# Notes for Class 12 Chemistry Chapter 2 Solutions in pdf

### Notes for Class 12 Chemistry Chapter 2 Solutions in pdf

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#### Notes Chapter 2 Solutions

A solution is a homogenous mixture of two or more non-reacting compounds whose composition can be varied within certain limits.

Depending upon the total components present in the solution, A solution may be binary solution (Two component), ternary solution (Three components), quaternary solution (Four components)

The components of binary solution are generally referred to as solute and solvent. The component which is present in large quantity is called Solvent while the component which is present in lesser quantity is called solute.

##### Types of Solutions

Depending on the physical states of solute and solvent, the solution can be of following types:

##### Methods for expressing the concentration of a Solution

The concentration of a solution may be defined as the amount of solute present in the given quantity of the solution. The concentration of solution can be expressed in various common ways as discussed below :

###### 1. Mass percentage (w/w) :

It may be defined as mass of a component in grams present in 100 g of the solution.

$$\frac{Mass~of~component~in~solution~\times 100}{Total~mass~of~solution}$$ =  $$\frac{W_{A}\times 100}{W_{sol}}$$

###### 2. Volume percentage (v/v) :

It may be defined as volume of a component in ml present in 100 ml of the solution.

$$\frac{Volume~of~component~in~solution~\times 100}{Total~volume~of~solution}$$  = $$\frac{V_{A}\times 100}{V_{sol}}$$

###### 3. Mass by volume percentage (w/v) :

It can be defined as the mass of component in grams dissolved in 100 ml of solution.

$$\frac{Mass~of~component~in~solution~\times 100}{Total~volume~of~solution~in~ml}$$ = $$\frac{W_{A}\times 100}{V_{sol}}$$

###### 4. Parts per million (ppm) :

It can be defined as the part of a component per million parts of the solution.

$$\frac{Parts~of~component~in~solution~\times 10^{6}}{Total~parts~of~solution}$$ = $$\frac{W_{A}\times 10^{6}}{W_{sol}}$$

###### 5. Strength (gm/L) :

It may be defined as the amount of solute in grams present in 1 litre of the solution.

$$S=\frac{Weight~of~the~solute~in~grams}{Volume~of~solution~in~litres}$$

$$S = \frac{W_{B}}{V_{sol~in~L}} = \frac{M_{B}\times 1000}{V_{sol~in~ml}}$$

###### 6. Molarity (mol/L) :

It may be defined as number of moles of solute dissolved in 1 litre of the solution. It is denoted by ‘M’.

M =  $$\frac{Number~of~moles~of~solute}{Volume~of~the~solution~in~Litres}$$

M = $$\frac{n_{B}}{V_{sol~in~L}}=\frac{n_{B}\times 1000}{V_{sol~in~ml}}~~~~\left [ \because ~n_{B}=\frac{W_{B}}{M_{B}} \right ]$$

M = $$\frac{W_{B}\times 1000}{M_{B}\times V_{sol~in~ml}}~=~\frac{Strength}{M_{B}}$$

###### 7. Molality (mol/Kg) :

It may be defined as the number of moles of solute dissolved in 1000 g (or 1 Kg) of the solvent. It is denoted by ‘m’.

m =  $$\frac{Number~of~moles~of~solute}{Weight~of~the~solvent~in~Kg}$$

m = $$\frac{n_{B}}{W_{A~in~Kg}}=\frac{n_{B}\times 1000}{W_{A~in~gm}}~~~~\left [ \because ~n_{B}=\frac{W_{B}}{M_{B}} \right ]$$

Molality (m) = $$\frac{W_{B}\times 1000}{M_{B}\times W_{A~in~gm}}$$

###### 8. Mole fraction : 'χ' (chi)

Mole fraction of solvent (χA)  =  $$\frac{n_{A}}{n_{A}~+~n_{B}}$$

Mole fraction of solute (χB)  =  $$\frac{n_{B}}{n_{A}~+~n_{B}}$$

The sum of mole fraction of all components in solution is always equal to 1,

χA + χB = 1

$$\frac{n_{A}}{n_{A}~+~n_{B}}~+~\frac{n_{A}}{n_{A}~+~n_{B}}$$ = 1

###### 9. Normality (gm eq.wt./L) :

It may be defined as the number of gram equivalents of the solute dissolved in 1 litre of the solution. It is denoted by ‘N’.

N =  $$\frac{Number~of~gram~equivalent~of~solute}{Volume~of~solution~in~Litre}$$

N =  $$\frac{Number~of~gram~equivalent~of~solute~\times ~1000}{Volume~of~solution~in~ml}$$

Gram equivalent of solute =  $$\frac{W_{B}}{Eq.Wt._{B}}$$

Normality (N) =  $$\frac{W_{B}~\times ~1000}{Eq.wt._{B}~\times ~V_{sol~in~ml}}$$

Equivalent weight of solute =  $$\frac{M_{B}}{‘n’~factor}$$

Normality (N) =  $$\frac{W_{B}~\times ~1000 ~\times ~’n’~factor}{M_{B}~\times ~V_{sol~in~ml}}$$

Normality (N) = Molarity $$\times$$ ‘n’ factor

Normality (N) =  $$\frac{Strength~\times ~’n’~factor}{M_{B}}$$

How to determine ‘n’ factor of solute ?

‘n’ factor of acid  = Basicity of the acid (or number of H+)

‘n’ factor of base = Acidity of the base (or number of OH)

‘n’ factor of ion = Charge on the ion

‘n’ factor of salt  =  Total cationic charge or Total anionic charge per molecule

‘n’ factor of an element  = Valency of the element

‘n’ factor of oxidising or reducing agent  = No. of  gain or loss by 1 molecule

##### Factors Affecting the Solubility of a Solid in a Liquid
###### a.) Nature of the solute and the solvent

A solid dissolve in a liquid (solvent) if the intermolecular interactions are similar in solute and solvent. This is expressed by saying “Like dissolves like”. This means that, non-polar substances are more likely to be soluble in non-polar solvents while ionic (or polar) substances are more likely to be soluble in polar solvents.

###### b.) Effect of Temperature :

The solubility may increase or decrease or show irregular behaviour with increase in temperature. We observe following three behaviours:

• The solubility of solids increases with increase in temperature: The solubility of most of substances such as NaNO3 , NH4Cl , KCl , AgNO3 , NaCl, KI and sugar etc. increases with rise in temperature. This is because the dissolution process for these substances is endothermic.
• The solubility of solids decreases with increase in temperature:  The solubility of some substances like Li2SO4 , Ce2(SO4)3 , Na2CO3 and CaO etc. decrease with rise in temperature. This is because dissolution process is exothermic
• The solubility shows irregular behavior with increase in temperature: The solubility of some substances (like Na2SO4) increases up to a certain temperature and then decreases as the temperature is further raised. The temperature at which solubility starts decrease is called transition temperature.
###### c.) Effect of Pressure :

The effect of pressure on the solubility of solids in liquid is very small or insignificant. This is because solids and liquids are incompressible and practically remain unaffected by changes in pressure.

##### Factors Affecting the Solubility of a Gas in a Liquid
###### a.) Nature of the gas and the solvent

Gases that can react chemically or are capable of forming ions in aqueous solution are more soluble in water than in any other solvent. Example:

• Gases like H2, O2, N2 dissolve in water in very small amounts whereas the gases like SO2, H2S, HCl, NH3 etc. are highly soluble.
• O2, N2 and CO2 are more soluble in ethyl alcohol than in water while gases such as NH3 and H2S are more soluble in water than in ethyl alcohol.
###### b.) Effect of Temperature :

Gases dissolve in a liquid by exothermic process. Therefore, according to Le-Chatelier’s principle, an increase in temperature will result in a decrease in the solubility of a gas if the process is exothermic.

###### c.) Effect of Pressure :

The solubility of gases increases with increase of pressure.

William Henry gave quantitative relationship between the solubility of a gas in a solvent and the pressure of a gas on a solution, known as Henry’s law.

The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas on that solution at equilibrium.

If we express the solubility of a gas in terms of mole fractions, then Henry’s law can be written as:

Thus, most commonly used form of Henry’s law may be defined as :

The pressure of a gas over a solution in which the gas is dissolved is directly proportional to the mole fraction of the gas dissolved in the solution.

Mathematical Modifications to Henry’s Law

According to Henry’s Law :  Pgas = KH . χgas

χgas  =   $$\frac{n_{gas}}{n_{solution}}~=~\frac{n_{gas}}{n_{solvent}~+~n_{gas}}$$

[nsolvent >> ngas  , So, nsolvent + ngas ≅ nsolvent ]

Pgas =  $$\frac{K_{H}~\times ~n_{gas}}{n_{solvent}}$$

Pgas =  $$\frac{K_{H}~\times ~n_{gas}~\times ~M_{A}}{W_{A}}$$

Pgas =  $$\frac{K_{H}~\times ~W_{gas}~\times ~M_{A}}{M_{gas}~\times ~W_{A}}$$

Properties of Henry’s constant (KH) :

1. KH is a function of nature of gas. Different gases have different KH values at same temperature.
2. Greater the value of KH, lower the solubility of the gas.
3. The value of KH increases with increase in temperature.
###### Applications of Henry’s law :

a.) In the production of carbonated beverages:

To increase the solubility of CO2 in soft drinks, soda water, beer or champagne, the bottles are sealed under high pressure. When the bottle is opened under normal atmospheric conditions, the pressure inside the bottle falls and the excess CO2 bubbles out of the solution.

b.) In deep sea diving (Scuba diving):

Scuba divers depends upon compressed air for breathing under water. According to Henry’s law, increased pressure increases the solubility of atmospheric gases in blood. Oxygen is used up for metabolism, but nitrogen will remain dissolved in blood. When the divers come towards the surface, the pressure gradually decreases. N2 comes out of the body quickly forming bubbles in the blood stream. These bubbles restrict blood flow and can even burst or blocks capillaries, which cause painful and dangerous condition called the bends or decompression sickness.

To avoid bends as well as the toxic effects of high concentration of nitrogen in blood, most divers these days use air diluted with helium gas (about 11.7 % He, 56.2 % N2 and 32.1 % O2).

c.)  At high altitudes:

At high altitudes the partial pressure of O2 is less than that of the ground level. This results in low concentration of O2 in the blood and tissue of the climbers. They feel weak and are unable to think properly, a disease called  anoxia.

d.)  Combination of haemoglobin and oxygen in lungs:

Haemoglobin combines with oxygen to form oxyhaemoglobin, which is under high partial pressure in the lungs. In various tissues, where the partial pressure of oxygen is low, this oxyhaemoglobin releases oxygen for use in cellular activities.

##### Vapour Pressure of Liquid Solutions

When a liquid is allowed to evaporate in a closed vessel, a part of the liquid evaporates. It fills the available space with vapours and level of liquid decreases. This process is called vaporisation. Some molecules of vapour strike the surface of liquid and get condensed. The process of condensation acts in opposite direction to the process of evaporation. Ultimately, a stage is reached when the rate of evaporation becomes equal to rate of condensation and equilibrium gets established between liquid and vapour phase.

The pressure exerted by the vapours above the liquid surface at the time of equilibrium at a given temperature is called vapour pressure.

###### Factors Affecting Vapour Pressure :

(A)  Nature of liquid :   The liquids, which have weaker intermolecular forces, have greater vapour pressure because more molecules tend to escape into vapour phase. Example :  Dimethyl ether and ethyl alcohol have greater vapour pressure than water because of weaker intermolecular forces as compared to water

(B)  Temperature :  The vapour pressure of liquid increases with increase in temperature. This is because more molecules will have larger kinetic energies and escape readily into vapour phase resulting in higher vapour pressure.

###### Vapour Pressure of Liquid-Liquid Solutions

When a binary solution of two volatile liquids is placed in a closed vessel, both components would evaporate and eventually an equilibrium would be established between liquid and vapour phase.

The total vapour pressure (PTotal) in the container will be the sum of the partial vapour pressures (PA and PB) of both components of the solution.

###### PTotal  =  PA  +  PB

According to Raoult’s law : For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to their mole fraction in solution.

Then, according to Raoult’s law :

$$P_{A} \propto \chi _{A}~~~and~~~P_{A}=P_{A}^{o}.\chi _{A}$$

$$P_{B} \propto \chi _{B}~~~and~~~P_{B}=P_{B}^{o}.\chi _{B}$$

Where:

$$P_{A}~and~P_{B}$$ are the partial vapour pressures of components in solution.

$$P_{A}^{o}~and~P_{B}^{o}$$ are the vapour pressures of components in Pure state.

$$\chi _{A}~and~\chi _B$$ are the mole fractions of components in solution

According to Dalton’s law of partial pressures, the total vapour pressure (PTotal) over the solution in a closed vessel will be sum of the partial vapour pressures (PA and PB) of both components of solution. This is given as :

###### PTotal  =  PA + PB

Substituting the values of PA and PB, we get

$$P_{Total}=\left [ P_{A}^{o}.\chi _{A} \right ]~+~\left [ P_{B}^{o}.\chi _{B} \right ]$$

As we know,  $$\chi _{A}=1-\chi _{B}$$

$$P_{Total}=\left [ P_{A}^{o}.\left ( 1-\chi _{B} \right ) \right ]~+~\left [ P_{B}^{o}.\chi _{B} \right ]$$

$$P_{Total}=P_{A}^{o}-P_{A}^{o}.\chi _{B}~+~P_{B}^{o}.\chi _{B}$$

$$P_{Total}=P_{A}^{o}~+~\left ( P_{B}^{o}-P_{A}^{o} \right )\chi _{B}$$

The composition of vapour phase in equilibrium can be determined by the partial pressure of components. If YA and YB are the mole fractions of components A and B respectively in vapour phase then, using Dalton’s law of partial pressures :

$$Y_{A}=\frac{P_{A}}{P_{A}+P_{B}}=\frac{P_{A}}{P_{Total}}$$

and

$$Y_{B}=\frac{P_{B}}{P_{A}+P_{B}}=\frac{P_{B}}{P_{Total}}$$

###### Vapour Pressure of Solid-Liquid Solutions

When the solute is non-volatile, only the solvent molecules are present in the vapour phase. Therefore, the vapour pressure of solution is due to solvent only.

###### PTotal  =  PA

According to Raoult’s law, the partial vapour pressure of volatile component in the solution is directly proportional to the mole fraction in it.

$$P_{A}=P_{A}^{o}.\chi _{A}$$

Now:        $$P_{Total}=P_{A}=P_{A}^{o}.\chi _{A}$$

The above relationship may also be put forward in different ways as :

$$P_{A}=P_{A}^{o}.\chi _{A}~~~\left ( \therefore \chi _{A}=1-\chi _{B} \right )$$

$$P_{A}=P_{A}^{o}\left (1-\chi _{B} \right )=P_{A}^{o}-P_{A}^{o}\chi _{B}$$

$$P_{A}^{o}\chi _{B}=P_{A}^{o}-P_{A}~~,~~\chi _{B}=\frac{P_{A}^{o}-P_{A}}{P_{A}^{o}}$$

In this expression, $$\Delta P_{A}=P_{A}^{o}-P_{A}$$ expresses the lowering of vapour pressure while $$\frac{P_{A}^{o}-P_{A}}{P_{A}^{o}}$$ is called relative lowering in vapour pressure and χB represents mole fraction of the solute in solution.

##### Ideal and Non-Ideal Solutions :
###### A.) Ideal Solutions

An ideal solution may be defined as the solution which obeys Raoult’s law under all conditions of temperatures and concentrations. Such solutions are formed by mixing the two components which are identical in molecular size and have almost identical intermolecular forces.

The ideal solutions have the following characteristics :

1. Heat change on mixing is zero (ΔHmixing = 0)
2. Volume change on mixing is zero (ΔVmixing = 0)
3. Obeys Raoult’s law

Examples of ideal solution :

• Benzene + Toluene
• n-Hexane + n-Heptane
• Ethyl chloride + Ethyl bromide
• Chlorobenzene + Bromobenzene
###### B.) Non-Ideal Solutions

The solutions which do not obey Raoult’s law under all conditions of temperatures and concentrations are called non-ideal solutions.

The non-ideal solutions have the following characteristics :

1. Heat change on mixing is not equal to zero (ΔHmixing ≠ 0)
2. Volume change on mixing is not equal to zero (ΔVmixing ≠ 0)
3. Do not obey Raoult’s law

Types of non-ideal solution :

1. Non-ideal solution showing positive deviations from Raoult’s law.
2. Non-ideal solution showing negative deviations from Raoult’s law.
###### Non-ideal solution showing positive deviation from Raoult’s law

When the partial pressure of components is found to be more than expected on the basis of Raoult’s law. As a result, total vapour pressure of the solution is greater than expected.

The intermolecular interactions between components A – B are of less magnitude as in pure components A – A and B – B.

Characteristics :

1. Heat change on mixing is always positive (ΔHmixing = positive)
2. Volume change on mixing is also always positive (ΔVmixing = positive)
3. Partial pressure is found to be more than expected
4. Behave as minimum boiling azeotropes

Examples :

• Acetone + Carbon disulphide
• Acetone + Ethyl alcohol
• Acetone + Benzene
• Methyl alcohol + Water
###### Non-ideal solution showing negative deviation from Raoult’s law

When the partial pressure of components is found to be less than expected on the basis of Raoult’s law. As a result, total vapour pressure of the solution is lower than expected.

The intermolecular interactions between components A – B are of greater magnitude as in pure components A – A and B – B.

Characteristics :

1. Heat change on mixing is always negative (ΔHmixing = negative)
2. Volume change on mixing is also always negative (ΔVmixing = negative)
3. Partial pressure is found to be less than expected
4. Behave as maximum boiling azeotropes.

Examples :

• Chloroform + Acetone
• Acetone + Aniline
• Chloroform + Benzene
• HCl + Water
• Chloroform + Diethyl ether
##### Azeotropes

Azeotropes are the binary solution having the same composition in both liquid and vapour phase, and boil at a constant temperature. In such cases, it is not possible to separate the components by fractional distillation method.

###### There are two types of Azeotropes :

Minimum boiling azeotropes or positive azeotropesWhen the boiling point of the azeotrope is less than the boiling points of the pure components, then it is called minimum boiling azeotropesExample :  95.4 % solution of ethanol with water boil at 351.15 K

Maximum boiling azeotropes or negative azeotropesWhen the boiling point of the azeotrope is more than the boiling point of the pure components, then it is called maximum boiling azeotropesExample :  68 % solution of nitric acid with water boil at 393.5 K.

##### Colligative Properties

Those properties of ideal solutions which depend only on the number of solute particles but do not depend on the nature of solute are called colligative properties.

The important colligative properties are :

###### 1. Relative lowering of vapour pressure :

$$~\chi _{B}=\frac{P_{A}^{o}-P_{A}}{P_{A}^{o}}$$

In this expression, $$\Delta P_{A}=P_{A}^{o}-P_{A}$$ expresses the lowering of vapour pressure while $$\frac{P_{A}^{o}-P_{A}}{P_{A}^{o}}$$ is called relative lowering in vapour pressure and χB represents mole fraction of the solute in solution.

$$\frac{W_{B}~\times ~M_{A}}{M_{B}~\times ~W_{A}}=\frac{P_{A}^{o}-P_{A}}{P_{A}^{o}}$$

###### 2. Elevation in Boiling Point :

The boiling point of liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure.

The elevation in boiling point (ΔTb) depends on molal concentration (m) of solute in solution.

###### ΔTb ∝ m     or    ΔTb = Kb × m

Where Kb is called the molal elevation constant or ebullioscopic constant and ‘m’ is the molality.

$$\Delta T_{b}=K_{b}\times \frac{W_{B}~\times ~1000}{M_{B}~\times W_{solvent~in~gm}}$$

Note: $$\Delta T_{b}=T_{b}-T_{b}^{o}$$

###### 3. Depression in Freezing Point :

Freezing point of a substance is the temperature at which the solid and liquid forms of the substance have same vapour pressure. It is found that the freezing point of the solution is always lower than that of pure of solvent.

The depression in freezing point (ΔTf) depends on molal concentration (m) of solute in solution.

###### ΔTf ∝ m     or    ΔTf = Kf × m

Where Kf is called molal depression constant or cryoscopic constant and ‘m’ is the molality.

$$\Delta T_{f}=K_{f}\times \frac{W_{B}~\times ~1000}{M_{B}~\times W_{solvent~in~gm}}$$

Note: $$\Delta T_{f}=T_{f}^{o}-T_{f}$$

###### Applications of Depression of freezing point :
• In making antifreeze solutions:

The running of a vehicle (car or bus) in sub-zero weather even when the radiator is full of water, which has freezing point 0 ˚C or 273 K, has been possible due to the fact that depression in freezing point of water takes place by addition of some quantity of ethylene glycol (called antifreeze) in water.

• In melting of ice on roads:

In winter, where snowfall occurs heavily, the ice deposited on roads is molten by scattering common salt (NaCl) or CaCl2 on the roads. This is because salt-ice mixture has very low freezing point and hence ice keeps on melting.

###### 4. Osmotic Pressure :

Osmosis is the spontaneous process of movement of solvent only, from its higher concentration to lower concentration through semi-permeable membrane.

Osmotic Pressure :

It can be defined as the additional pressure applied on the solution to prevent the entry of the solvent into the solution through semi-permeable membrane. It is represented by ‘π.

Reverse Osmosis (R.O.) :

If a pressure higher than the osmotic pressure is applied on the solution, the solvent will flow from the solution into the pure solvent through the semi-permeable membrane. The process is called reverse osmosis (RO).

Osmotic pressure (π) of a solution is found to be directly proportional to molar concentration (C) of the solution and its temperature (T).

π ∝ C —(i)                 π ∝ T —(ii)

Compare the equation (i) and (ii)

π ∝ CT        or         π = RCT

Where ‘R’ is a constant and its value is found to be same as that of the ‘Gas constant’. ‘C’ is molar concentration (n/V). ‘n’ is number of moles of solute and ‘V’ is the volume of solution in litre. The above equation is usually written as :

$$\pi = \frac{n}{V}.RT~~~or~~~\pi V=nRT$$

Determination of Molecular mass from Osmotic Pressure :

$$\left [ n=\frac{W_{B}}{M _{B}} \right ] ~~~~~~~ \pi V = \frac{W_{B}}{M _{B}}.RT$$

or

$$M _{B}=\frac{W_{B}\times RT}{\pi~ \times V}$$

Hypertonic, Hypotonic and Isotonic solutions :

• The solution which has high osmotic pressure with respect to another solution is called hypertonic solution. It contains relatively large quantity of solute.
• The solution which has low osmotic pressure with respect to another solution is called hypotonic solution. It contains relatively small quantity of solute.
• The solution which has same osmotic pressure with respect to another solution is called isotonic solution. It has equimolar concentration of solute.

Biological Importance of Osmosis :

• Plants absorb water from the soil through their roots due to osmosis.
• In animals, water moves into different parts of the body due to osmosis.
• Bursting of RBCs when placed in water is also due to osmosis.
• Opening and closing of stomata of plant leaves is also due to osmosis.
• Raw mangoes shrink to pickle when placed in common salt is due to osmosis.
• Wilted flowers revive when placed in fresh water is due to osmosis.

By study of osmosis, it helps us to conclude the following :

• When RBCs are placed in hypotonic solution, having salt concentration less than 0.9 %, they will start swell up due to endosmosis.
• When RBCs are placed in hypertonic solution, having salt concentration more than 0.9 %, they will start shrink due to exosmosis.
• When RBCs are placed in isotonic solution, having salt concentration 0.9 %, the cell never swell and nor shrink because no osmosis takes place.

Remember that osmosis takes place in the direction of :

• High solvent concentration to Low solvent concentration
• Low solute concentration to high solute concentration
• Less concentrated solution to More concentrated solution
• Low osmotic pressure to high osmotic pressure
##### Abnormal Molecular Masses

When the molecular mass of a substance determined by studying any of the colligative properties comes out to be different than the theoretically expected value, then the substance is said to show abnormal molecular mass.

The abnormal molecular masses are observed in any one of the following cases :

• When the solute undergoes dissociation in the solution.
• When the solute undergoes association in the solution.
• When the solution is non-ideal.

To calculate the extent of association or dissociation, Van’t Hoff introduced a factor ‘i’ called Van’t Hoff factor. It is defined as the ratio of the observed or experimental value of the colligative property to the calculated or normal value of the colligative property.

$$i=\frac{Observed~Colligative~Property}{Calculated~Colligative~property}=\frac{C_{o}}{C_{c}}$$

As we know that, colligative property is inversely proportional to the molecular mass of the solute, we can also write:

$$i=\frac{Calculated~Molecular~Mass}{Observed~Molecular~Mass}=\frac{M_{c}}{M_{o}}$$

Note:

• i > 1  ::  Dissociation
• i < 1  ::  Association
• i = 1  ::  No association nor dissociation
###### Modified Expressions of Colligative Properties :

Relative Lowering in Vapour pressure:

$$\frac{\Delta P_{A}}{P_{A}^{o}}=i.\chi _{B}~~~OR~~~\frac{\Delta P_{A}}{P_{A}^{o}}=~\frac{i~\times W_{B}\times M_{A}}{M_{B}\times W_{A}}$$

Elevation in Boiling point:

$$\Delta T_{b}=i.K_{b}.m$$

Or

$$\Delta T_{b}=\frac{i~\times K_{b}\times W_{B}\times ~1000}{M_{B}\times W_{A}}$$

Depression in Freezing point:

$$\Delta T_{f}=i.K_{f}.m$$

Or

$$\Delta T_{f}=\frac{i~\times K_{f}\times W_{B}\times ~1000}{M_{B}\times W_{A}}$$

Osmotic pressure:

$$\pi =\frac{i~\times W_{B}\times RT\times ~1000}{M_{B}\times V_{sol~in~ml}}$$

###### 1. Calculation of the degree of Dissociation :

When an electrolyte is dissolved in a solvent, it dissociates into ions. The fraction of the total number of electrolyte molecules that undergoes dissociation is called degree of dissociation (α).

$$\alpha =\frac{Number~of~moles~dissociated}{Total~number~of~moles~taken}$$

Observed colligative property = (1-α)+nα = 1+(n-1)α

Calculated colligative property = 1

$$i=\frac{C_{o}}{C_{c}}=\frac{1+(n-1)\alpha }{1}~~OR~~\alpha =\frac{i-1}{n-1}$$

As we know that,  $$i=\frac{M_{c}}{M_{o}}$$

$$\therefore~~ \alpha =\frac{\frac{M_{c}}{M_{o}}-1}{n-1}=\frac{M_{c}-M_{o}}{M_{o}(n-1)}$$

###### 2. Calculation of the degree of Association :

In some cases, ‘n’ molecules of the solute ‘A’ associate to form large associated molecule ‘An’. The fraction of the total number of solute molecules which exist in the form of associated molecule is called degree of association (α).

$$\alpha =\frac{Number~of~moles~associated}{Total~number~of~moles~taken}$$

Observed colligative property =  $$(1-\alpha )+\frac{\alpha }{n}$$

Calculated colligative property = 1

$$i=\frac{C_{o}}{C_{c}}=\frac{(1-\alpha )+\frac{\alpha }{n}}{1}$$

$$\alpha =(1-i) \frac{n}{n-1}$$

As we know that,  $$i=\frac{M_{c}}{M_{o}}$$

$$\therefore ~~~\alpha = \left ( 1-\frac{M_{c}}{M_{o}} \right )\frac{n}{n-1}= \left ( \frac{M_{o}-M_{c}}{M_{o}} \right )\frac{n}{n-1}$$