Class 12 Chemistry Chapter 2 Notes – Solution

The Class 12 Chemistry Chapter 2 Notes deal with the topic of Solutions. In everyday life, we come across pure substances on some occasions. Furthermore, most of these mixtures contain multiple pure substances whose utility is dependent on their composition. For Example, the properties of Brass are different in comparison to that of Bronze. Liquid solutions and their formations certainly play an essential role in Chemistry.

The Class 12 Chemistry Chapter 2 Notes covers each topic and subtopic related to solid, liquid, and gaseous solutions. It is a scoring topic, meaning it has good marks weightage, and the topic is easy to understand, in turn enabling students to score well. Extramarks Class 12 Chemistry Chapter 2 notes are the finest study materials that students refer to instead of turning pages of their textbooks again and again.

Students may register at Extramarks to refer to Class 12 Chemistry Chapter 2 Notes for free.

Key Topics Covered Under Class 12 Chemistry Chapter 2 Notes

Solutions – chapter 2 Chemistry class 12 notes deal with types of solutions and their properties. It can be classified into three types:

• Solid solutions
• Gaseous solutions
• Liquid solutions

Some of the key topics covered in class 12 Chemistry chapter 2 notes are Raoult’s Law and Henry’s Law, the concentration of solutions, the vapour pressure of liquid solutions, abnormal molar masses, and colligative properties.

The major topics covered in Chemistry chapter 2 class 12 notes are briefly described below:

Introduction 

The solution is a homogeneous mixture of 2 or various substances in the same and different physical phases. The substances forming the solution are known as components of the Solution. A solution of two components is called a binary solution based on the number of components. Under this section of class 12 Chemistry chapter 2 notes, we will focus on liquid solutions and their various properties in this unit.

Solute and Solvent

In a binary solution, the solvent is the component present in large quantities, while the other component is known as Solute. For Example, salt solution. 

• Solution: It is a mixture of two or more components that is homogeneous—for Example, ordinary salt in water.
• Solute: The component present in a lesser amount or whose physical state is changed during the formation of the Solution is called Solute.
• Solvent: The component present in more significant amounts and determines the physical state of the Solution is called a solvent.

The Class 12 Chemistry Chapter 2 notes explain in detail Solvent and Solute. Students can refer to the study material in addition to important questions to understand the topic better. 

Solutions can be broadly categorised in 2 ways:

• Homogeneous solutions:  Solutions with the same composition and properties throughout the Solution are homogeneous solutions. E.g. Solution of NaCl (salt) or sugar in water, cough syrup, a cup of coffee, Mouthwash, perfume is a homogeneous mixture of chemicals and dyes, etc.
• Heterogeneous solutions:  Solutions with not the same composition and properties throughout the Solution are heterogeneous solutions. E.g., solution of oil and water, water and sand,  water and chalk powder, etc.

Classification of Solutions

Under Class 12 Chemistry Chapter 2 notes, the classification of solutions is as follows.

If H2O is used as a solvent, the solution is known as an aqueous solution, and if not, the solution is called a non-aqueous solution. Apart from the class 12 Chemistry chapter 2 notes, students may also refer to NCERT books for more information.

Expressing Concentration Of Solutions

Under this section of class 12 Chemistry chapter 2 notes, students learn about the concentration of solutions. The solution concentration refers to the amount of Solute dissolved per unit of solution. 

• The strength of a solution is the measure of the composition of a solution.
• A solution with a relatively massive quantity of Solute is called a concentrated solution.
• A solution with a relatively minimal quantity of Solute is called a dilute solution.

There are various methods for determining the strength of a solution which are as follows.

1. Mass Percentage (%w/w): “It represents the mass of a component in 100 g of Solution.

Mass % of a component = Mass of the component in the solution  X 100

                                             The total mass of solutions 

For Example Mass percentage of glucose in a 100g solution containing 10 g of glucose dissolved in 90 g of water is 10%.

Mass % of glucose= Mass of glucose in the solution x 100 

                                    The total mass of the Solution

 =10/100×100 = 10 %

The second Example of Commercial bleaching solution contains a 3.62 mass percentage of sodium hypochlorite in water.

2. Volume Percentage (%v/v): As represented in class 12 Chemistry chapter 2 notes, Volume percentage can be defined as “Represents the volume of a component in a solution of 100 mL.

Volume % of a component = Volume  of component X 100

                                                 Total volume of solution

For Example, a 35.5 % (v/v) solution of ethylene glycol, an antifreeze, is utilised in cars and buses for cooling the engine. At this concentration, the antifreeze reduces the freezing point of H2O to 255.4K or –17.6°C.

3. Mass by Volume Percentage %( w / v): “It represents the mass of the solute in grams dissolved  in 100 mL of solutions.”

Mass by volume % = Mass of Solute in g  X 100

                                  The volume of Solution in mL

For Example, Rashida dissolved 40g of sugar in 600 mL of sugar solution. Calculate the mass by volume percentage.

Solution :  Mass by volume% =Mass of volume/Volume of solutions x 100 = 40g/600 x 100 = 6.66%

Students can refer to the short notes, MCQ questions and different solutions in addition to class 12 Chemistry chapter 2 notes for a clear understanding.

4. Parts Per Million (ppm): It can be defined as the ratio of several parts to the total number of elements of each component of the solution multiplied by 106. It shows the concentration of a solution where Solute is present in the trace. For instance, the concentration of pollutants in water or the atmosphere is often described in terms of g mL–1 or ppm.

Mathematically,

Parts per million = No. of parts of the component  X 106

                                The total number of all the components of sol.

For example, let 0.0005g of sodium chloride  dissolved in 100mL of water then parts per million = 

= 0.0005/100 x 106= 0.000005 x 106 = 5 ppm

Volume to volume, Mass to mass,  and mass to the volume are all ways to express concentration in parts per million. Students may refer to class 12 Chemistry chapter 2 notes for more information.

5. Mole Fraction (x): As per Chemistry chapter 2 class 12 notes, “It represents the number of moles of a solute in one mole of solution.”

Mole fraction = No. of moles of the component

                           Total no. of moles all the component

If the amount of moles A and B in the binary mixture is nA and  nB, respectively, the mole fraction of A will be 

      xA =              nA

                _____________

                      nA + nB

6. Molarity, M: It denotes the number of moles of Solute in 1 litre of Solution.

Molarity, M = Moles of Solute 

                          Vol. of sol in L

Molarity is measured in milligrams per litre and is expressed by the letters ‘M ‘or ‘Molar.’ “The density of a solution is its mass per unit volume.”

Density = Mass of Solution

                  Vol. of Solution

Students may refer to class 12 Chemistry chapter 2 notes provided by Extramarks to get a clear understanding of Molarity. In addition, students may also refer to study notes, essential questions, and sample question papers specific to Solutions – chapter 2 Chemistry class 12 notes.

7. Molality, m: It represents the number of moles of solute present per kilogram of solvent

Molality, m = Moles of Solute

                       Mass of solvent in kg

Molality is calculated in mol/kg, which can also be expressed as ‘m’ or ‘molal.’

For instance, the molality of a solution containing 74.5 g (1 mol) of potassium chloride dissolved in one kg of water is 1.00 m or 1.00 mol kg–1.

8. Normality, N: It represents the number of solute equivalents in 1 litre of Solution.

Normality, N = No. of Equivalents of Solute

                              Vol. of sol. in L

No. of equivalents, eq =     Weight_____  

                                       Equivalent weight(W/E)

E = M, where Z is the valency factor.

      Z

9. Formality: Ionic solutes do not exist in the form of molecules. This molecular mass is expressed as Gram formula mass. Molarity for ionic compounds is called formality.

Extramarks provides detailed study material and revision notes for class 12 Chemistry. Students may refer to class 12 Chemistry chapter 2 notes to understand the concentration of the Solution.

Solubility

As defined under the class 12 Chemistry chapter 2 notes, Solubility is a physical property of a solution. The solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent.

Solubility property of Solution depends upon the factors like

• Nature of solute and solvent and interaction between them: The strong attraction between solute and solvent molecules, the maximum will be the solubility.
• The molecular size of Solute: Larger molecules are more difficult to dissolve in solvents, whereas the smaller molecules dissolve quickly and are more soluble.
• Polarity: Polar solvents (having bonds with different electronegativities) like Water, ethanol, and formaldehyde dissolve polar solutes, whereas nonpolar solvents (having bonds with similar electronegativities) like pyridine, toluene, and hexane dissolve nonpolar solutes.
• Temperature: Rise in temperature increases the solubility.
• Pressure: Rise in pressure increases solubility.

For more information on Solubility, students may refer to NCERT books in addition to Class 12 Chemistry Chapter 2 notes.

Solubility of a Solid in Liquid

The solution of sugar or salt dissolved is a typical example of the solubility of a solid in a liquid. But not all solids need to dissolve in liquid. For Example, solids like naphthalene and anthracene do not mix in water but dissolve readily in benzene, but sugar and salt do not dissolve in benzene. This is because it is a property of Solution that polar solutes dissolve in polar solvents like water, ethanol, formaldehyde and chloroform, whereas nonpolar solutes in nonpolar solvents like toluene, pyridine, as well as hexane.

Some vital concepts as defined in class 12 Chemistry chapter 2 notes include

• The phenomenon of an increase in the concentration of Solution due to the dissolving of solid solute to the solvent is called dissolution.
• The process of collision of solute particles in a solution resulting in the separation of the solute particles from the solution is called crystallisation.
• A phase when the number of solute particles going into the solution is equal to the solute particles separating from the solution is called equilibrium. The strength of the solute remains the same at this phase.
• A solution in which no more solute can be dissolved into the solvent at 
• constant pressure and temperature are called a Saturated solution.
• According to Le Chatelier’s Principle in a saturated solution, if the dissolution process is endothermic (Δsol H > 0), the solubility should increase with a temperature rise. Still, if the dissolution process is exothermic (Δsol H > 0), the solubility should decrease.
• Pressure does not have a crucial effect on the solubility of solids in liquids because they remain unaffected by pressure due to the high incompressibility of solids and liquids.

Solubility of a gas in a liquid:

Aerated drinks and dissolved oxygen in water bodies that support aquatic life are natural instances of solubility of a gas in a liquid. HCl is also highly soluble in water. The phenomenon of an increase in the concentration of solution due to the dissolving of solid solute to the solvent is called dissolution.

Dissolution of salt in water: The process of collision of solute particles in a solution resulting in the separation of the solute particles from the solution is called crystallisation. An example includes the Crystal of copper sulphate obtained by the crystallisation of Copper Sulphate solution.

• A phase when the number of solute particles going into the solution is equal to the number of solute particles separating from the solution is called equilibrium. The strength of the solute remains constant at this stage of dynamic equilibrium.
• The solubility of a gas in a liquid is highly affected by temperature and pressure.
• An increase in temperature reduces the solubility of gases in liquids because in dissolving gas molecules in a liquid, dissolution is similar to condensation leading to the evolution of heat. Hence, dissolution is an exothermic process, due to which the solubility decreases with an increase in temperature.
• An increase in pressure by compressing the gas to a minor volume increase the number of
• gaseous particles per unit volume above the solution and the rate at which the gaseous particles are striking the solution’s surface to enter it, leading to the increase in the solubility of gases.

Students can refer to detailed information on the solubility of a gas in liquid in NCERT books in addition to class 12 Chemistry chapter 2 notes.

Henry’s Law

Henry’s Law establishes a quantitative relation between pressure and solubility of a gas in a solvent. This Law is for gas-liquid Solutions. The solubility of a gas in a liquid is directly proportional to the pressure of the gas at a fixed temperature, according to Henry’s Law.

As defined in class 12 Chemistry chapter 2 notes, “The partial pressure of the gas in the vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution,” according to the most frequent version of Henry’s Law. 

This is written as:

p = KH x

Here KH is the henry law constant.

At the constant temperature, different gasses have different KH. It is a role of the nature of the gas. At a given pressure increasing the value of KH implies lower solubility of the gas in the liquid. The value of KH increases with the increase in temperature; Hence solubility of gases increases with decreasing temperature. Because of this reason, cold water is more sustainable for aquatic life than warm water.

Henry’s Law has a Variety of Applications, as mentioned below.

• Sealing soft drink bottles at high pressure increases the solubility of CO2 in soft drinks and soda water.
• When a diver goes underwater, his body is subjected to high pressure. Due to this, the body tissues absorb more gases. The breathed-in oxygen is required for cellular respiration to provide energy to the diver. But the nitrogen gets absorbed into the body tissues.
• The pressure increases with the increase in depth. Therefore as the diver approaches the bottom of the water body, the pressure increases. With this increase in pressure, more nitrogen gets absorbed by the body tissues.
•  They are challenging and can even cause a threat to their life. The water utilised by scuba jumpers is loaded with contaminated air containing 11.7% helium, 56.2% nitrogen, and 32.1% oxygen to protect the divers from such a situation.
• At high heights, the fractional weight of oxygen is not as much as that at the ground level. This stimulates low oxygen concentrations in the blood and tissues of individuals living at high heights or climbers. Low blood oxygen causes climbers to feel feeble and not able to think plainly, indications of a condition known as anoxia.

Students may refer to CBSE revision notes, CBSE sample papers and important questions in addition to class 12 Chemistry chapter 2 notes while preparing for the board examination.

Henry’s Law as a Specific Example of Raoult’s Law

With reference to class 12 Chemistry chapter 2 notes, Raoult’s Law builds a quantitative relationship between the partial vapour pressure and the mole fraction of a solution. This Law is only for liquid-liquid solutions. The Law states that for a solution of volatile liquids, the partial vapour pressure (p) of each component in the solution is directly proportional to its mole fraction (x). As a result, Raoult’s Law becomes a specific case of Henry’s Law, in which

Mathematically, p ∝ x

            Or p = p 0x 

Where p0 is the vapour pressure of a pure component at the constant temperature.

Hence, for component 1, p1 = p1 0 x1 and for component 2 p2 = p20 x

Let the solvent be represented by one and solute be represented by 2. At this point, if the solute is non-volatile, just the solvent molecules are present in the vapour phase and add to vapour pressure.

so p1 = vapour pressure of the solvent,

x1 = mole fraction,

pi0 = vapour pressure in the pure state.

According to Raoult’s Law

p1 ∝ x1

And p1 = p1 0 x 1

The class 12 Chemistry chapter 2 notes are very detailed. But when students are preparing for the examination, they may also want to consider the CBSE syllabus and CBSE past years’ question papers to understand the weightage Henry’s Law has in the examination.

Vapour Pressure of liquid solutions

A liquid solution is formed when a solvent is liquid. The solute can be gas, solid, or liquid. Generally, In a solution, a liquid solvent is volatile. The solute may or may not. Here we can discuss the properties of only binary solutions.

Vapour Pressure of Liquid Solutions and Raoult’s Law

(Raoult’s Law for volatile solutes)

Raoult’s Law states that the partial vapour pressure of each component in a solution of volatile liquids is directly proportional to its mole fraction.

p1 ∝ x1

p1=p10x1 

Where p10 is the vapour pressure of pure component  1 at the same temperature

Similarly for Component 2

p2=p20x2

Where p20 represents the vapour pressure of the pure component 2. 

The total pressure p(total)  over the solution phase in the container, according to Dalton’s partial pressure law, is equal to the sum of the partial pressures of the Solution’s components and is given as:

p(total) = p1 + p2

We get by substituting the values of

p(total )=x1p10 and x2p20

=p10+(p20−p10)x2

The minimum value of ptotal is p10, and the maximum value is p20, assuming p10 < p20.

Let y1 = Mole fractions of component 1

y2 = Mole fractions of component 2

According to Dalton’s Law of partial pressures:

P1 = y1 ptotal

P2 = y2 ptotal

It is advised that while studying the class 12 Chemistry chapter 2 notes, students make revision notes that will help them during the examination.

Vapour Pressures of Solutions of Solids in Liquids and Raoult’s Law

(Raoult’s Law for non-volatile solutes)

When a non-volatile solute is added to a solvent to make a solution, the number of solvent molecules leaving the surface reduces the vapour pressure. Regardless of its composition, the amount of non-volatile Solute present in the Solution determines the drop in solvent vapour pressure. In its most general form, Raoult’s Law states that the partial vapour pressure of each volatile component in a solution is directly proportional to its mole fraction for every Solution.

p 1 ∝ x1

p1  = x1p10 = ptotal 

The proportionality constant equals the vapour pressure of pure solvent, p10.

If Raoult’s Law obeys true for all concentrations, a solution’s vapour pressure will vary linearly from zero to the pure solvent’s vapour pressure. 

Ideal and Non-Ideal Solutions

Ideal Solutions: An ideal solution is one in which each component follows Raoult’s rule over the entire range of temperature and concentration circumstances known as the Ideal Solution.

Properties of Ideal Solutions: Under class 12 Chemistry chapter 2 notes, the properties of the ideal solution are given as

ΔmixH = 0

ΔmixV = 0

It means no heat is absorbed or evolved when the components are mixed.

• The enthalpy of mixing the pure components for the formation of the solution is 0. 

So, Δmix H = 0.

• This means that heat is neither absorbed nor evolved during mixing components in a solution.
• When pure components are mixed for the formation of the solution, then the volume is also 0. Mathematically, Δmix V = 0
• The volume of the entire solution is = the sum of the volumes of the components.

Suppose the attractive intermolecular forces between the A-A and B-B molecules are approximately comparable to those between the A-B molecules. An ideal solution is rare, but some solutions are suitable for behaviour. For Example, n-hexane and n-heptane solution, benzene and toluene solution fall into this category.

Non–Ideal solutions: When a solution deviates from Raoult’s Law over a broad concentration range, it is known as a non-ideal solution. The solutions that don’t follow Raoult’s Law at every concentration range and all temperatures are Non-Ideal solutions. 

Non-ideal solutions illustrate the below characteristics:

• The solute-solute and solvent-solvent interaction differ from that of the Solute-solvent interaction
• The enthalpy of mixing i.e. is Δmix H ≠ 0, which means that heat might have been evolved if the enthalpy of mixing is negative  (Δmix H < 0), or the heat might have been observed if the enthalpy of mixing is positive (Δmix H > 0)
• The mixing volume is Δmix V ≠ 0, which depicts that there will be some expansion or contraction in the dissolution of liquids. 

For more details of an ideal and non-ideal solution, students may refer to various other study materials over and above Extramarks class 12 Chemistry chapter 2 notes.

Non-ideal solutions are categorised into two types:

i)Non-ideal solutions showing positive deviation from Raoult’s Law

1. ii) Non-ideal solutions showing negative deviation from Raoult’s Law

i) Positive Deviation from Raoult’s Law: It occurs when the component’s vapour pressure is more significant than what is expected in Raoult’s Law. For Example, we consider two components, A and B, to create non-ideal solutions. So the vapour pressure, pure vapour pressure, and mole fraction of component A is PA, PA0 and xA, and that of component B be PB, PB0 and xB, respectively. These Liquids will present a positive deviation when Raoult’s Law:

• PA > PA0 xA and PB > P0B  xB, as the total vapour pressure (PA0 xA + P0B xB), is more significant than what it should be according to Raoult’s Law.
• The solute-solvent forces of attraction are weaker than solute-solute and solvent-solvent interaction, i.e., is, A – B < A – A or B – B
• The mixing enthalpy is +ve, which is Δmix H > 0. The heat absorbed to develop new molecular interaction is less than the heat released to break the original interaction.
• The volume of mixing is +ve, that is, Δmix V > 0, as the volume expands on the dissolution of components A and B.

 Positive Deviation examples:

• Acetone and Carbon disulphide
• Acetone and Benzene
• Carbon Tetrachloride and Toluene or Chloroform
• Ethanol and Water
• Acetone and Ethanol
• Methyl alcohol and Water

ii) Negative Deviation from Raoult’s Law: It occurs when the total vapour pressure is less than what it should be according to Raoult’s Law. We consider the same A and B components to form a non-ideal solution. It will present a negative deviation only from Raoult’s Law when:

• PA < PA0 xA and PB < P0B xB as the total vapour pressure (PA0 xA + P0B xB) is less than what it should be concerning Raoult’s Law
• The solute-solvent interaction is more vital than the solute-solute and solvent-solvent interaction, i.e. is, A – B > A – A or B – B
• The enthalpy of mixing is -ve, that is, Δmix H < 0 because more heat is released when new molecular interactions are created.
• The volume of mixing is -ve, that is,  Δmix V < 0 as the volume decreases on the dissolution of components A and B.

Examples of solutions showing negative deviation from Raoult’s Law.

• Acetone and AnilineChloroform and Benzene
• Chloroform and Diether
• Nitric Acid ( HNO3) and water
• Acetic Acid and pyridine
• Hydrochloric Acid ( HCl) and Water

Students may refer to various other study materials in addition to class 12 Chemistry chapter 2 notes to know more about Raoult’s Law.

Azeotropes

Azeotropes are binary mixes that boil at the same temperature and have the same composition in the liquid and vapour phases.

Minimum Boiling Azeotrope: Minimum boiling azeotrope at a specific composition is formed by solutions that demonstrate a considerable positive departure from Raoult’s Law.

For Example, an azeotrope with a boiling point of 351.15 K is formed by a mixture of ethanol and water containing roughly 95% ethanol. The temperature at which water boils – composition Large positive deviations are depicted in this diagram for solutions (Minimum boiling azeotrope).

Maximum Boiling Azeotrope: Maximum boiling azeotrope at a specific composition is formed by solutions that demonstrate a substantial negative divergence from Raoult’s Law. A 68 per cent nitric acid and water mixture produces an azeotrope with a boiling point of 393.5 K. The temperature at which water boils – composition Large negative variances are depicted in this diagram for solutions. (Azeotrope with the highest boiling point)

Colligative properties and determination of molar mass

Colligative is a Latin word that means ‘together binds’. Colligative properties depend upon the number of solute particles in a solution, irrespective of their nature. More details of colligative properties are shown in class 12 Chemistry chapter 2 notes NCERT solution.

There are four colligative properties: a relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmosis and osmotic pressures. 

Related to these properties, the following deductions can be made:

1. Relative Lowering of Vapour Pressure

When a non-volatile solute is mixed with a solvent, the vapour pressure lowers. This process is the lowering of vapour pressure. Raoult’s Law states that for a solution of volatile liquid, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. As per the laws created by scientists at that time, lowering vapour pressure = vapour pressure of pure solvent – vapour pressure of the solvent. We can determine the total molar mass of a solute through this equation.

(Psolvent = Xsolvent Posolvent)

1. Elevation of Boiling Point

The vapour pressure increases when a non-volatile solute is added with a solvent. The boiling point of this solution is always more significant than the pure solvent to which it is added. It is because the vapour pressure is indirectly proportional to the solution temperature. If the solution needs to be boiled, the temperature of such a solution has to be increased. This process comes to be identified as the elevation of the boiling point. The solute particles and the vapour pressure simultaneously play a role in this phenomenon.

Formula: ΔT = iKbm

Here ΔT= change in temperature

i = the van’t Hoff factor, which is the number of particles into which the solute dissolved

m = the molality, which is the moles of Solute per kilograms of solvent

Kb = the molal boiling point constant (for water, Kb = 0.5121oC/m)

1. Depression of Freezing Point

When the vapour pressure of a solution decreases, it reduces the freezing point of the solution. The solution’s freezing point can be identified as a point where the vapour pressure of a given substance is equal to liquid and the vapour state. As per Raoult’s Law, the freezing point for a given dilute solution also stays directly proportional to the molality of the solute, the same as the boiling elevation point.

Formula: ΔT = iKfm

Here ΔT= change in temperature

i = the van’t Hoff factor

m = the molality, which is the moles of Solute per kilograms of solvent

Kf = the molal freezing point constant (for water, Kf = -1.86 oC/m)

1. Osmosis and Osmotic Pressure

You may have observed raw mangoes shrivelling when placed in brine water or wilted flowers reviving when placed in freshwater. Osmosis is the flow of solvent molecules from pure solvent to the solution through a  semipermeable membrane. The flow continues until it reaches equilibrium.

The process of osmosis happens through a membrane that looks continuous but contains tiny pores, which allow small solvent molecules like water to pass through, but not bigger solute molecules. Such membranes are known as semipermeable membranes (SPM). This pressure is known as the osmotic pressure of the Solution. Osmotic pressure depends on the concentration of the Solution. The osmotic pressure is the excess pressure applied to the Solution to prevent osmosis. It is also a colligative property and depends on the number of solute molecules and not their identity.

Osmosis

When a semipermeable membrane separates a pure solvent and a solution, the solvent particles move through the membrane from the solvent side to the solution side. “Osmosis” is the name given to this phenomenon. The semipermeable membrane allows only small molecules to pass through while blocking bigger solute molecules.

Osmotic Pressure: The excess pressure that must be given to a solution to prevent osmosis, or the flow of solvent molecules through a semipermeable barrier into the Solution, is known as the osmotic pressure of a solution. The osmotic pressure must be equal to the excess pressure. To prevent osmosis, a coating should be put on the solution.

Expression: At a given temperature T, osmotic pressure is proportional to the molarity, C, of a dilute solution.

Thus:

π=C R T

Where  π=osmotic pressure, C=molarity of the Solution, R=Universal Gas constant, T=Temperature

Molarity can be calculated as the number of moles of solute per litre of the Solution. Suppose w2 is the mass of Solute having molar mass M2 in a solution of volume V, then we can write

π=w2RT/M2V.

We use the equation extensively to determine the molar mass of complex molecules of proteins, polymers and other macromolecules. Scientists use this preferably for bio-molecules as it shows a significant value even at room temperature.

Isotonic Solutions: Isotonic solutions are two liquids that have the same osmotic pressure at the same temperature.

Compared to a more concentrated solution, a solution with a lower concentration or lower osmotic pressure is “Hypotonic.”

Compared to a dilute solution, a solution with a higher concentration or higher osmotic pressure is “hypertonic.”

Reverse Osmosis: The solvent will flow from the Solution into the pure solvent through the semipermeable membrane if a pressure more significant than the osmotic pressure is applied to the solution side. The process is known as reverse osmosis.

Application: Desalination of seawater: Pure water is squeezed out of the seawater through the membrane when pressure greater than the osmotic pressure is applied. Cellulose acetate is used as a permeable membrane to water but impermeable to impurities, and ions are placed over support.

Abnormal Molar Masses

The molecular mass is said to have aberrant molar mass when it is determined by investigating any of the colligative qualities and differs from the theoretically expected value. Some molecules of solvents with a low dielectric constant, like molecules of ethanoic acid, dimerise into benzene due to a hydrogen bond. Association of all the ethanoic acid molecules into benzene means Tb or Tf for ethanoic acid will be half the standard value.

The molar mass calculated based on this ΔTb or ΔTf will be twice the expected molar mass. This molar mass value is lower or higher than the expected or normal value and is known as abnormal molar mass. In 1880 Jacobus Henricus van ‘t Hoff introduced factor i, which was named the van’t Hoff factor. The van’t Hoff factor accounts for the degree of dissociation or association. 

The van’t Hoff factor  ‘i’ can be explained as the ratio of Normal molar mass to the abnormal molar mass. 

Mathematically,

i= Normal molar mass / Abnormal molar mass

i = Observed colligative property / Calculated colligative property

i= Total number of moles of particles after association or dissociation/Number of moles of particles before association or dissociation.

In addition, as explained in class 12 Chemistry chapter 2 notes,

• In the case of an association, the value of i is less than unity.
• In the case of dissociation, the value of i is greater than unity.
• Relative lowering of the vapour pressure of solvent =p10– p1 / p10   = i n2 / n1
• Elevation of Boiling point, ΔTb = i Kb m
• Depression of Freezing point, ΔTf = i Kf m
• The osmotic pressure of the Solution, ð  = i n2RT/V 

In addition to class 12 Chemistry chapter 2 notes, students may refer to Extramarks important questions, revision notes, CBSE extra questions, and CBSE previous year question papers to explore more about this chapter. 

Some Important Relationships

Some of the essential relationships defined in class 12 Chemistry chapter 2 notes solutions are

Dilution Law: When we dilute a solution with solvent, the amount of Solute remains constant, and we can write:

M1 V1 = M2V2 &  N1 V1= N2 V2

Molarity and Normality: Normality =Z x Molarity

Important Note: The temperature does not affect the mass percent, ppm, mole fraction, or molality; however, temperature influences molarity and normality. This is because temperature affects volume, whereas mass is not.

Solubility: The maximum or the highest amount of solute that can be dissolved in a given amount of solvent given temperature is its solubility.

The solubility of a solute in a liquid strongly depends upon the factors shown below.

• Nature of Solute
• Nature of the solvent
• The temperature of the Solution
• Pressure(Gases)

For a detailed explanation of solubility, students may refer to CNERT solutions class 12 Chemistry Chapter 2 notes.

NCERT Exemplar Class 12 Chemistry Chapter 2

NCERT Exemplar for Class 12 Chemistry Chapter 2 Notes Solutions is an extended way of revision notes. This will guide students to learn about the topic and brief concepts involved in Class 12 Chemistry Chapter 2 Notes. The Exemplar includes extra questions with their respective answers. It also covers previous year’s question papers, sample papers, worksheets, MCQs, and all Class 12 Chemistry Chapter 2 Notes questions and answers.

There is a lot of vital information covered in the Exemplar pertaining to the Class 12 Chemistry Chapter 2, Solutions which can be overwhelming to students if they are not structured properly. Thus, to reduce the burden on students, Extramarks offers a well-structured Exemplar pertaining to Class 12 Chemistry Chapter 2 Notes that are also recommended to be used during the examination.

Key features of  NCERT Solutions Class 12 Chemistry Chapter 2 Notes

The chapter Solutions covers many different topics making it complicated for many students. By referring to Class 12 Chemistry Chapter 2 Notes, students can understand the various topics and be prepared. 

The Class 12 Chemistry Chapter 2 Notes aims to enhance the concept for students and suggest a different approach to understanding the topic covered in chapter 2 – solution. It helps students clarify their doubts on various topics and solve problems using a logical method. Chemistry class 12 chapter 2 notes help students understand the chapter, study the question paper patterns and prepare for board exams.

The Class 12 Chemistry Chapter 2 Notes can help students get an in-depth understanding of all topics under Solutions. They are also helpful for all types of school examinations or competitive examinations.