Vidyarthi Academy

CBSE NOTES CLASS 12 CHEMISTRY CHAPTER 2

SOLUTIONS

Solution

Solution is a homogeneous mixture of two or more substances in same or different physical phases.

The substances forming the solution are called components of the solution.

A solution of two components is called binary solution.

Solute and solvent

In a binary solution, solvent is the component which is present in large quantity while the other component is known as solute.

Classification of solutions

(A) Following types of solutions are seen on the basis of physical state of solute and solvent.

Aqueous Solution

If water is used as a solvent, the solution is called aqueous solution and if not, the solution is called non-aqueous solution.

(B) Depending upon the amount of solute dissolved in a solvent we have the following types of solutions:

(i) Unsaturated solution

A solution in which more solute can be dissolved without raising temperature is called an unsaturated solution.

(ii) Saturated solution

A solution in which no solute can be dissolved further at a given temperature is called a saturated solution.

(iii) Supersaturated solution

A solution which contains more solute than that would be necessary to saturate it at a given temperature is called a supersaturated solution.

Solubility

The maximum amount of a solute that can be dissolved in a given amount of solvent (generally 100g) at a given temperature is termed as its solubility at that temperature.

• Polar solutes dissolve in polar solvents and non polar solutes in nonpolar solvents.

  Like dissolves like.

Concentration of Solutions

The concentration of a solution is defined as the relative amount of solute present in a solution. On the basis of concentration of solution there are two types of solutions.

(i) Dilute solution (ii) Concentrated solution.

Methods of Expressing Concentration of Solutions :

(i) Percentage by mass (w/w %): It is defined as the amount of solute present in 100 g of solution.

$$\frac{\mathrm{w}}{\mathrm{w}}\mathrm{\%}\mathrm{ }=\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}\times 100$$

(ii) Volume percentage (V/V %) is defined as:

$$\frac{\mathrm{V}}{\mathrm{V}}\mathrm{\%}=\frac{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{ }\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}\times 100$$

(iii) Mass by volume (w/V) is defined as the weight of solute present in 100 mL of solution.

$$\frac{\mathrm{w}}{\mathrm{V}}\mathrm{\%}\mathrm{ }=\mathrm{ }\left(\frac{\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }}\right)\times 100$$

(iv) Parts per million (ppm) is defined as the parts of a component per million parts (10⁶) of the solution. It is widely used when a solute is present in trace quantities.

$$\mathrm{p}\mathrm{p}\mathrm{m}\mathrm{ }=\frac{\mathrm{n}\mathrm{o}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{n}\mathrm{o}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{ }}\times {10}^6$$

(v) Mole fraction (χ) is defined as the ratio of the number of moles of a component to the total number of moles of all the components. For a binary solution, if the number of moles of A and B are n_A and n_B respectively, the mole fraction of A will be

$${\mathrm{\chi }}_{\mathrm{A}}\mathrm{ }=\frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{B}}}\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }{\mathrm{\chi }}_{\mathrm{B}}\mathrm{ }=\frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{B}}}$$

$$\Rightarrow \mathrm{ }{\mathrm{\chi }}_{\mathrm{A}}+{\mathrm{\chi }}_{\mathrm{ }\mathrm{B}}=1$$

(vi) Molarity (M) is the number of moles of solute present in 1L (dm³) of the solution.

$$\mathrm{M}\mathrm{ }=\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{L}}$$

$$=\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{ }\left(\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}\right)\times 1000}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\times \mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{m}\mathrm{L}}$$

Also,

$$\mathrm{M}\mathrm{ }=\frac{\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{ }\mathrm{b}\mathrm{y}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\times \mathrm{ }\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{ }\times \mathrm{ }10}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}$$

• Molarity varies with temperature due to change in volume of solution.

When molarity of a solution is 1 M, it is called a molar solution. 0.1 M solution is called a decimolar solution while 0.5 M solution is known as semi molar solution

(vii) Molality (m) is the number of moles of solute per kilogram of the solvent.

$$\mathrm{m}=\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{k}\mathrm{g}}$$

$$=\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\times 1000}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\times \mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{g}}$$

• Molality is independent of temperature.

When solvent used is water, a molar (1 M) solution is more concentrated than a molal (1 m) solution. (Why?)

(viii) Normality (N) is the number of gram equivalents of solute present in 1 L of solution.

$$\mathrm{N}=\mathrm{ }\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}\mathrm{ }\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{L}}$$

Equivalent weight (also known as gram equivalent) is the mass of one equivalent, that is, the mass of a given substance which will

• combine or displace directly or indirectly with 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine – these values correspond to the atomic weight divided by the usual valence; or


• supply or react with one mole of hydrogen cations (H+) in an acid–base reaction; or


• supply or react with one mole of electrons (e⁻) in a redox reaction.


• Like molarity, normality relates the amount of solute to the total volume of solution; however, normality is specifically used for acids and bases.

The mole equivalents of an acid or base are calculated by determining the number of H⁺ or OH^- ions per molecule:

$$\mathrm{E}\mathrm{q}.\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{d}=\mathrm{ }\frac{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{d}}{\mathrm{B}\mathrm{a}\mathrm{s}\mathrm{i}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}}$$

$$\mathrm{E}\mathrm{q}.\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{a}\mathrm{ }\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}=\mathrm{ }\frac{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}}{\mathrm{A}\mathrm{c}\mathrm{i}\mathrm{d}\mathrm{i}\mathrm{t}\mathrm{y}}$$

$$\mathrm{E}\mathrm{q}.\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{a}\mathrm{ }\mathrm{s}\mathrm{a}\mathrm{l}\mathrm{t}=\mathrm{ }\frac{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{s}\mathrm{a}\mathrm{l}\mathrm{t}}{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{p}\mathrm{o}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{ }\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}}$$

$$\mathrm{Number\; of\; g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{ }\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{ }=\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}}{\mathrm{E}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}$$

Relationship between molarity and normality

N = n × M (where n is an integer)

For an acid solution, n is the number of H⁺ ions provided by a formula unit of acid (i.e. basicity).

For a basic solution, n is the number of OH⁺ ions provided by a formula unit of base (i.e. acidity).

Also,

N × Eq. mass = M × mol. Mass

(ix) Formality (F) is the number of formula weights of solute present per litre of the solution.

$$\mathrm{F}\mathrm{ }=\frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{L}}$$

($\mathrm{x})\mathrm{ }$Mass fraction of any component in the solution is the mass of that component divided by the total mass of the solution.

$$\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{f}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }=\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}$$

• Molality, mole fraction and mass fraction are preferred over molarity, normality, etc., because former involve weights which do not change with temperature, whereas later involves volumes, which are temperature dependent.

Dilution Law

• For dilution from a solution of molarity M₁ and volume V₁ to molarity M₂ and volume V₂

  M₁V₁ = M₂V₂

  If a solution with volume V₁ and normality N₁ is diluted to volume V₂ with normality N₂, then,

  N₁ × V₁ = N₂ × V₂

• For a balanced chemical equation, if n₁ moles of reactant reacts with n₂ moles of reactant 2, then,

  $$\frac{{\mathrm{M}}_1{\mathrm{V}}_1}{{\mathrm{n}}_1}\mathrm{ }=\mathrm{ }\frac{{\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{n}}_2}$$

  Where n₁ and n₂ are stoichiometric coefficient in balanced equation



• If two solutions of the same solute having volumes and molarities V₁, M₁ and V₂, M₂ are mixed, the molarity of the resulting solution is,

  $$\mathrm{M}=\frac{{\mathrm{M}}_1{\mathrm{V}}_1+{\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{V}}_1+{\mathrm{V}}_2}$$

  And

  $$\mathrm{N}=\frac{{\mathrm{N}}_1{\mathrm{V}}_1+{\mathrm{N}}_2{\mathrm{V}}_2}{{\mathrm{V}}_1+{\mathrm{V}}_2}$$



• To dilute V₁ mL of a solution having molarity M₁ to molarity M₂ up to the final volume V₂ mL, the volume of water added is

  $${\mathrm{V}}_2-{\mathrm{V}}_1=\left(\frac{{\mathrm{M}}_1-{\mathrm{M}}_2}{{\mathrm{M}}_2}\right){\mathrm{V}}_1=\left(\frac{{\mathrm{N}}_1-{\mathrm{N}}_2}{{\mathrm{N}}_2}\right){\mathrm{V}}_1$$



• For a chemical reaction involving an acid and a base,

  n_a × M_a × V_b = n_b × M_b × V_b

  where,

  n_a = Basicity of acid

  n_b = Acidity of base

  V_a = Volume of acid

  V_b = Volume of base

  M_a = Molarity of acid

  M_b = Molarity of base

Solubility of a Solid in a Liquid

Effect of temperature

The solution being in dynamic equilibrium, must follow Le Chateliers Principle.

• If in a nearly saturated solution, the dissolution process is endothermic (Δ_solH > 0), the solubility should increase with rise in temperature


• If it is exothermic (Δ_solH < 0), the solubility should decrease.

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.

Solubility of a Gas in a Liquid

Effect of Pressure – Henry’s Law

The law states that at a constant temperature, 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.

χ ∝ p

The most commonly used form of Henry’s law states

“the partial pressure (p) of the gas in vapour phase is proportional to the mole fraction (χ) of the gas in the solution”

and is expressed as

p = K_H χ .

Where, K_H. is a proportionality constant called Henery’ constant.

• Greater the value of K_H, lower the solubility of the gas.


• K_H is characteristic of gas-liquid pair.


• The value of K_H increases with increase in the temperature.


• SI unit of K_H is bar

Applications of Henry’s Law

1. In manufacture of soft drinks and soda water, CO₂ is passed at high pressure to increase its solubility.


2. In case of deep sea diving, 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 condition known as bends, which are painful and dangerous to life.

   To minimise the painful effects (bends) accompanying the decompression of deep sea divers, O₂ is diluted with less soluble He gas as breathing gas.



3. At high altitudes, the partial pressure of O₂ is less than that at the ground level. This leads to low concentrations of O₂ in the blood of climbers which causes anoxia.

Effect of Temperature

Dissolution of gases in liquids is an exothermic process. Hence the solubility of gases in liquids decreases with rise in temperature. Thus, aquatic species are more comfortable in cold water [more dissolved O₂] rather than warm water.

Vapour Pressure of Liquid-Liquid Solutions – Raoult’s Law

For a solution of two volatile liquids, the vapour pressure of each liquid in the solution is less than the respective vapour pressure of the pure liquids and the equilibrium partial vapour pressure of the liquid is directly proportional to its mole fraction.

In case of a binary solution, containing two liquids 1 and 2 with mole fractions χ₁ and χ_2, for component 1 and 2 respectively,

p₁ ∝ χ₁ ⇒ p₁ = k χ₁

The proportionality constant is obtained by considering the pure liquid.

When χ₁ = 1 then k = p^o₁, where p^o₁ is the vapour pressure of pure component 1 at the same temperature.

Hence, p₁ = p^o₁ χ₁

Similarly, for component 2

P₂ ∝ χ₂ ⇒ p₂ = p^o₂ χ₂

where p^o₂ is the vapour pressure of pure component 2 at the same temperature.

According to Dalton’s law of partial pressures, the total pressure (p_total) over the solution phase in the container will be the sum of the partial pressures of the components of the solution. Or,

p_total = p₁ + p₂

Substituting the values of p₁ and p₂, we get

p_total = p^o₁ χ₁ + p^o₂ χ₂

= (1-χ₂) p^o₁ + p^o₂ χ₂

= p^o₁ + (p^o₂ - p^o₁)χ₂

Konowaloff Rule

At any fixed temperature, the vapour phase is always richer in the more volatile component as compared to the solution phase.

In other words, mole fraction of the more volatile component is always greater in the vapour phase than in the solution phase.

The composition of vapour phase in equilibrium with the solution is determined by the partial pressure of components. If y₁ and y₂ are the mole fractions of the component 1 and 2 respectively in the vapour phase then, using Dalton’s law of partial pressure, then

p₁ = y₁ × p_total

p₂ = y₂ × p_total

Raoult’s Law and Henry’s Law

Raoult’s Law - The vapour pressure of a volatile component in a given solution is given by,

p₁ = p^o₁ χ₁

[Raoult’s law gives the partial vapour pressures of components in terms of solutbility of components]

Henry’s Law - Solubility of gas is given by,

p = K_Hχ.

Hence Raoult’s law is a special case of Henry’s law in which K_H becomes p^o₁.

Raoult’s Law for Solutions of Solids in Liquids

Raoult’s law in its general form 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.

If component 1 is the solvent then,

p₁ = p^o₁ χ₁

Since the other component is solid p₂ = 0

Ideal solutions

The solutions which satisfy the following conditions called ideal solutions.

1. Solution must obey Raoult‟s law, i.e.,

   p₁ = p^o₁ χ₁ and p₂ = p^o₂ χ₂



2. Δ_mixH = 0 (No energy evolved or absorbed)


3. Δ_mixV = 0 (No expansion or contraction on mixing)

Some solutions behave like nearly ideal solutions are,

• benzene + toluene


• n-hexane + n-heptane


• ethyl iodide + ethyl bromide


• chlorobenzene + bromobenzene.

Non-ideal solutions

Those solutions which show deviation from Raoult’s law, are called non-ideal solution.

For such solutions, Δ_mixH ≠ 0 and Δ_mixV ≠ 0

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 if it is lower, it exhibits negative deviation from Raoult’s law.

Non-ideal solutions showing positive deviation

In such a case, the A – B interactions are weaker than A – A or B – B interactions and the observed vapour pressure of each component and the total vapour pressure are greater than that predicted by Raoult‟s law.

For such solutions

p₁ > p^o₁ χ₁

p₂ > p^o₂ χ₂

p_total > p^o₁ χ₁ + p^o₂ χ₂

ΔH_mix > 0 and ΔV_mix > 0

Examples

• Ethnol + Water,


• CS₂ + Acetone,


• CCL₄ + C₆H₆,


• CCL₄ + C₆H₅CH₃,


• Ethnol + Cyclohexane,


• CCL₄ + CHCl₃

Explanation

• 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, more vapour is formed and the solution shows positive deviation from Raoult’s law.



• 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.

Non-ideal solution showing negative deviation

In such a case, the A – B interactions are stronger than A – A or B – B interactions and the observed vapour pressure of each component and the total vapour pressure are lesser than that predicted by Raoult’s law.

For such solutions

p₁ < p^o₁ χ₁

p₂ < p^o₂ χ₂

p_total <p^o₁ χ₁ + p^o₂ χ₂

ΔH_mix < 0 and ΔV_mix < 0

Examples

• CHCl₃ + CH₃COCH₃


• CHCl₃ + C₆H₆


• H₂O + HCl


• H₂O + HNO₃


• Methanol + acetic acid

Explanation

• 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, because chloroform molecule is able to form hydrogen bond with acetone molecule.

  

  This decreases the escaping tendency of molecules for each component and consequently the vapour pressure decreases resulting in negative deviation from Raoult’s law.

Azeotropic mixture

The binary mixtures which are having the same composition in liquid and vapour phase and boil at a constant temperature are called Azeotropic mixtures.

In such cases, it is not possible to separate the components by fractional distillation.

1. Minimum boiling azeotropes are formed by those liquid pairs which show large positive deviation from ideal behaviour. Such azeotropes have boiling points lower than either of the components, e.g., C₂H₅OH (95.57%) + H₂O (4.43%) by mass.


2. Maximum boiling azeotropes are formed by those liquid pairs; which show negative deviation from ideal behaviour. Such azeotropes have boiling points higher than either of the components. e.g., H₂O (20.22%) + HCl (79.78%) and HNO₃ (68%) + H₂O (32%) by mass.

Colligative properties

(colligative: Latin - co means together, ligare means to bind)

Colligative properties are those properties of solution which depend only on the number of solute particles in a solution irrespective of their nature.

Explanation

The vapour pressure of solution decreases when a non-volatile solute is added to a volatile solvent.

We know that evaporation is a surface phenomenon. The fraction of molecules changing from liquid phase to gaseous phase depends on the number of molecules of the solvent at the surface.

When a non-volatile solid is added to a solvent, the concentration of solvent molecules, at the surface, decreases. As a result, the number of molecules changing from liquid phase to gas phase decreases and vapour pressure decreases.

Some of the important colligative properties of solutions

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.

Relative lowering of vapour pressure

It is the ratio of lowering in vapour pressure to vapour pressure of pure solvent. The relative lowering in vapour pressure of solution containing a nonvolatile solute is equal to the mole fraction of solute in the solution.

For a solute (2) in solvent (1), we have,

p₁ = p^o₁ χ₁

Hence

p^o₁ - p₁ = (p^o₁ - p^o₁ χ₁)

= p^o₁ (1- χ₁) = p^o₁ χ₂

$$\Rightarrow \mathrm{ }\frac{{\mathrm{p}}_1^{\mathrm{o}}\mathrm{ }-\mathrm{ }{\mathrm{p}}_1}{{\mathrm{p}}_1^{\mathrm{o}}}\mathrm{ }=\mathrm{ }{\mathrm{\chi }}_2=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}$$

For dilute solutions n₂ << n₁

Therefore,

$$\frac{{\mathrm{p}}_1^{\mathrm{o}}\mathrm{ }-\mathrm{ }{\mathrm{p}}_1}{{\mathrm{p}}_1^{\mathrm{o}}}=\mathrm{ }\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}$$

Or,

$$\frac{{\mathrm{p}}_1^{\mathrm{o}}\mathrm{ }-\mathrm{ }{\mathrm{p}}_1}{{\mathrm{p}}_1^{\mathrm{o}}}=\mathrm{ }\frac{{\mathrm{W}}_2\times {\mathrm{M}}_1}{{\mathrm{W}}_1\times {\mathrm{M}}_2}$$

Where, W₂ and W₁ = mass of solute and solvent respectively. M₂ and M₁ = molecular weight of solute and solvent respectively.

$$\Rightarrow \mathrm{ }{\mathrm{M}}_2=\mathrm{ }\frac{{\mathrm{W}}_2\times {\mathrm{M}}_1}{{\mathrm{W}}_1}\times \frac{{\mathrm{p}}_1^{\mathrm{o}}}{{\mathrm{p}}_1^{\mathrm{o}}\mathrm{ }-\mathrm{ }{\mathrm{p}}_1}$$

From the above expression we can find the molecular weight of an unknown solute dissolved in a given solvent.

Elevation in boiling point (ΔT_b)

Boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure. As the vapour pressure of a solution containing a nonvolatile solute is lower than that of the pure solvent, its boiling point will be higher than that of the pure solvent.

The increase in boiling point is known as elevation in boiling point and is given by,

ΔT_b = T_b – T°_b

ΔT_b = K_b m

where; m = molality of the solution.

K_b is called the molal elevation constant or ebullioscopic constant. The unit of K_b is K kg mol^-1

Calculation of Molecular Mass of Solute

If W₂ gram of solute of molar mass M₂ is dissolved in W₁ gram of solvent, then molality, m of the solution is given by

$$\mathrm{m}=\frac{{\mathrm{W}}_2/{\mathrm{M}}_2}{{\mathrm{W}}_1/1000}=\frac{{\mathrm{W}}_2\times 1000}{{\mathrm{M}}_2\times {\mathrm{W}}_1}\mathrm{ }$$

$$\mathrm{S}\mathrm{o},\mathrm{ }\mathrm{ }\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}\mathrm{ }=\mathrm{ }{\mathrm{K}}_{\mathrm{b}}\mathrm{m}=\frac{{\mathrm{K}}_{\mathrm{b}}\times {\mathrm{W}}_2\times 1000}{{\mathrm{M}}_2\times {\mathrm{W}}_1}$$

$$\mathrm{O}\mathrm{r},\mathrm{ }\mathrm{ }{\mathrm{M}}_2\mathrm{ }=\frac{{\mathrm{K}}_{\mathrm{b}}\times {\mathrm{W}}_2\times 1000}{\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}\times {\mathrm{W}}_1}$$

For water, K_b = 0.52 K kg mol^-1

Depression in Freezing Point (ΔTf)

Freezing point of a liquid is the temperature at which vapour pressure of the solvent in its liquid and solid phase become equal. As the vapour pressure of solution containing non-volatile solute is lower than that of pure solvent, solid form gets separated out at a lower temperature.

This decrease in freezing point of a liquid is known as depression in freezing point.

ΔT_f = T°_f – T_f

• It has been found experimentally that for dilute solutions the elevation of boiling point (ΔT_f) is directly proportional to the molal concentration of the solute in a solution.

  ΔT_f = K_f m

  where; m = molality of solution

K_f is called the molal depression constant or Cryoscopic Constant constant. The unit of K_f is K kg mol^-1

Calculation of molecular mass using depression in freezing point

$$\mathrm{m}=\frac{{\mathrm{W}}_2/{\mathrm{M}}_2}{{\mathrm{W}}_1/1000}=\frac{{\mathrm{W}}_2\times 1000}{{\mathrm{M}}_2\times {\mathrm{W}}_1}\mathrm{ }$$

$$\mathrm{S}\mathrm{o},\mathrm{ }\mathrm{ }\mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}\mathrm{ }=\mathrm{ }{\mathrm{K}}_{\mathrm{f}}\mathrm{ }\mathrm{m}=\frac{{\mathrm{K}}_{\mathrm{f}}\times {\mathrm{W}}_2\times 1000}{{\mathrm{M}}_2\times {\mathrm{W}}_1}$$

$$\mathrm{O}\mathrm{r},\mathrm{ }\mathrm{ }{\mathrm{M}}_2\mathrm{ }=\mathrm{ }{\mathrm{K}}_{\mathrm{f}}\mathrm{ }\mathrm{m}=\frac{{\mathrm{K}}_{\mathrm{f}}\times {\mathrm{W}}_2\times 1000}{\mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}\times {\mathrm{W}}_1}$$

For water, K_f = 1.86 K kg mol^-1

$${\mathrm{K}}_{\mathrm{f}}=\frac{\mathrm{R}\times {\mathrm{M}}_1\times {{\mathrm{T}}_{\mathrm{f}}}^2}{{\mathrm{\Delta }}_{\mathrm{f}\mathrm{u}\mathrm{s}}\mathrm{H}\times 1000}$$

$${\mathrm{K}}_{\mathrm{b}}=\frac{\mathrm{R}\times {\mathrm{M}}_1\times {{\mathrm{T}}_{\mathrm{b}}}^2}{{\mathrm{\Delta }}_{\mathrm{v}\mathrm{a}\mathrm{p}}\mathrm{H}\times 1000}$$

Where ${\mathrm{\Delta }}_{\mathrm{f}\mathrm{u}\mathrm{s}}\mathrm{H}$ and$\mathrm{ }{\mathrm{\Delta }}_{\mathrm{v}\mathrm{a}\mathrm{p}}\mathrm{H}$ are molar enthalpies of fusion and vapourisation of the solvent, respectively.

Ethylene glycol is usually added to water in the radiator to lower its freezing point. It is called antifreeze solution.

[Common salt (NaCl) and anhydrous CaCl₂ are used to clear snow on the roads because they depress the freezing point of water.]

Osmotic pressure (π)

Osmosis is the phenomenon of spontaneous flow of the solvent molecules through a semipermeable membrane from pure solvent to solution or from a dilute solution to concentrated solution.

Some natural semipermeable membranes are animal bladder, cell membrane etc. Cu₂[Fe(CN)₆] is an artificial semipermeable membrane which does not work in non-aqueous solutions as it dissolves in them.

Osmosis may be

1. Exosmosis - outward flow of water or solvent from a through semipermeable membrane.


2. Endosmosis - inward flow of water or solvent from a through a semipermeable membrane.

The hydrostatic pressure developed on the solution which just prevents the osmosis of pure solvent into the solution through a semipermeable membrane is called osmotic pressure.

For dilute solutions, it has been found experimentally that osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Or,

π = C R T

Here π is the osmotic pressure and R is the gas constant.

$$\mathrm{\pi }\mathrm{ }=\mathrm{ }\left(\frac{{\mathrm{n}}_2}{\mathrm{V}}\right)\mathrm{R}\mathrm{ }\mathrm{T}\mathrm{ }$$

Here V is volume of a solution in litres containing n₂ moles of solute.

If W₂ grams of solute, of molar mass, M₂ is present in the solution, then

$${\mathrm{n}}_2\mathrm{ }=\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2}$$

Hence,

$$\mathrm{\pi }\mathrm{ }=\mathrm{ }\left(\frac{{\mathrm{W}}_2}{\mathrm{V}\times {\mathrm{M}}_2}\right)\mathrm{R}\mathrm{ }\mathrm{T}\mathrm{ }$$

$$\Rightarrow {\mathrm{ }\mathrm{M}}_2=\mathrm{ }\mathrm{ }\frac{{\mathrm{W}}_2\mathrm{R}\mathrm{ }\mathrm{T}}{\mathrm{\pi }\mathrm{V}}$$

where R = gas constant, T = temperature

Note: Osmotic pressure method is the best method for determining the molecular masses of polymers since observed value of any other colligative property is too small to be measured with reasonable accuracy.

Types of solutions based on osmosis

On the basis of osmotic pressure, the solution can be

1. Hypertonic solution

   A solution is called hypertonic if its osmotic pressure is higher than that of the solution from which it is separated by a semipermeable membrane.

   When a plant cell is placed in a hypertonic solution, the fluid from the plant cell comes out and cell shrinks, this phenomenon is called plasmolysis.

(ii) Hypotonic solution

A solution is called hypotonic if its osmotic pressure is lower than that of the solution from which it is separated by a semipermeable membrane.

(iii) Isotonic solution

Two solutions having same osmotic pressure at a given temperature are called isotonic solutions. These solutions have same molar concentration. 0.91% solution of pure NaCl is isotonic with human RBC‟s.

Isotonic solutions have same molar concentration, e.g., if x % solution of X is isotonic with y % solution of Y, this means molar concentration of X = Molar concentration of Y,

$$\frac{\mathrm{x}}{100}\times \frac{1000}{{\mathrm{M}}_{\mathrm{x}}}=\frac{\mathrm{y}}{100}\times \frac{1000}{{\mathrm{M}}_{\mathrm{y}}}$$

$$\Rightarrow \frac{\mathrm{x}}{{\mathrm{M}}_{\mathrm{x}}}=\mathrm{ }\frac{\mathrm{y}}{{\mathrm{M}}_{\mathrm{y}}}$$

In case the solution is not dilute,

$$\frac{\mathrm{x}/(1-\mathrm{x})}{{\mathrm{M}}_{\mathrm{x}}}=\mathrm{ }\frac{\mathrm{y}/(1-\mathrm{y})}{{\mathrm{M}}_{\mathrm{y}}}$$

Reverse osmosis

When the external pressure applied on the solution is more than osmotic pressure, the solvent flows from the solution to the pure solvent, which is called reverse osmosis. Desalination of sea water is done by reverse Osmosis.

Cellulose acetate is permeable to water but impermeable to impurities and ions present in sea water.

Abnormal molecular masses

When there is dissociation of solute into ions, the experimentally determined molar mass is always lower than the true value.

Similarly, using freezing point and boiling point method also, the experimentally determined molar mass is again lower than the true value.

van’t Hoff factor (i)

These observed values are corrected by multiplying the experimentally determined values by van‟t Hoff factor (i).

It is the ratio of observed value of colligative property to the calculated value of colligative property.

$$\mathrm{i}\mathrm{ }=\frac{\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{e}\mathrm{d}\mathrm{ }\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{g}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{ }\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{y}}{\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d}\mathrm{ }\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{g}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{ }\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{y}}$$

or

$$\mathrm{i}\mathrm{ }=\frac{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{e}\mathrm{d}\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$$

or

$$\mathrm{i}\mathrm{ }=\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{ }}{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{y}}$$

So to correct the observed value of molar mass, van‟t Hoff factor (i) must be included in different expressions for colligative properties.

$$\frac{{\mathrm{p}}_1^{\mathrm{o}}\mathrm{ }-\mathrm{ }{\mathrm{p}}_1}{{\mathrm{p}}_1^{\mathrm{o}}}\mathrm{ }=\mathrm{ }\mathrm{i}\mathrm{ }{\mathrm{\chi }}_2$$

ΔT_b = i K_b m

ΔT_f = i K_f m

π = i C R T

Degree of dissociation (α) and van’t Hoff factor (i)

1. If one molecule of a substance A gets dissociated into n particles or molecules

   A → nP

   And if α is the degree of dissociation then

   Total number of moles	A	P
   Initially	1	0
   At equilibrium	1 - α	n α

   Therefore

   $$\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{ }\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{l}\mathrm{i}\mathrm{b}\mathrm{r}\mathrm{i}\mathrm{u}\mathrm{m}=\mathrm{ }1\mathrm{ }–\mathrm{ }\mathrm{\alpha }\mathrm{ }+\mathrm{ }\mathrm{n}\mathrm{\alpha }\mathrm{ }=\mathrm{ }1+\left(\mathrm{n}-1\right)\mathrm{\alpha }$$

   $$\mathrm{i}\mathrm{ }=\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{ }}{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{y}}$$

   $$\mathrm{i}\mathrm{ }=\frac{1+\left(\mathrm{n}-1\right)\mathrm{\alpha }}{1}$$

   $$\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{\alpha }\mathrm{ }=\frac{\mathrm{i}-1}{\mathrm{n}-1}$$

Degree of association (α) and van’t Hoff factor (i)

If n molecules of a substance A associate to form A_n

$$\mathrm{A}\mathrm{ }\mathrm{\to }\frac{{\mathrm{A}}_{\mathrm{n}}}{\mathrm{n}}$$

And if α is the degree of association then

Total number of moles	A	An
Initially	1	0
At equilibrium	1 - α	$\frac{\mathrm{\alpha }}{\mathrm{n}}$

Therefore

$$\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{ }\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{l}\mathrm{i}\mathrm{b}\mathrm{r}\mathrm{i}\mathrm{u}\mathrm{m}=\left(1-\mathrm{\alpha }+\frac{\mathrm{a}}{\mathrm{n}}\right)$$

$$\mathrm{i}\mathrm{ }=\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{ }}{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{y}}$$

$$\mathrm{i}\mathrm{ }=\frac{1-\mathrm{\alpha }\mathrm{ }+\mathrm{\alpha }/\mathrm{n}}{1}$$

$$\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{\alpha }\mathrm{ }=\frac{\mathrm{i}-1}{1/\mathrm{n}-1}$$

Note:

• van’t Hoff factor i > 1 for solutes undergoing dissociation




• it is < 1 for solutes undergoing association.