Solutions Class 12

 Solutions Class 12 

Homogenous mixture 

The homogenous mixture we mean that its composition and properties are uniform throughout the mixture. 


Solutions are homogeneous mixtures of two or more than two components. 


The component that is present in the largest quantity in the solution is known as solvent. Solvent determines the physical state in which solution exists.


The component present in a smaller quantity in the solution other than solvent is called the solute. 

 Binary solutions 

 Binary solutions  i.e., consisting of two components.

Types of Solutions

  • Gaseous Solutions
  • Liquid Solutions
  • Solid Solutions
Types of Solutions

Composition of a Solution

The composition of a solution can be described by expressing its concentration.  It can be expressed as 
i) qualitatively
ii)  quantitatively

i) Qualitatively
 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).

ii)  Quantitatively
There are several ways by which we can describe the concentration of the solution quantitatively.
a) Mass percentage (w/w)
b) Volume percentage (V/V)
c) Mass by volume percentage (w/V)
d)  Parts per million
e)  Mole fraction
f) Molarity
g) Molality

a) Mass percentage (w/w): The mass percentage of a component of a solution is defined as:
 Mass % of a component

b) Volume percentage (V/V): The volume percentage is defined as:

c) Mass by volume percentage (w/V): 

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

d)  Parts per million:

It is the parts of a component per million `(10^6)` parts of the solution.

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 `10^6`g (5.8 ppm) of sea water. 

Mass % of  `O_2 = \frac {6\times 10^{-3}}{1030}\times 100 = 5.8 \times 10^{-4} %` 

ppm of `O_2 =   \frac {6\times 10^{-3}}{1030} \times 10^6 = 5.8` ppm

e)  Mole fraction
The symbol for mole fraction is `x` and subscript used on the right hand side of `x` denotes the component. It is defined as: 

πŸ‘‰ In a binary mixture, if the number of moles of `A` and `B` are `n_A` and `n_B` respectively, the mole fraction of `A` will be

`x_A = \frac{n_A}{n_A + n_B}`

πŸ‘‰ For a solution containing  `i`  number of components, we have:

`x_i = \frac{n_i}{n_1 + n_2 + n_3 ..+ n_i} =  \frac{n_i}{\sum n_i}`

f) Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution

For 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.

g) Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as: 


πŸ‘‰The solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature. 
It depends upon the nature of solute and solvent as well as temperature and pressure. 

Solubility of a Solid in a Liquid

Every solid does not dissolve in a given liquid. 

πŸ‘‰ Polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. 

In general, a solute dissolves in a solvent if the intermolecular interactions are similar in the two or we may say like dissolves like.

πŸ‘‰ Sodium chloride and sugar dissolve readily in water
πŸ‘‰ Naphthalene and anthracene do not dissolve in water 

πŸ‘‰ Naphthalene and anthracene dissolve readily in benzene.
πŸ‘‰ Sodium chloride and sugar do not.


πŸ‘‰When a solid solute is added to the solvent, some solute dissolves and its concentration increases in the solution. This process is known as dissolution.

Saturated solution

πŸ‘‰Such a solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution.

Unsaturated solution

An unsaturated solution is one in which more solute can be dissolved at the same temperature 

Effect of temperature on solubility 

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

Effect of pressure 

Pressure does not have any significant effect on the solubility of solids in liquids. It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.

 Solubility of a Gas in a Liquid

Solubility of gases in liquids is greatly affected by pressure and temperature. The solubility of gases increase with increase of pressure

Consider a system as shown in Fig. (a). 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 [Fig. (b)]. This will increase the number of gaseous particles per unit volume over the solution.

 πŸ‘‰ Quantitative relation between pressure and solubility of a gas in a solvent which is known as Henry’s law.

Henry's Law

The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution. 


The partial pressure of the gas in vapour phase `(p)` is proportional to the mole fraction of the gas `(x)` in the solution.

`p = K_H  x`

Here `K_H`  is the Henry’s law constant. 

πŸ‘‰ Different gases have different `K_H`  values at the same temperature . This suggests that `K_H`  is a function of the nature of the gas. 

πŸ‘‰ It is obvious from equation `p = K_H  x` that higher the value of `K_H`  at a given pressure, the lower is the solubility of the gas in the liquid.

Vapour Pressure of Solutions

The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure 

Raoult’s law

For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. 

Thus, for component 1

`p_1 ∝ x_1`
and   `p_1 = p_1^0 x_1` 

where `p_1^0 ` is the vapour pressure of pure component 1 at the same temperature 

Similarly, for component 2

`p_2 = p_1^0 x_2` 

where `p_2^0 ` is the vapour pressure of pure component 2 at the same temperature 

According to Dalton’s law of partial pressures, the total pressure `( p_t )` 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:

`p_t = p_1 +p_2`

Substituting the values of `p_1`  and `p_2`, we get

`p_t  = x_1p_1^0 + x_2 p_2^0` 

`= (1 - x_2)p_1^0  + x_2 p_2^0`

` = p_1^0 + (p_2^0 - p_1^0)x_2` 

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 Fig. 

These lines (I and II) pass through the points for which `x_1`  and `x_2`  are equal to unity. Similarly, the plot (line III) of `p_t` versus `x_2`  is also linear (Fig. ). The minimum value of  `p_t`  is  `p_1^0` and the maximum value is `p_2^0` , assuming that component 1 is less volatile than component 2, i.e.,   `p_1^0 < p_2^0`

πŸ‘‰ 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_1p_t`
`p_2 = y_2p_t`

Raoult’s Law as a special case of Henry’s Law

In the solution of a gas in a liquid

According to  Raoult’s Law 
`p_1 = x_1p_1^0`

According to  Henry’s Law
`p = K_H x`

Only the proportionality constant `K_H` differs from `p_1^0` . Thus, Raoult’s law becomes a special case of Henry’s law in which `K_H = p_1^0` .

Vapour Pressure of Solutions of Solids in Liquids

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 present in vapour phase and contribute to vapour pressure.

Let `p_1` be the vapour pressure of the solvent, `x_1`  be its mole fraction, `p_1^0`  be its vapour pressure in the pure state. Then according to Raoult’s law 

`p_1 \propto x_1`

The proportionality constant is equal to the vapour pressure of pure solvent, `p_1^0` . 

A plot between the vapour pressure and the mole fraction of the solvent is linear.

Ideal Solutions

The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.

The ideal solutions have two other important properties.
πŸ‘‰The enthalpy of mixing of the pure components to form the solution is zero

`Delta_{mix} H = 0`

πŸ‘‰ The volume of mixing is also zero.

`Delta_{mix} V = 0`

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

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.

πŸ‘‰Examples: Solution of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene, etc. fall into this category.

Non-ideal Solutions 

When a solution does not obey Raoult’s law over the entire range of concentration, then it is called non-ideal solution.
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 exhibits positive deviation and if it is lower, it exhibits negative deviation from Raoult’s law. 

The cause for these deviations

Positive Deviation

The cause for these deviations lie in the nature of interactions at the molecular level.
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 pressure and result in positive deviation.
 Mixtures of ethanol and acetone behave in this manner.

Negative 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.

 Colligative Properties

πŸ‘‰The properties of solutions which depend on the number of solute particles and are independent of their chemical identity are called colligative properties.

Four important colligative propertie 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 the solution.

(1) Relative lowering of vapour pressure of the solvent 

Non-volatile solute in a solution.

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

`p_1 = x_1p_1^0`

The reduction in the vapour pressure of solvent  `(∆p_1)` is given as:

`∆p_1 = p_1^0 - p_1 = p_1^0 - p_1^0 x_1`

`= p_1^0(1-x_1)` 

`\Delta p_1 =  x_2 p_1^0`

Equation  can be written as

`\frac {\Delta p_1}{p_1^0} = \frac {p_1^0 - p_1}{p_1^0}  = x_2`

Relative lowering of vapour pressure and is equal to the mole fraction of the solute.

`\frac {p_1^0 - p_1}{p_1^0} = \frac {n_2}{n_1 +n_2}  (∵ x_2 = \frac {n_2}{n_1 +n_2})`  

Here `n_1` and `n_2` are the number of moles of solvent and solute respectively present in the solution. For dilute solutions `n_2 < < n_1`, hence neglecting `n_2` in the denominator we have

`frac {p_1^0 - p_1}{p_1^0} = \frac {n_2}{n_1}`


`frac {p_1^0 - p_1}{p_1^0} = \frac {w_2 \times M_1}{M_2\times w_1}`

From this equation , knowing all other quantities, the molar mass of solute `(M_2)` can be calculated.

Elevation of Boiling Point

The vapour pressure of a liquid increases with increase of temperature. It boils at the temperature at which its vapour pressure is equal to the atmospheric pressure. For example, water boils at `373.15 K` `(100° C)` because at this temperature the vapour pressure of water is `1.013` bar (1 atmosphere). 

The vapour pressure of the solvent decreases in the presence of a non-volatile solute. 

For 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 `1.013` 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 Fig. 

Let  `T_b^0`  be the boiling point of pure solvent and `T_b`  be the boiling point of solution.

πŸ‘‰ The increase in the boiling point  `∆T_b = T_b - T_b^0` is known as elevation of boiling point.

For dilute solutions the elevation of boiling point `(∆Tb)` is directly proportional to the molal concentration of the solute in a solution. Thus
 `∆Tb \propto m`
 `∆Tb = K_b m`

Here `m` (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, `K_b`  is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant).  The unit of `K_b`  is `K  kg  mol^{-1}`.


The phenomenon of the flow of solvent through a semi-permeable membrane from pure solvent (or dilute solution) to the solution (or concentrated solution) is called osmosis.  


Semipermeable Membranes (SPM)

Small solvent molecules, like water, can pass through these holes but the passage of bigger molecules like solute is hindered. Membranes having this kind of properties are known as semipermeable membranes (SPM).

Osmotic Pressure `(\Pi)`

The osmotic pressure of a solution is the excess pressure that must be applied to a solution to prevent osmosis, i.e., to stop the passage of solvent molecules through a semipermeable membrane into the solution.
Osmotic pressure is a colligative property as it depends on the number of solute molecules and not on their identity. 

πŸ‘‰ Osmotic pressure is proportional to the molarity, `C` of the solution at a given temperature `T`. Thus: 

`\Pi = CRT`

Here `Pi` is the osmotic pressure and `R` is the gas constant

`\Pi = (\frac {n_2}{V})RT`

Here `V` is volume of a solution in litres containing `n_2`  moles of solute. 

If `w_2`  grams of solute, of molar mass, `M_2`  is present in the solution, then  `n_2 = w_2 / M_2`  and we can write,

`\Pi V = \frac {w_2RT}{M_2}`

or `M_2= \frac {w_2RT}{\Pi V}`

Thus, knowing the quantities `w_2 , T, Ξ ` and `V` we can calculate the molar mass of the solute.

Isotonic Solutions

Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.
When such solutions are separated by semipermeable membrane no osmosis occurs between them. 
For example, the osmotic pressure associated with the fluid inside the blood cell is equivalent to that of 0.9% (mass/ volume) sodium chloride solution, called normal saline solution and it is safe to inject intravenously. 

Hypertonic Solutions

If we place the cells in a solution containing more than 0.9% (mass/volume) sodium chloride, water will flow out of the cells and they would shrink. Such a solution is called hypertonic. 

Hypotonic solutions

If the salt concentration is less than 0.9% (mass/volume), the solution is said to be hypotonic. In this case, water will flow into the cells if placed in this solution 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. That is, now the pure solvent flows out of the solution through the semipermeable membrane. This phenomenon is called reverse osmosis
 Reverse osmosis is used in the desalination of sea water. A schematic set up for the process is shown in Fig.

When pressure more than osmotic pressure is applied, pure water is squeezed out of the sea water through the membrane.
The pressure required for reverse osmosis is quite high. A workable porous membrane is a film of cellulose acetate placed over a suitable support. Cellulose acetate is permeable to water but impermeable to impurities and ions present in sea water. 


Previous Post Next Post