tf) is called depression of
freezing point (∆Tf).
i.e. ∆Tf = Tf0 - Tf
The vapour pressure – temperature graph representing
the freezing point of pure solvent and solution is as
follows:

For dilute solutions, it is found that the depression

of freezing point (∆Tf) is directly proportional to molality
(m) of the solution.
Thus ∆Tf α m
Or, ∆Tf = Kf.m
Where Kf is a constant called Freezing point
depression constant or Molal depression constant or
Cryoscopic constant. It is defined as the depression of
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freezing point for 1 molal solution. i.e. it is the depression in freezing point of a solution prepared by
dissolving 1 mol of solute in 1 kg of solvent.
The unit of Kf is K kg/mol. For water, Kf = 1.86 K kg/mol.
w2 x 1000
We know that molality, m =
M2 x w1
Kf x w2 x 1000
Therefore, ∆Tf =
M2 x w1
𝟏𝟎𝟎𝟎 𝐊 𝐟 .𝐰𝟐
Or, ∆Tf =
𝐌𝟐 .𝐰𝟏
Where w1 = mass of solvent, w2 = mass of solute, M2 = molar mass of solute. By using this equation, we
can calculate the molar mass of an unknown solute.
4. Osmosis and Osmotic Pressure
Osmosis is the process of flow of solvent molecules from pure solvent to the solution through a
semi-permeable membrane. Or, it is the flow of solvent molecules from lower concentration side to a
higher concentration side through a semi-permeable membrane (SPM).
A membrane that allows the passage of only solvent molecules is called a semi-permeable
membrane. E.g. egg membrane, parchment, pig’s bladder, all animal and plant membrane. Cellulose
acetate is an example for artificial semi-permeable membrane.
Osmotic pressure is defined as the excess pressure that must be applied on solution side to stop
osmosis. Or, it is the pressure that just stops the flow of solvent molecules. It is denoted by π. It is a
colligative property, since it depends on the number of solute molecules and not on their nature.
For dilute solutions, osmotic pressure is proportional to the molarity (C) and temperature (T).
i.e. π α CT
Or, π = CRT
Here R is the universal gas constant (R= 0.0821 Latm/K/mol or R = 0.083 Lbar/K/mol).
But C = n2/V, the concentration of the solution.
n2 RT
Therefore, π =
V
Or, πV = n2RT
w2 RT
Or, πV =
M2
Where V is the volume of the solution, w2 is the mass of solute and M2 is the molar mass of solute.
Thus by knowing all other values, we can calculate the molar mass of the unknown solute by the equation:
𝐰𝟐 𝐑𝐓
M2 =
𝛑𝐕

Advantages of osmotic pressure measurement over other colligative property measurement

1. Osmotic pressure measurement can be done at room temperature.
2. Here molarity of the solution is used instead of molality, which can be determined easily.
3. The magnitude of osmotic pressure is large even for very dilute solutions.
4. This method can be used for the determination of molar masses of Biomolecules (which are generally
not stable at higher temperatures) and for polymers (which have poor solubility).
Examples for osmosis:
a) Raw mango placed in concentrated salt solution loses water and shrink.
b) Wilted flowers revive when placed in fresh water
c) Blood cells collapse when suspended in saline water.
d) The preservation of meat by salting and fruits by adding sugar protect against bacterial action. Through
the process of osmosis, a bacterium on salted meat or candid fruit loses water, shrinks and dies.

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Isotonic, hypertonic and hypotonic solutions
Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.
When such solutions are separated by a semi-permeable membrane, no osmosis occurs.
For e.g. our blood cells are isotonic with 0.9% (mass/volume) sodium chloride solution, called normal
saline solution. So it is safe to inject intravenously.
A solution having higher osmotic pressure than another is called hypertonic solution. While a solution
having lower osmotic pressure than another is called hypotonic solution.
If we place our blood cells in a solution containing more than 0.9% (mass/volume) sodium chloride
solution, water will flow out of the cells and they would shrink. On the other hand, if they are placed in a
solution containing less than 0.9% (mass/volume) NaCl, water will flow into the cells and they would swell.
Reverse osmosis and water purification
The direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to
the solution side. Now the pure solvent flows out of the solution through the semi permeable membrane.
This phenomenon is called reverse osmosis. It is used in desalination of sea water.
When pressure more than osmotic pressure is applied, pure water is squeezed out of the sea water
through the membrane. Commonly used semi-permeable membrane is cellulose acetate. A schematic
representation of reverse osmosis is as follows:

Reverse osmosis is also used in the purification of water.

ABNORMAL MOLARMASS
The molar mass obtained by colligative property measurement is incorrect, if there is association or
dissociation of particles. Such molar masses are called abnormal molar masses.
For e.g. acetic acid dimerises in benzene due to hydrogen bonding. So the number of molecules in
solution decreases and hence the colligative property decreases and molecular mass increases.
In order to correct the abnormal molar masses, van’t Hoff introduced a factor called van’t Hoff factor (i).
It is defined as:
Normal molar mass
i=
Abnormal molar mass
Observed colligative property
Or, i =
Calculated colligative property
Number of moles of particles after association/dissociation
Or, i=
Number of moles of particles before association/dissociation
In the case of association, the value of i < 1 and in dissociation, the value of i > 1.
Thus for NaCl, i =2, for K2SO4, i = 3, for CaCl2, i = 3 and for acetic acid in benzene, i = ½
Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows:
∆𝐩 𝐰 𝐌
1. Relative lowering of vapour pressure, 𝒑𝟎 = 𝑖. 𝐰𝟐 x 𝐌𝟏
𝟏 𝟏 𝟐
2. Elevation of Boiling point (∆Tb) = i.Kb.m
3. Depression of freezing point (∆Tf) = i.Kf.m
4. Osmotic Pressure (π) = i.CRT

Solutions - Notes by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 10

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