After studying this Unit, you will be

able to

types of solutions;

in different units;

Raoult?s law;

non-ideal solutions;

from Raoult?s law;

solutions and correlate these with

molar masses of the solutes;

properties exhibited by some

solutes in solutions.

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. For example, the properties of

brass (mixture of copper and zinc) are quite different

from those of German silver (mixture of copper, zinc

and nickel) or bronze (mixture of copper and tin);

1 part per million (ppm) of fluoride ions in water

prevents tooth decay, while 1.5 ppm causes the tooth

to become mottled and high concentrations of fluoride

ions can be poisonous (for example, 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. This will be followed by

studying the properties of the solutions, like vapour

pressure and colligative properties. We will begin with

types of solutions and then various alternatives in

which concentrations of a solute can be expressed in

liquid solution.

Solutions

Objectives

2.1 Types of

Solutions

2

Unit

2

components. By homogenous mixture we mean that its composition

and properties are uniform throughout the mixture. Generally, the

Solvent determines the physical state in which solution exists. One or

more components present in the solution other than solvent are called

Chemistry 34

Liquid Gas Chloroform mixed with nitrogen gas

Solid Gas Camphor in nitrogen gas

Liquid Liquid Ethanol dissolved in water

Solid Liquid Glucose dissolved in water

Liquid Solid Amalgam of mercury with sodium

Solid Solid Copper dissolved in gold

consisting of two components). Here each component may be solid,

liquid or in gaseous state and are summarised in Table 2.1.

Composition of a solution can be described by expressing its

concentration. The latter can be expressed either qualitatively or

quantitatively. For example, qualitatively we can say that the solution

is dilute (i.e., relatively very small quantity of solute) or it is concentrated

(i.e., relatively very large quantity of solute). But in real life these kinds

of description can add to lot of confusion and thus the need for a

quantitative description of the solution.

There are several ways by which we can describe the concentration

of the solution quantitatively.

a solution is defined as:

Mass % of a component

100

Total mass of the solution

(2.1)

For example, if a solution is described by 10% glucose in water

by mass, it 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. For example, commercial bleaching solution contains

3.62 mass percentage of sodium hypochlorite in water.

100

Total volume of solution

(2.2)

2.2 Expressing

Concentration

of Solutions

35 Solutions

For example, 10% ethanol solution in water means that 10 mL of

ethanol is dissolved in water such that the total volume of the

solution is 100 mL. Solutions containing liquids are commonly

ethylene glycol, an antifreeze, is used in cars for cooling the engine.

At this concentration the antifreeze lowers the freezing point of

water to 255.4K (?17.6°C).

commonly used in medicine and pharmacy is mass by volume

percentage. It is the mass of solute dissolved in 100 mL of the

solution.

and is defined as:

Parts per million =

×10

Total number of parts of all components of the solution

(2.3)

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

water. The concentration of pollutants in water or atmosphere is

It is defined as:

Mole fraction of a component =

Number of moles of the component

For example, in a binary mixture, if the number of moles of A and

+

A

A B

(2.5)

For a solution containing i number of components, we have:

+ + +

i

1 2 i

.......

i

i

(2.6)

It can be shown that in a given solution sum of all the mole

fractions is unity, i.e.

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.

Chemistry 36

Assume that we have 100 g of solution (one can start with any amount of

solution because the results obtained will be the same). Solution will

contain 20 g of ethylene glycol and 80 g of water.

20 g

62 g mol

= 0.322 mol

80 g

18 g mol

= 4.444 mol

=

+

2 6 2

glycol

2 6 2 2

moles of C H O

x

moles of C H O moles of H O

=

+

0.322mol

4.444 mol

0.932

0.322 mol 4.444 mol

Mole fraction of water can also be calculated as: 1 ? 0.068 = 0.932

Example 2.1

dissolved in one litre (or one cubic decimetre) of solution,

Molarity

Volume of solution in litre

(2.8)

0.25 mol of NaOH has been dissolved in one litre (or one cubic decimetre).

solution.

5 g

40 g mol

= 0.125 mol

Using equation (2.8),

Molarity =

450 mL

= 0.278 M

Solution

Solution

37 Solutions

Example 2.3

Solution

solute per kilogram (kg) of the solvent and is expressed as:

Molality (m) =

Moles of solute

Mass of solvent in kg

(2.9)

1 mol (74.5 g) of KCl is dissolved in 1 kg of water.

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 is a function of

temperature. This is because volume depends on temperature

and the mass does not.

Solubility of a substance is its maximum amount that can be dissolved

in a specified amount of solvent. It depends upon the nature of solute

and solvent as well as temperature and pressure. Let us consider the

effect of these factors in solution of a solid or a gas in a liquid.

2.3 Solubility

2.5 g

kg of benzene

=

75 g

Intext Questions

mass in carbon tetrachloride.

500 mL.

0.25 molal aqueous solution.

Chemistry 38

Every solid does not dissolve in a given liquid. While sodium chloride

and sugar dissolve readily in water, naphthalene and anthracene do

not. On the other hand, naphthalene and anthracene dissolve readily in

benzene but sodium chloride and sugar do not. It is observed that

polar solutes dissolve in polar solvents and non polar solutes in nonpolar

solvents. In general, a solute dissolves in a solvent if the

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

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.

At this stage the concentration of solute in solution will remain

constant under the given conditions, i.e., temperature and pressure.

Similar process is followed when gases are dissolved in liquid solvents.

Such a solution in which no more solute can be dissolved at the same

the same temperature. The solution which is in dynamic equilibrium

with undissolved solute is the saturated solution and contains the

maximum amount of solute dissolved in a given amount of solvent.

Thus, the concentration of solute in such a solution is its solubility.

Earlier we have observed that solubility of one substance into

another depends on the nature of the substances. In addition to these

variables, two other parameters, i.e., temperature and pressure also

control this phenomenon.

The solubility of a solid in a liquid is significantly affected by temperature

changes. Consider the equilibrium represented by equation 2.10. This,

decrease. These trends are also observed experimentally.

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.

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

39 Solutions

temperature. The solubility of gases increase with increase of pressure.

For solution of gases in a solvent, consider a system as shown in

Fig. 2.1 (a). The lower part is solution and the upper part is gaseous

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 [Fig. 2.1 (b)]. 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 to give a

quantitative relation between

pressure and solubility of a gas

in a solvent which is known as

contemporary of Henry, also

concluded independently that the

solubility of a gas in a liquid

solution is a function of partial

pressure of the gas. If we use the

mole fraction of a gas in the

solution as a measure of its

graph between partial pressure of the gas versus mole

fraction of the gas in solution, then we should get a plot

of the type as shown in Fig. 2.2.

function of the nature of the gas.

It is obvious from equation (2.11) that higher the

of the gas in the liquid. It can be seen from Table 2.2

of temperature indicating that the solubility of gases

Chemistry 40

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.

Henry?s law finds several applications in industry and explains some

biological phenomena. Notable among these are:

bottle is sealed under high pressure.

while breathing air at high pressure underwater. Increased pressure

increases the solubility of atmospheric gases in blood. When the

divers come towards surface, the pressure gradually decreases. This

releases the dissolved gases and leads to the formation of bubbles

of nitrogen in the blood. This blocks capillaries and creates a medical

The solubility of gas is related to the mole fraction in aqueous solution.

The mole fraction of the gas in the solution is calculated by applying

Henry?s law. Thus:

H

=

0.987bar

76,480 bar

mol

=

55.5

=

1 mol

?

= 0.716 mmol

Example 2.4

Solution

He 293 144.97

Argon 298 40.3

Methane 298 0.413

Vinyl chloride 298 0.611

41 Solutions

To avoid bends, as well as, the toxic effects of high concentrations

of nitrogen in the blood, the tanks used by scuba divers are filled

with air diluted with helium (11.7% helium, 56.2% nitrogen and

32.1% oxygen).

the ground level. This leads to low concentrations of oxygen in the

blood and tissues of people living at high altitudes or climbers. Low

blood oxygen causes climbers to become weak and unable to think

Solubility of gases in liquids decreases with rise in temperature. When

dissolved, 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. We have learnt in the last Section that

dissolution process involves dynamic equilibrium and thus must follow

Le Chatelier?s Principle. As dissolution is an exothermic process, the

solubility should decrease with increase of temperature.

Liquid solutions are formed when solvent is a liquid. The solute can be

a gas, a liquid or a solid. Solutions of gases in liquids have already

been discussed in Section 2.3.2. In this Section, we shall discuss the

solutions of liquids and solids in a liquid. Such solutions may contain

one or more volatile components. Generally, the liquid solvent is volatile.

The solute may or may not be volatile. We shall discuss the properties

of only binary solutions, that is, the solutions containing two

components, namely, the solutions of (i) liquids in liquids and (ii) solids

in liquids.

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 an equilibrium would be

established between vapour phase and the liquid phase. Let the total

vapour pressures of the two components 1 and 2 respectively. These

components 1 and 2 respectively.

The French chemist, Francois Marte Raoult (1886) gave the

quantitative relationship between them. The relationship is known as

2.4 Vapour

Pressure of

Liquid

Solutions

Intext Questions

Chemistry 42

Thus, for component 1

temperature.

Similarly, for component 2

partial pressures of the components of the solution and is given as:

0

Following conclusions can be drawn from equation (2.16).

(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 2.

(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.

linear plot as shown in Fig. 2.3. These

lines (I and II) pass through the points

0

and the

component 1 is less volatile than

The composition of vapour phase in

equilibrium with the solution is determined

by the partial pressures of the components.

43 Solutions

components 1 and 2 respectively in the vapour phase then, using Dalton?s

law of partial pressures:

In general

at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate

the vapour pressure of the solution prepared by mixing 25.5 g of

component in vapour phase.

40 g

25.5 g

Total number of moles = 0.47 + 0.213 = 0.683 mol

0.47 mol

0.683 mol

= 0.688

Using equation (2.16),

= 200 + 147.9 = 347.9 mm Hg

Example 2.5

Solution

Chemistry 44

According to Raoult?s law, the vapour pressure of a volatile component

liquid, one of the components is so volatile that it exists as a gas and

we have already seen that its solubility is given by Henry?s law which

states that

If we compare the equations for Raoult?s law and Henry?s law, it

can be seen that the partial pressure of the volatile component or gas

is directly proportional to its mole fraction in solution. Only the

Another important class of solutions consists of solids dissolved in

liquid, for example, sodium chloride, glucose, urea and cane sugar in

water and iodine and sulphur dissolved in carbon disulphide. Some

physical properties of these solutions are quite different from those of

pure solvents. For example, vapour pressure. We have learnt in Unit 5,

Class XI, that liquids at a given temperature vapourise and under

equilibrium conditions the pressure exerted

by the vapours of the liquid over the liquid

phase is called vapour pressure [Fig. 2.4 (a)].

In a pure liquid the entire surface is

occupied by the molecules of the liquid. If a

non-volatile solute is added to a solvent to

give a solution [Fig. 2.4.(b)], the vapour

pressure of the solution is solely from the

solvent alone. This vapour pressure of the

solution at a given temperature is found to

be lower than the vapour pressure of the

pure solvent at the same temperature. In

the solution, the surface has both solute and

solvent molecules; thereby the fraction of the

surface covered by the solvent molecules gets

reduced. Consequently, the number of

solvent molecules escaping from the surface

is correspondingly reduced, thus, the vapour

pressure is also reduced.

The decrease in the vapour pressure of solvent depends on the

quantity of non-volatile solute present in the solution, irrespective of

its nature. For example, decrease in the vapour pressure of water by

adding 1.0 mol of sucrose to one kg of water is nearly similar to that

produced by adding 1.0 mol of urea to the same quantity of water at

the same temperature.

In a binary solution, let us denote the solvent by 1 and solute by

2. When the solute is non-volatile, only the solvent molecules are

45 Solutions

in the pure state. Then according to

Raoult?s law

0

The proportionality constant is equal

1

A plot between the vapour pressure and

the mole fraction of the solvent is linear

(Fig. 2.5).

Liquid-liquid solutions can be classified into ideal and non-ideal

solutions on the basis of Raoult?s law.

The solutions which obey Raoult?s law over the entire range of

two other important properties. The enthalpy of mixing of the pure

components to form the solution is zero and the volume of mixing is

also zero, i.e.,

It means that no heat is absorbed or evolved when the components

are mixed. Also, the volume of solution would be equal to the sum of

volumes of the two components. At molecular level, ideal behaviour of

the solutions can be explained by considering two components A and

B. In pure components, the intermolecular attractive interactions will

be of types A-A and B-B, whereas in the binary solutions in addition

to these two interactions, A-B type of interactions will also be present.

If the intermolecular attractive forces between the A-A and B-B are

nearly equal to those between A-B, this leads to the formation of ideal

solution. A perfectly ideal solution is rare but some solutions are nearly

ideal in behaviour. Solution of n-hexane and n-heptane, bromoethane

and chloroethane, benzene and toluene, etc. fall into this category.

When a solution does not obey Raoult?s law over the entire range of

of such a solution is either higher or lower than that predicted by

law. The plots of vapour pressure as a function of mole fractions

for such solutions are shown in Fig. 2.6.

The cause for these deviations lie in the nature of interactions at the

molecular level. In case of positive deviation from Raoult?s law, A-B

interactions are weaker than those between A-A or B-B, i.e., in this case

the intermolecular attractive forces between the solute-solvent molecules

are weaker than those between the solute-solute and solvent-solvent

molecules. This means that in such solutions, molecules of A (or B) will

find it easier to escape than in pure state. This will increase the vapour

2.5 Ideal and Nonideal

Solutions

Chemistry 46

pressure and result in positive deviation. Mixtures 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. Due to weakening of

interactions, the solution shows positive deviation from Raoult?s law

[Fig. 2.6 (a)]. In a solution formed by adding carbon disulphide to

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.

In case of negative deviations from Raoult?s law, 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. An example of this type is a 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. Similarly, a 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.

This decreases the escaping tendency of molecules for each

component and consequently the vapour pressure decreases resulting

in negative deviation from Raoult?s law [Fig. 2.6. (b)].

having the same composition in liquid and vapour phase and boil at

a constant temperature. In such cases, it is not possible to separate the

components by fractional distillation. There are two types of azeotropes

Raoult?s law form minimum boiling azeotrope at a specific composition.

47 Solutions

For example, ethanol-water mixture (obtained by fermentation of sugars)

on fractional distillation gives a solution containing approximately 95%

by volume of ethanol. Once this composition, known as azeotrope

composition, has been achieved, the liquid and vapour have the same

composition, and no further separation occurs.

The solutions that show large negative deviation from Raoult?s law

form maximum boiling azeotrope at a specific composition. Nitric acid

and water is an example of this class of azeotrope. This azeotrope has

the approximate composition, 68% nitric acid and 32% water by mass,

with a boiling point of 393.5 K.

2.6 Colligative

Properties and

Determination

of Molar Mass

We have learnt in Section 2.4.3 that the vapour pressure of solution

decreases when a non-volatile solute is added to a volatile solvent.

There are many properties of solutions which are connected with this

decrease of vapour pressure. These are: (1) relative lowering of vapour

pressure of the solvent (2) depression of freezing point of the solvent

(3) elevation of boiling point of the solvent and (4) osmotic pressure of

means to bind). In the following Sections we will discuss these properties

one by one.

We have learnt in Section 2.4.3 that the vapour pressure of a solvent in

solution is less than that of the pure solvent. Raoult established that the

lowering of vapour pressure depends only on the concentration of the

solute particles and it is independent of their identity. The equation (2.20)

given in Section 2.4.3 establishes a relation between vapour pressure of

the solution, mole fraction and vapour pressure of the solvent, i.e.,

In a solution containing several non-volatile solutes, the lowering of the

vapour pressure depends on the sum of the mole fraction of different solutes.

Equation (2.24) can be written as

1

0

1

=

0

1 1

0

1

Intext Question

respectively, at 350 K . Find out the composition of the liquid mixture if total

vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Chemistry 48

The expression on the left hand side of the equation as mentioned

0

1 1

0

1

1 2

+

2

2

1 2

0

1 1

0

1

?

1

(2.27)

or

0

1 1

0

1