SOLUTIONS

 1. Solution: A solution is a homogeneous mixture of two or more pure substances, the relative ratio of which can be changed within certain limits.

When the constituents of the solution are two it is called binary, if three then ternary, if four then quaternary and so on.

The two constituents of the solution are solvent and solute.

Solvent (A): It is the component of the solution:

(i) which is present in a relatively large proportion in the solution, and

(ii) whose physical state is same as that of the resulting solution.

Solute (B): It is the component of a solution which is present in relatively small proportion.

2. Types of Solutions

Table 2.1

Type of Solution	Solute	Solvent	Examples
Gaseous Solutions	Gas	Gas	Mixture of oxygen and nitrogen gases
	Liquid	Gas	Chloroform mixed with nitrogen gas
	Solid	Gas	Camphor in nitrogen gas
Liquid Solutions	Gas	Liquid	Oxygen dissolved in water
	Liquid	Liquid	Ethanol dissolved in water
	Solid	Liquid	Glucose dissolved in water
Solid Solutions	Gas	Solid	Solution of hydrogen in palladium
	Liquid	Solid	Amalgam of mercury with sodium
	Solid	Solid	Copper dissolved in gold

Amongst the nine types of solutions, the widely studied ones are:

(a) Solid–liquid (b) Liquid–liquid and (c) Gas–liquid solutions.

(a) Solid–liquid solutions: A small amount of solute (usually ionic solids) is dissolved in a large quantity of solvent. If the amount of solvent is large as compared to the solute, the solution is referred to as a dilute solution. If the amount of solvent is small as compared to the solute, the solution is referred to as a concentrated solution.

Saturated solution: A solution is said to be saturated if it holds the maximum amount of solute at a given temperature in a given quantity of the solvent.

Solubility: It may be defined as the maximum amount of solute that can be dissolved in 100 g of solvent at a specified temperature. The solubility of solid into liquid depends upon the following factors:

(i) Nature of solute (ii) Nature of solvent (iii) Temperature

Causes of solubility: The following types of forces of attraction are operated when a solute is mixed with a solvent:

(i) Inter-ionic attraction in the solute molecules: Ions are held together in the lattice due to electrostatic forces. Due to these forces molecules are stabilised and the energy released is called lattice energy. This is defined as the energy released when 1 g mole of the compound is formed due to electrostatic attraction between the ions.

(ii) Inter-molecular attraction between solvent molecules: Water is a polar solvent because of the difference in electronegativity between hydrogen and oxygen atoms constituting water molecule. This difference gives rise to the development of a slight negative charge on oxygen and equal positive charge on hydrogen. A dipole is thus created giving rise to dipole–dipole attraction between water molecules.

(iii) Solvation: It represents force of attraction between solute and solvent molecules. If the solvent is water then the energy released is called hydration energy.

If hydration energy > lattice energy, then solution is easily formed. Both the ions of the solute get hydrated to overcome the lattice energy of the solute.

(iv) Temperature: Saturated solution represents equilibrium between undissolved solute and dissolved solute.

Undissolvedsolute+Solvent→SolutionΔsolH=±x

If Dsol H < 0, i.e., (–ve), the dissolution is exothermic. In this case, as the temperature increases, solubility decreases (Le Chatelier’s principle).

If Dsol H > 0, i.e., (+ve), there is endothermic dissolution. In this case, increase in temperature increases the solubility (Le Chatelier’s principle).

(b) Liquid–liquid solutions: When two liquids are mixed, three different situations result:

(i) Miscible liquids: The two components are completely soluble. They are miscible only when they have similar nature or belong to the same homologous series. Example: water and alcohol (both polar), benzene–toluene (both belong to the same homologous series). There is a rule:

Like dissolves like – Polar solute is soluble in polar solvent and a non-polar one in a non-polar solvent.

(ii) Partially miscible liquids: This happens only when the intermolecular forces of one liquid is greater than that of the other is. Solubility, however, increases with increasing temperature. Examples: aniline-water, phenol-water.

(iii) Immiscible liquids: Two components are completely insoluble. This happens when one liquid is polar and the other is non-polar. Examples: Carbon tetrachloride-water; chloroform-water.

(c) Gas–liquid solutions: The gases are generally soluble in water and to a limited extent in other solvents too. Solubility, however, depends on the following factors:

(i) Nature of gas: Easily liquefiable gases are generally more soluble in common solvents.

(ii) Nature of liquid: Those gases which easily form ions in solution are more soluble in water than in other solvents. Ion formation in other solvents is not an easy process.

HCl(g) + H2O(l)  H3O+(aq) + Cl–(aq)

(iii) Pressure: Pressure is an important factor affecting the solubility of gas in liquids. This is governed by Henry’s law.

(iv) Temperature: With rise in temperature, the solubility generally decreases because gas is expelled. Some gases, however, find their solubility increased at a higher temperature.

 Henry’s law: It states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

The most commonly used form of Henry’s law states that the partial pressure (p) of a gas in vapour phase is proportional to the mole fraction of the gas (x) in the solution and is expressed as

p = KH x

Here KH is the Henry’s law constant and x is the mole fraction of the gas.

Note: Mole fraction is discussed in detail further in this chapter.

Limitations of Henry’s Law:

Henry’s law is applicable only when

l The pressure of the gas is not too high and temperature is not too low.

l The gas should not undergo any chemical change.

l The gas should not undergo association or dissociation in the solution.

Applications of Henry’s Law:

l To increase the solubility of CO2 in soda water and soft drinks, the bottle is sealed under high pressure.

l To avoid the toxic effects of high concentration 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).

l At high altitudes, low blood oxygen causes climbers to become weak and make them unable to think clearly, which are symptoms of a condition known as anoxia.

3. Methods of Expressing Concentration of Solutions:

The concentration of solution is the amount of the solute present in the given quantity of solution or solvent. It can be expressed in any of the following ways:

(a) Mass percentage (w/w): It may be defined as mass of solute per 100 g of solution.

Mass% of solute = MassofsoluteTotalmassofthesolution × 100

(b) Volume percentage (V/V): It may be defined as volume of solute per 100 mL of solution.

Volume % of solute = VolumeofsoluteTotalvolumeofsolution × 100

(c) Mass by volume percentage (w/V): It may be defined as the mass of solute per 100 mL of the solution.

Mass by volume % = MassofsoluteVolumeofsolution × 100

(d) Parts per million (ppm): It is the parts of a component per million (106) parts of the solution.

ppm = NumberofpartsofthecomponentTotalnumberofpartsofallcomponentsofthesolution×106

It is convenient to express concentration in ppm when a solute is present in trace quantities.

(e) Mole fraction (x): It may be defined as the ratio of the number of moles of one component (solute or solvent) to the total number of moles of all the components present in the solution.

If in a solution, nA and nB are the number of moles of solvent and solute, respectively, then

Mole fraction of solvent in the solution,

xA = nAnA+nB

Mole fraction of solute in the solution,

xB = nBnA+nB

In a solution, sum of the mole fractions of all the components is unity. For example, in a binary solution (having two components), xA + xB = 1.

(f) Molarity (M): It is defined as number of moles of solute dissolved in one litre of solution.

Molarity = MolesofsoluteVolumeofsolution(inlitre)

Unit of molarity is mol L–1 or M (molar). Molarity changes with change in temperature as volume changes with change in temperature.

(g) Molality (m): It is defined as the number of moles of the solute dissolved in one kilogram (kg) of the solvent and is expressed as:

Molality = MolesofsoluteMassofsolvent(inkg)=NumberofmolesofsoluteMassofsolvent(ingrams)×1000

Unit of molality is mol kg–1 or molal (m).

Molality is independent of temperature.

4. Vapour Pressure of Pure Liquid and Solution

Vapour pressure: When a liquid is taken in a closed vessel, a part of the liquid evaporates and its vapours occupy the available empty space. These vapours cannot escape as the vessel is closed. They would rather have a tendency to condense into liquid form. In fact an equilibrium is established between vapour phase and liquid phase and the pressure that its vapour exert is termed as vapour pressure. Thus, vapour pressure of a liquid may be defined as the pressure exerted by the vapours above the liquid surface in equilibrium with the liquid phase at a given temperature.

The vapour pressure of a liquid depends on the following factors:

Nature of the liquid: Liquids having weak intermolecular forces are volatile and therefore have greater vapour pressure.

Temperature: Vapour pressure of a liquid increases with increase in temperature. This is because with increase in temperature, the kinetic energy of the molecules increases and therefore large number of molecules are available for escaping from the surface of the liquid.

(a) Vapour pressure of liquid–liquid solution:

Raoult’s Law for solutions of volatile liquids: It states that for a solution of volatile liquids the partial pressure of each component of the solution is directly proportional to its mole fraction present in a solution. Mathematically,

pA xA	pB xB
pA = poA xA	pB = poB xB

where pA and pB are partial vapour pressures, xA and xB are mole fractions, poA and poB are the vapour pressure of pure components A and B respectively.

If p is total vapour pressure then according to Dalton’s law of partial pressure,

p = pA + pB

= poA xA + poB xB

= poA (1 – xB) + poB xB

= poA + (poB – poA) xB

As poA and poB are constants at a given temperature, it is evident from the above equations that the total vapour pressure varies linearly with the mole fraction xB (or xA since xA = 1 – xB).

The composition of the vapour phase in equilibrium with the solution can be determined from the partial pressure of the two components. If yA and yB are the mole fractions of components A and B respectively in the vapour phase, then

pA = yA ptotal,

pB = yB ptotal

In general, pi = yi ptotal

 Raoult’s law as a special case of Henry’s law.

In the solution of a gas in a liquid, if one of the components is so volatile that it exists as a gas, then it can be said that Raoult’s law becomes a special case of Henry’s law in which KH becomes equal to poA.

(b) Vapour pressure of solutions of solids in liquids

Raoult’s law for a solution containing a non-volatile solute and volatile solvent: It states that the relative lowering of vapour pressure is equal to mole fractions of solute which is non-volatile.

Mathematically, p = pA + pB

or, p = pA (Since solute B is non-volatile)

p = poA xA

p = poA (1 – xB) = poA – poA xB

poA xB = poA – p

or poA–ppoA = xB

or Relative lowering of vapour pressure = Mole fraction of solute

5. Ideal and Non-Ideal Solutions

(a) Ideal solution: A solution is called an ideal solution if it obeys Raoult’s law over a wide range of concentration (Fig. 2.1) at a specified temperature.

For an ideal solution,

p = pA + pB = poA xA + poB xB

Liquids having similar nature and structure are likely to form ideal solutions. Examples are:

l Mixture of methanol and ethanol

l Mixture of n-hexane and n-heptane

l Mixture of benzene and toluene.

Reasons for formation of ideal solutions: A solution of two miscible liquids A and B will be ideal if two essential conditions are fulfilled.

(i) If FA – A is the force of attraction between molecules of A and FB – B is that of molecules of B, then A and B will form an ideal solution only if,

FA – B = FA – A = FB – B

(ii) The solution of A and B liquids will be ideal if A and B have similar structures and polarity. Methanol and ethanol have the same functional group and almost same polarity and therefore, form ideal solutions.

For an ideal solution

(i) Raoult’s law is obeyed, i.e., pA = poA xA and pB = poB xB

(ii) Dmix H = 0 and

(iii) Dmix V = 0.

(b) Non-ideal solution: A solution which does not obey Raoult’s law for all concentrations is called a non-ideal solution.

For a non-ideal solution

(i) Raoult’s law is not obeyed, i.e., pA ≠ poA xA and pB ≠ poB xB

(ii) Dmix H ≠ 0 and

(iii) Dmix V ≠ 0.

A non-ideal solution can show either positive or negative deviation from Raoult’s law.

(A) Positive Deviation: The deviation will be called positive when the partial pressure of each component and the resultant total pressure are greater than the pressure expected on the basis of Raoult’s law (Fig. 2.2).

In such cases, the intermolecular forces between solvent–solute molecules (FA – B) are weaker than those between solvent–solvent (FA – A) and solute–solute (FB – B) molecules. That is,

FA – B < FA – A and FB – B

This shows that the molecules of A or B will escape more easily from the surface of the solution, i.e., the vapour pressure of solution will be higher.

Characteristics of a Solution Showing Positive Deviation

l pA > poA xA ; pB > poB xB

l Dmix H > 0, i.e., + ve,

l Dmix V > 0, i.e., + ve,

Some examples of the solution exhibiting positive deviations are:

(i) Ethyl alcohol and water

(ii) Acetone and carbon disulphide

(iii) Carbon tetrachloride and benzene

(iv) Acetone and benzene

(B) Negative Deviation: The deviation is called negative deviation, if the partial pressure of each component (A and B) and resultant total vapour pressure are less than the pressure expected on the basis of Raoult’s law. (Fig. 2.3)

This type of deviation is shown by the solutions in which

FA – B > FA – A and FB – B

Due to this, there is decrease in the escaping tendency of A or B molecules from the surface of solution.

Consequently, the vapour pressure of the solution will be lower.

Characteristics of a Solution Showing Negative Deviation

l pA < poA xA ; pB < poB xB

l Dmix H < 0, i.e., –ve, because weak A–A and B–B bonds are broken and strong A–B bond is formed. Heat is consequently released.

l Dmix V < 0, i.e., –ve,

Some examples of the solution showing negative deviations are

(i) HNO3 and water

(ii) Chloroform and acetone

(iii) Acetic acid and pyridine

(iv) Hydrochloric acid and water

Table 2.2: Characteristics of Ideal and Non-ideal Solutions

S.No.	Ideal Solutions	Non-ideal Solutions
		Positive Deviation	Negative Deviation
(i)	FA – A = FB – B = FA – B	FA – B < FA – A and FB – B	FA – B > FA – A and FB – B
(ii)	pA = poAxA ; pB = poBxB	pA > poAxA ; pB > poBxB	pA < poAxA ; pB < poBxB
(iii)	Dmix H = 0	Dmix H > 0, i.e., +ve	Dmix H < 0, i.e., –ve
(iv)	Dmix V = 0	Dmix V > 0, i.e., +ve	Dmix V < 0, i.e., –ve

Azeotropes or Azeotropic mixture: Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature.

Types of Azeotropes:

(i) Minimum boiling azeotropes: These are the binary mixtures whose boiling point is less than either of the two components. The non-ideal solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a specific composition, e.g., a mixture of 94.5% ethyl alcohol and 4.5% water by volume.

(ii) Maximum boiling azeotropes: These are the binary mixtures whose boiling point is more than either of the two components. The solutions that show large negative deviation from Raoults’s law form maximum boiling azeotrope at a specific composition, e.g., a mixture of 68% HNO3 and 32% H2O by mass.

6. Colligative Properties

Those properties which depend on the number of solute particles (molecules, atoms or ions) but not upon their nature are called colligative properties. The following are the colligative properties:

(a) Relative lowering of vapour pressure of the solvent,

(b) Elevation of boiling point of the solvent,

(c) Depression of freezing point of the solvent,

(d) Osmotic pressure of the solution.

(a) Relative lowering of vapour pressure: The addition of a non-volatile solute to a volatile solvent decreases the escaping tendency of the solvent molecules from the surface of solutions as some of the surface area is occupied by non-volatile solute particles. According to Raoult’s law, the relative lowering of vapour pressure is equal to mole fraction of solute.

Thus, poA–ppoA = xB = nBnA+nB

For a dilute solution, nB << nA, hence neglecting nB in the denominator, we have

poA–ppoA = nBnA=WB×MAMB×WA

MB = (poApoA–p)×WB×MAWA

(b) Elevation of boiling point: Boiling point is the temperature at which the vapour pressure of a liquid becomes equal to the atmospheric pressure. When a non-volatile solute is added to a volatile solvent, the vapour pressure of the solvent decreases. In order to make this solution boil, its vapour pressure must be increased by raising the temperature above the boiling point of the pure solvent. The difference in the boiling point of solution (Tbo) and that of pure solvent (Tb) is called elevation of boiling point (DTb). Thus,

Elevation of boiling point = DTb = Tb – Tbo (Fig. 2.4)

For dilute solution, it has been found that the elevation of boiling point is directly proportional to the molal concentration of the solute in the solution. Thus,

DTb m or DTb = Kb m

where Kb is a constant called molal elevation constant or ebullioscopic constant.

When m = 1 mol kg–1, DTb = Kb

Hence, molal elevation constant may be defined as the elevation in boiling point when one mole of a non-volatile solute is dissolved in one kilogram (1000 g) of solvent. The unit of Kb is K kg mol–1.

As m = WB×1000MB×WA

Therefore, DTb = Kb×WB×1000MB×WA

MB = Kb×WB×1000ΔTb×WA

(c) Depression of freezing point: Freezing point is defined as the temperature at which the vapour pressure of a substance in its liquid phase is equal to its vapour pressure in the solid phase. A solution freezes when its vapour pressure equals the vapour pressure of the pure solid solvent. Whenever a non-volatile solute is added to the volatile solvent, its vapour pressure decreases and it would become equal to that of solid solvent at a lower temperature. The difference in the freezing point of pure solvent (T fo) and that of the solution (Tf ) is known as depression of freezing point (DTf ). Thus,

Depression of freezing point = DTf = T fo – Tf .

It has been found that for a dilute solution, depression in freezing is directly proportional to molality of the solution.

DTf m or DTf = Kf m

where Kf is a constant called molal depression constant or cryoscopic constant.

When m = 1 mol kg–1, DTf = Kf

Hence, molal depression constant may be defined as the depression in freezing point when one mole of non-volatile solute is dissolved in one kilogram (1000 g) of solvent. The unit of Kf is K kg mol–1.

Since, m = WB×1000MB×WA

\ DTf = Kf×WB×1000MB×WA

or, MB = Kf×WB×1000ΔTf×WA

The values of Kb and Kf which depend upon the nature of the solvent and concentration of the solution, can be ascertained from the following relations:

Kb = R×MA×(Tob)2△vapH×1000

Kf = R×MA×(Tof)2△fusH×1000

where, R = Universal gas constant

MA = Molecular mass of solvent

Tbo = Boiling point of pure solvent

Dvap H = Enthalpy of vapourisation of solvent

T fo = Freezing point of pure solvent

Dfus H = Enthalpy of fusion of solid solvent

(d) Osmosis: When a solution is separated from its solvent by a semipermeable membrane (SPM) there is a spontaneous flow of solvent molecules from solvent compartment to solution compartment. The phenomenon is called osmosis. This movement of solvent is only in one direction. In diffusion, however, movement takes place in both the directions.

 Semipermeable membrane (SPM): A membrane through which only solvent molecules can pass but not the solute ones. Cellophane, parchment paper and the wall of living cell are the examples of SPM. It is also made of inorganic material, copper ferrocyanide [Cu2[Fe(CN)6]].

 Osmotic pressure (p): 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 into it through semipermeable membrane (Fig. 2.6).

Osmotic pressure (p) is proportional to molarity (C) of the solution at a given temperature T. Thus,

p = CRT

or, p = nBVRT=WB×R×TMB×V

or, MB = WB×R×Tπ×V

 Reverse osmosis: If a pressure larger than the osmotic pressure is applied to the solution side, the pure solvent (or water) flows out of the solution through the semipermeable membrane. In this way the direction of osmosis is reversed and so the process is called reverse osmosis (Fig. 2.7).

Thus, we can say that reverse osmosis is just opposite to the osmosis when an extra pressure is applied. Reverse osmosis is used in desalination to get pure water from sea water.

 Isotonic solutions: Two solutions are said to be isotonic when they exert the same osmotic pressure because they have the same molar concentration. All intravenous injections must be isotonic with body fluids.

 Isosmotic solutions: When two isotonic solutions are separated by a semipermeable membrane, no osmosis occurs. The solutions are called isosmotic solutions.

 Hypotonic solutions: A solution having lower osmotic pressure than the other solution is said to be hypotonic with respect to the other solution.

 Hypertonic solution: A solution having higher osmotic pressure than the other solution is said to be hypertonic with respect to other solution.

 Plasmolysis: When the cell is placed in a hypertonic solution, the fluid comes out of the cell due to osmosis and as a result cell material shrinks gradually. This process is called plasmolysis.

7. Abnormal Molar Masses

 Association: Association of molecules leads to decrease in the number of particles in the solution resulting in a decrease in the value of colligative property. As colligative property is inversely related to the molecular mass. Therefore, higher value is obtained for molecular mass than normal values.

For example, when ethanoic acid is dissolved in benzene it undergoes dimerisation and shows a molecular mass of 120 (normal molecular mass is 60).

 Dissociation: Dissociation leads to increase in the number of solute particles in the solution resulting in an increase in the value of colligative property. Since colligative property is inversely related to the molecular mass, therefore, molecular mass of such a substance as calculated from colligative property will be less than its normal value. For example, KCl is an electrolyte. When it is dissolved in water it dissociates into K+ and Cl– ions and there would be double the number of particles if complete dissociation takes place. Hence, it is expected to have molecular mass 37.25 g or (74.52)g.

KCl  K+ + Cl–

 van’t Hoff Factor (i): It may be defined as the ratio of normal molecular mass to the observed molecular mass of the solute.

i = NormalmolecularmassObservedmolecularmass

or, i = ObservedcolligativepropertyCalculatedvalueofcolligativeproperty

or, i = Totalnumberofmolesofparticlesafterassociation/dissociationNumberofparticlesbeforeassociation/dissociation

van’t Hoff factor (i) expresses the extent of association or dissociation of the solute particles in the solutions.

In case of association, i < 1

In case of dissociation, i > 1

When there is neither association nor dissociation, i = 1

8. Modified Form of Colligative Properties

Inclusion of ‘i’ modifies the equation for colligative properties as follows:

 Relative lowering of vapour pressure of solvent

poA–pApoA = i nBnA

 Elevation of boiling point, DTb = i Kb m

 Depression of freezing point, DTf = i Kf m

 Osmotic pressure, p = i CRT

Important Formulae

In the formulae given below, A represents solvent and B represents solute, also

MA = Molar mass of solvent

MB = Molar mass of solute

WA = Mass of solvent

WB = Mass of solute

V = Volume of solution

d = Density of solution

GEM = Gram Equivalent Mass

GMM = Gram Molecular Mass

1. Mass percentage (w/w) = WBWA+WB × 100

Volume percentage (V/V) = VBVA+VB × 100

Mass by volume percentage (wV)=WB×100V(mL)

Parts per million (ppm) = WBWA+WB × 106

2. Mole fraction of A, xA = nAnA+nB

Mole fraction of B, xB = nBnA+nB

xA + xB = 1

3. Molarity (M) = MolesofsoluteVolumeofsolutioninlitre=nBV(inL)=WBMB×V(inL)

4. Molality (m) = MolesofsoluteMassofsolventinkg = nBWA(inkg) or m = WB×1000MB×WA(ing)

5. Normality (N) = GramequivalentsofsoluteVolumeofsolutioninlitre = WBGEMofsolute×V(inL)

6. Relationship between Molarity and Normality

The normality (N) and molarity (M) of a solution are related as follows:

Normality × Equivalent mass (solute) = Molarity × Molar mass (solute)

7. Relationship between Molarity and Normality with Mass percentage (p)

If p is the mass percentage and d is the density of the solution then

Molarity = p×d×10Molecularmass(solute);

Normality = p×d×10Equivalentmass(solute)

8. Relationship between Molarity (M) and Molality (m)

m = 1000×M(1000×d)–(M×GMMB)

9. Relationship between Molality (m) and Mole fraction of solute (xB)

xB = m×GMMA1000+m×GMMA

Also, m = 1000xBxA×GMMA

10. Dilution formula: If the solution of some substance is diluted by adding solvent from volume V1 to volume V2 then

M1V1 = M2V2

Similarly, N1V1 = N2V2

11. Molarity of a mixture: If V1 mL of a solution of molarity M1 is mixed with another solution of same substance with volume V2 and molarity M2 then molarity of the resulting mixture of solution (M) can be obtained as:

M = M1V1+M2V2V1+V2

12. Relationship between molarity (M) and mole fraction of solute (xB)

xB = M×GMMAM(GMMA–GMMB)+1000d

Also, M = 1000×d×xBxA×GMMA+xB×GMMB

13. Raoult’s law for volatile solute

pA = pAoxA and pB = pBoxB

where pA and pB are partial vapour pressures of component ‘A’ and component ‘B’ in the solution. pAo and pBo are vapour pressures of pure components ‘A’ and ‘B’ respectively.

Total vapour pressure, p = pA + pB = pAoxA + pBoxB

14. Raoult’s law for non-volatile solute

poA–ppoA = xB = nBnA+nB=nBnA=WBMB×MAWA (For a dilute solution nB <<nA).

MB = (poApoA–pA)WB×MAWA

where xB is mole fraction of solute and poA–pApoA is relative lowering of vapour pressure.

15. Elevation in boiling point:

DTb = Kb × m

DTb = Kb×WB×1000MB×WA

or, MB = Kb×WB×1000ΔTb×WA

where, DTb = Tb – Tbo

16. Depression in freezing point:

DTf = Kf × m

DTf = Kf×WB×1000MB×WA

or, MB = Kf×WB×1000△Tf×WA

17. Osmotic pressure (p)

pV = nBRT

pV = WBMB×R×T,MB=WB×R×Tπ×V

p = nBV × R × T or p = CRT where ‘C’ is molarity.

Osmotic pressure is related to the relative lowering of vapour pressure, elevation in boiling point and depression in freezing points according to the following relations:

p = (poA–pApoA)×d×R×TMB

p = ΔTb×d×R×T1000×Kb

p = ΔTf×d×R×T1000×Kf

where d is the density of solution at temperature ‘T’.

18. van’t Hoff factor

i = NormalmolecularmassObservedmolecularmass

or, i = ObservedcolligativepropertyCalculatedcolligativeproperty

or, i = Totalnumberofmolesofparticlesafterassociation/dissociationNumberofmolesofparticlesbeforeassociation/dissociation

Modified forms of colligative properties:

(a) poA–pApoA = i xB

(b) DTb = i Kb m

(c) DTf = i Kf m

(d) pV = i nB RT

19. α=i–1n–1, where a is degree of dissociation, ‘i’ is van’t Hoff factor, ‘n’ is number of ions produced per formula of the compound.

20. α=1–i1–1n or α=i–11n–1

where a is degree of association, n is the number of molecules of solute that associate to form an associated molecule, 1n < 1.

21. Molal elevation constant, Kb = R×MA×(Tob)21000×△vapH

22. Molal depression constant, Kf = R×MA×(Tof)21000×△fusH