CBSE Solutions Class 12 Notes Chemistry Chapter 2

PDF Summary - Chemistry Chapter 2 Solutions Class 12 Notes

Introduction

We rarely come with pure substances in everyday life. The majority of them are blends of two or more pure chemicals. Their usefulness or significance in life is determined by their composition. The air we breathe is mostly composed of oxygen and nitrogen, and the water we drink has trace amounts of different salts dissolved in it. Our blood is made up of a variety of components. Brass, bronze, stainless steel, and other alloys are examples of mixes. We'll focus on liquid solutions and their properties in this unit.

Solutions

1.1 Definition

A solution is a mixture of two or more components that is homogeneous.

Take, for instance, ordinary salt in water.

1.2 Classification

Binary solutions are solutions that have two components in them. Components are substances that are used to make a solution.

The solvent is the component that is present in the greatest quantity. The physical condition of a solution is determined by the solvent.

Solute refers to the other component in the solution that is present in smaller amounts.

Each element could be solid, liquid, or gaseous.

1.3 Strength of Solutions

Strength of solution refers to the amount of solute dissolved per unit of solution or solvent. There are several methods for determining the strength of a solution.

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

Mass $\%$ of a component $=\dfrac{\text { Mass of component in the sol. }}{\text { Total Mass of sol. }} \times 100$

2. Volume Percentage (%v/v): "It represents the volume of a component in a solution of $100 \mathrm{~mL}$."

Volume $\%$ of a component $=\dfrac{\text { Volume of component }}{\text { Total Volume of sol. }} \times 100$

3. Mass by Volume Percentage (% w / v): "It denotes the mass of the solute in grams in $100 \mathrm{~mL}$ of solution."

Mass by vol, percent $=\dfrac{\text { Mass of solute in } \mathrm{g}}{\text { Vol. of sol. in } \mathrm{mL}} \times 100$

4. Parts Per Million (ppm): 

Parts per million $=\dfrac{\text { No. of parts of the component }}{\text { Total no. of all the components of sol. }} \times 10^{6}$

Mass to mass, volume to volume, and mass to volume are all ways to express concentration in parts per million.

5. Mole Fraction (x): "It denotes the number of moles of a solute in one mole of solution."

Mole fraction $=\dfrac{\text { No. of molos of the component }}{\text { Total no. of moles all the components }}$

If the amount of moles of $A$ and $B$ in a binary mixture is $n_{A}$ and $n_{B}$, respectively, the mole fraction of $A$ will be

$x_{A}=\dfrac{n_{A}}{n_{A}+n_{B}}$

6. Molarity, M: "It represents the number of moles of solute in 1 litre of solution."

Molarity, $\mathrm{M}=\dfrac{\text { Moles of solute }}{\text { Vol. of sol in } \mathrm{L}}$

Molarity is measured in milligrams per litre and is denoted by the letters ' $\mathrm{M}$ ' or 'Molar.' "The density of a solution is its mass per unit volume."

Density, $\mathrm{d}=\dfrac{\text { Mass of sol. }}{\text { Vol. of sol. }}=\mathrm{m} / \mathrm{V}$

7. Molality, m: “It denotes the number of moles of solute present per kilogramme of solvent.”

$\text { Molality, } \mathrm{m}=\dfrac{\text { Moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}$

Molality is measured in mol/kg, which can also be written as 'm' or 'molal.'

8. Normality, N: It denotes the number of solute equivalents in 1 litre of solution.

Normality, $N=\dfrac{\text { No. of Equivalents of solute }}{\text { Vol. of sol. in } \mathrm{L}}$

No. of equivalents, $e q=\dfrac{\text { Weight }}{\text { Equivalent weight }(\mathrm{W} / \mathrm{E})}$

$\mathrm{E}=\dfrac{\mathrm{M}}{\mathrm{z}}(\mathrm{Z}$ is the valency factor).

Some Important Relationships

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

$\mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2}$ and $\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}$

Molarity and Normality

Normality $=\mathrm{z} \times$ Molarity

Important:

The temperature has no effect on the mass per cent, ppm, mole fraction, or molality, however temperature influences molarity and normality. This is because the volume is affected by temperature whereas mass is not.

2. Vapour Pressure

2.1 Definition

At a given temperature, the vapour pressure of a liquid/solution is the pressure exerted by vapours in equilibrium with the liquid/solution.

$\text { Vapour pressure } \propto \text { escaping tendency }$

2.2. 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.

$\mathrm{p}_{1} \propto \mathrm{x}_{1}$

and $\mathrm{p}_{1}=\mathrm{p}_{1}^{0} \mathrm{x}_{1}$

Component 2 is the same way.

$\mathrm{p}_{2}=\mathrm{p}_{2}{ }^{0} \mathrm{x}_{2}$

The total pressure $\mathrm{P}_{(\text {total) }}$ 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:

$\mathrm{p}_{\text {(iotal) }}=\mathrm{p}_{1}+\mathrm{p}_{2}$

We get by substituting the values of $p_{1}$ and $p_{2}$

$p_{(\text {total })}=x_{1} p_{1}^{0} \text { and } x_{2} p_{2}^{0}$

$=\left(1-x_{2}\right) p_{1}^{0}+x_{2} p_{2}^{0}$

$=p_{1}^{0}+\left(p_{2}^{0}-p_{1}^{0}\right) x_{2}$

At constant temperature, the plot of vapour pressure and mole fraction of an ideal solution. The partial pressures of the components are shown by the dashed lines $I$ and $\|. \mathrm{p}_{1}$ and $\mathrm{p}_{2}$ are directly proportional to $\mathrm{x}_{1}$ and $\mathrm{X}_{2}$, respectively, as can be seen from the plot. The total vapour pressure is represented in the figure by line III.

Mole Fraction in The Vapour Phase

Using Dalton's rule of partial pressures, if $y_{1}$ and $y_{2}$ are the mole fractions of components 1 and 2 in the vapour phase, then:

$\mathrm{p}_{1}=y_{1}\mathrm{p}_{\text {total }}$ 

$\mathrm{p}_{2}=y_{2}\mathrm{p}_{\text {total }}$ 

Generally

$\mathrm{p}_{i}=y_{i}\mathrm{p}_{\text {total }}$

2.3. 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 from the surface is reduced, lowering the vapour pressure.

The amount of non-volatile solute present in the solution, regardless of its composition, 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.

$\mathrm{p}_{1} \propto \mathrm{x}_{1}$

$\mathrm{p}_{1}=\mathrm{x}_{1} \mathrm{p}_{1}^{0}=\mathrm{p}_{\text {total }}$

If Raoult's law holds true for all concentrations, a solution's vapour pressure will vary linearly from zero to the pure solvent's vapour pressure.

2.4. Ideal and Non-Ideal Solutions

Ideal Solutions:

An ideal solution is one in which each component follows Raoult's rule under all temperature and concentration circumstances.

Properties of Ideal Solutions:

\[\Delta {{\text{H}}_{\text{MIXING}}}=0\]

\[\Delta {{\text{V}}_{\text{MIXING}}}=0\]

The attractive forces between the A-A and B-B molecules are approximately comparable to those between the A-B molecules.

For example, benzene and toluene solution, n-hexane and n-heptane solution

Non – Ideal Solutions:

When a solution deviates from Raoult's law over a wide concentration range, it is referred to as a non-ideal solution.

Solutions Showing Positive Deviation from Raoult’s Law:

• SoluteSolute(B-B) and Solvent-Solvent(A-A) forces are weaker than Solvent-Solute(A-B) forces.

• The vapour pressure is higher than the law predicts.

• \[\Delta {{\text{H}}_{\text{MIXING}}}>0\]

• \[\Delta {{\text{V}}_{\text{MIXING}}}>0\]

Ethanol with acetone, for example, or carbon disulphide and acetone.

Negative Deviations from Raoult's Law are Shown in the Following Solutions:

• The Solvent-Solute(A-B) force is more powerful than the other two.

• The vapour pressure is lower than the law predicts.

• \[\Delta {{\text{H}}_{\text{MIXING}}}<0\]

• \[\Delta {{\text{V}}_{\text{MIXING}}}<0\]

Examples are phenol and aniline, chloroform and acetone, and so on.

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Curves of pressure composition for a solution with a negative deviation

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2.5. 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 certain 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 certain composition is formed by solutions that demonstrate a substantial negative divergence from Raoult's law. A mixture of 68 per cent nitric acid and water 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)

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3. Solubility

3.1. Solubility of a Solid in Liquid

A substance's solubility refers to how much of it can be dissolved in each amount of solvent.

Factors Impacting a Solid's Solubility in a Liquid Include: 

1. Nature of solute and solvent: Like disintegrates into like. Sodium chloride and sugar, for example, dissolve quickly in water, but naphthalene and anthracene do not. Naphthalene and anthracene, on the other hand, dissolve quickly in benzene, but sodium chloride and sugar do not.

2. Temperature: If \[({{\Delta }_{\text{sol}}}\text{H}0)\] the solubility of a nearly saturated solution increases as the temperature rises, and if \[({{\Delta }_{\text{sol}}}\text{H}0)\], the solubility falls as the temperature rises.

3. Pressure's Effect:  Solids and liquids are highly incompressible, so, this has little effect.

3.2. Henry’s Law

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.

“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:

The constant \[{{\text{K}}_{\text{H}}}\] stands for Henry's law.

\[{{\text{K}}_{\text{H}}}\] Characteristics

• The type of the gas determines \[{{\text{K}}_{\text{H}}}\].

• The lower the solubility of the gas in the liquid, the higher the value of \[{{\text{K}}_{\text{H}}}\]at a given pressure.

• \[{{\text{K}}_{\text{H}}}\] values increase as temperature decreases, indicating that gas solubility increases as temperature decreases.

Henry's Law has a Variety of Applications.

• In the manufacture of carbonated drinks.

• During a deep-water dive.

• Climbers and those who live at high altitudes will benefit from this.

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

Raoult's law states that

\[{{\text{p}}_{\text{i}}}\text{ = }{{\text{x}}_{\text{i}}}{{\text{p}}_{\text{i}}}^{\text{0}}\]

One of the components of a gas in a liquid solution is so volatile that it exists as a gas. According to Henry's law, it is soluble in water.

\[{{\text{p}}_{\text{i}}}\text{ = }{{\text{K}}_{\text{H}}}\text{x}\]

As a result, Raoult's law becomes a specific case of Henry's law, in which \[{{\text{K}}_{\text{H}}}\] equals \[{{\text{p}}_{\text{i}}}^{\text{0}}\].

4. Colligative Properties

Colligative qualities are those that are dependent on the number of solute particles present in a solution, regardless of their type relative to the total number of particles present.

There are four qualities that are colligative:

1. A Relative Decrease in Vapor Pressure

2. Boiling Point Elevation

3. Freezing point depression

4. Osmotic pressure is the factor to consider.

4.1. Relative Lowering of Vapour Pressure

When a non-volatile solute is added to a solvent, the vapour pressure decreases. The lowering of vapour pressure with respect to, the vapour pressure of the pure solvent is called “Relative lowering in vapour pressure”.

Raoult's Law States That:

\[{{\text{p}}_{1}}\text{ = }{{\text{x}}_{1}}{{\text{p}}_{1}}^{0}\]

The reduction in the vapour pressure of solvent \[(\Delta {{\text{p}}_{1}})\] is given as:

$Delta {{\text{p}}_{1}}={{\text{p}}_{1}}^{0}-{{\text{p}}_{1}}={{\text{p}}_{1}}^{0}-{{\text{p}}_{1}}^{0}{{\text{x}}_{1}}$ 

$={{\text{p}}_{1}}^{0}(1-{{\text{x}}_{1}})$

Knowing that \[{{\text{x}}_{2}}=1-{{\text{x}}_{1}}\] equation then reduces to 

\[\Delta \text{p=}{{\text{x}}_{2}}{{\text{p}}_{1}}^{0}\]

This equation can be written as 

\[\dfrac{\Delta \text{p}}{{{\text{p}}_{1}}^{0}}=\dfrac{{{\text{p}}_{1}}^{0}-{{\text{p}}_{1}}}{{{\text{p}}_{1}}^{0}}={{\text{x}}_{2}}\]

As previously stated, the expression on the left-hand side of the equation is known as relative lowering of vapour pressure and equals the mole fraction of the solute. The equation above can be represented as:

\[\dfrac{{{\text{p}}_{1}}^{0}-{{\text{p}}_{1}}}{{{\text{p}}_{1}}^{0}}\text{ }\dfrac{{{\text{n}}_{2}}}{{{\text{n}}_{1}}+{{\text{n}}_{2}}}\left( \text{since }{{\text{x}}_{2}}\text{ }\dfrac{{{\text{n}}_{2}}}{{{\text{n}}_{1}}+{{\text{n}}_{2}}} \right)\]

The number of moles of solvent and solute present in the solution are denoted by \[{{\text{n}}_{1}}\] and \[{{\text{n}}_{2}}\], respectively. Because \[{{\text{n}}_{2}}\] << \[{{\text{n}}_{1}}\] for dilute solutions, we can ignore \[{{\text{n}}_{2}}\] in the denominator.

\[\dfrac{{{\text{p}}_{1}}^{0}-{{\text{p}}_{1}}}{{{\text{p}}_{1}}^{0}}\text{=}\dfrac{{{\text{n}}_{2}}}{{{\text{n}}_{1}}}\]

Or

\[\dfrac{{{\text{p}}_{1}}^{0}-{{\text{p}}_{1}}}{{{\text{p}}_{1}}^{0}}\text{=}\dfrac{{{\text{w}}_{2}}\times {{\text{M}}_{1}}}{{{\text{M}}_{2}}\times {{\text{w}}_{1}}}\]

The masses \[{{\text{w}}_{1}}\] and \[{{\text{w}}_{2}}\] are the masses of the solvent and solute, respectively, while \[{{\text{M}}_{1}}\] and \[{{\text{M}}_{2}}\] are the molar masses of the solvent and solute.

4.2. Elevation in Boiling Point

The temperature at which a liquid's vapour pressure equals atmospheric pressure is known as the boiling point of the liquid.

The vapour pressure of the solvent decreases when a non-volatile solute is added, hence the needed temperature to boil the solution will be higher. As a result, the boiling point of the solution will rise.

Elevation of boiling point is defined as an increase in boiling point \[\Delta {{\text{T}}_{\text{b}}}={{\text{T}}_{\text{b}}}-{{\text{T}}_{\text{b}}}^{0}\], where \[{{\text{T}}_{\text{b}}}^{0}\] is the boiling point of pure solvent and \[{{\text{T}}_{\text{b}}}\] is the boiling point of solution.

Expression:

\[\Delta {{\text{T}}_{\text{b}}}={{K}_{\text{b}}}m\]

The Boiling Point Elevation Constant, or \[{{K}_{\text{b}}}\], is also known as the Molal Elevation Constant (Ebullioscopic Constant).

Calculation of the Solute's Molar Mass:

\[\text{m=}\dfrac{{{\text{w}}_{2}}/{{\text{M}}_{2}}}{{{\text{w}}_{1}}/1000}=\dfrac{1000\times {{\text{w}}_{2}}}{{{\text{M}}_{2}}\times {{\text{w}}_{1}}}\]

By changing the molality value in the equation, we get

$\Delta {{\text{T}}_{\text{b}}}=\dfrac{{{\text{K}}_{\text{b}}}\times 1000\times {{\text{w}}_{2}}}{{{\text{M}}_{2}}\times {{\text{w}}_{1}}}$ 

${{\text{M}}_{2}}=\dfrac{1000\times {{\text{w}}_{2}}\times {{\text{K}}_{\text{b}}}}{\Delta {{\text{T}}_{\text{b}}}\times {{\text{w}}_{1}}}$

\[{{\text{K}}_{\text{b}}}\]

When the molality of the solution is unity, it is defined as an increase in boiling point.

The unit of \[{{\text{K}}_{\text{b}}}\] is \[\text{kg mo}{{\text{l}}^{-1}}\]

Determination of \[{{\text{K}}_{\text{b}}}\]:

\[{{\text{K}}_{\text{b}}}=\dfrac{\text{R}\times {{\text{M}}_{\text{1}}}\times {{\text{T}}_{\text{b}}}^{2}}{1000\times {{\Delta }_{\text{vap}}}}\]

Where, \[\text{R}\]= gas constant \[(8.314\text{ JK/mol)}\]

\[{{\text{T}}_{\text{f}}}\]= freezing temperature in \[\text{K}\]

\[{{\text{M}}_{\text{1}}}\]= Molar mass of solvent in \[\text{Kg/mol}\]

\[{{\Delta }_{\text{vap}}}\text{H}\]= enthalpy of vapourization in \[\text{J/mol}\]

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The solution's vapour pressure curve is lower than that of pure water. \[\Delta {{\text{T}}_{\text{b}}}\]denotes the elevation of a solvent's boiling point in solution, as seen in the diagram.

4.3. Depression in freezing point

A substance's freezing point is the temperature at which the vapour pressure of the substance in its liquid phase equals the vapour pressure of the substance in its solid phase.

When a non-volatile solid is introduced to a solvent, the vapour pressure drops until it equals that of a solid solvent at a lower temperature. As a result, the solvent's freezing point drops.

Depression in freezing point is defined as \[\Delta {{\text{T}}_{\text{f}}}={{\text{T}}_{\text{f}}}^{0}-{{\text{T}}_{\text{f}}}\], where \[{{\text{T}}_{\text{f}}}^{0}\] is the freezing point of pure solvent and \[{{\text{T}}_{\text{f}}}\] is the freezing point when a non-volatile solute is dissolved.

Expression

\[\Delta {{\text{T}}_{\text{f}}}={{\text{K}}_{\text{f}}}\text{m}\]

Freezing Point Depression Constant, Molal Depression Constant, or Cryoscopic Constant are all names for \[{{\text{K}}_{\text{f}}}\]

Calculation of molar mass of solute:

\[\text{m= }\dfrac{{{\text{w}}_{\text{2}}}/{{\text{M}}_{\text{2}}}}{{{\text{w}}_{\text{1}}}/1000}\]

By Substituting This Molality Value into the Equation, We Get:

\[\Delta {{\text{T}}_{\text{f}}}\text{= }\dfrac{{{\text{K}}_{\text{f}}}\times {{\text{w}}_{\text{2}}}/{{\text{M}}_{\text{2}}}}{{{\text{w}}_{\text{1}}}/1000}\]

\[\Delta {{\text{T}}_{\text{f}}}\text{= }\dfrac{{{\text{K}}_{\text{f}}}\times {{\text{w}}_{\text{2}}}\times 1000}{{{\text{M}}_{\text{2}}}\times {{\text{w}}_{\text{1}}}}\]

\[{{\text{M}}_{\text{2}}}\text{= }\dfrac{{{\text{K}}_{\text{f}}}\times {{\text{w}}_{\text{2}}}\times 1000}{\Delta {{\text{T}}_{\text{f}}}\times {{\text{w}}_{\text{1}}}}\]

When the molality of the solution is unity, it is defined as a decrease in freezing point. \[{{\text{K}}_{\text{f}}}\] is measured in \[\text{kg mo}{{\text{l}}^{-1}}\].

Determination of \[{{\text{K}}_{\text{f}}}\]

\[{{\text{K}}_{\text{f}}}\text{= }\dfrac{\text{R}\times {{\text{M}}_{\text{2}}}\times {{\text{T}}_{\text{f}}}^{0}}{1000\times {{\Delta }_{\text{fus}}}}\]

Where  \[\text{R}\]= gas constant \[(8.314\text{ JK/mol)}\]

\[{{\text{T}}_{\text{f}}}\]= freezing temperature in \[\text{K}\]

\[{{\text{M}}_{\text{1}}}\]= Molar mass of solvent in \[\text{Kg/mol}\]

\[{{\Delta }_{\text{fus}}}\text{H}\]= enthalpy of fusion in \[\text{J/mol}\]

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\[{{\text{T}}_{\text{f}}}\], or the depression of a solvent's freezing point in a solution, is depicted in this diagram.

4.4. 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 semi-permeable membrane allows only small molecules to pass through, while blocking bigger solute molecules.

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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.

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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:

\[\pi =\text{C R T}\]

The osmotic pressure is P, while the gas constant is R.

Calculation of Molar Mass:

\[\pi =\left( {{\text{n}}_{2}}/\text{V} \right)\text{RT}\]

\[{{\text{M}}_{2}}=\dfrac{{{\text{w}}_{2}}\text{RT}}{\pi \text{V}}\]

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

In comparison to a more concentrated solution, a solution with a lower concentration or lower osmotic pressure is referred to as "Hypotonic."

In comparison to a dilute solution, a solution with a higher concentration or higher osmotic pressure is referred to as "hypertonic."

Reverse Osmosis: The solvent will flow from the solution into the pure solvent through the semi-permeable membrane if a pressure greater 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.

4.5. Abnormal Molar Masses

The molecular mass of a substance 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.

Abnormal Molar Masses are Observed:

1. When the solute in the solution forms a connection.

2. When the solute in the solution undergoes dissociation.

Van’t Hoff Factor:

Van't Hoff proposed a factor \[i\] known as the van't Hoff Factor, to determine the level of association or dissociation.

\[i\text{=}\dfrac{\text{Normal molar mass}}{\text{Abnormal molar mass}}\]

\[i\text{=}\dfrac{\text{Observed colligative property}}{\text{Calculated colligative property}}=\dfrac{\text{Total no}\text{. of moles of particles after association(dissociation)}}{\text{No}\text{. of moles of particles before association(dissociation)}}\]

Association: the number of particles will always decrease because of connection, \[\text{i1}\]

\[\text{nA}\to {{\text{A}}_{\text{n}}}\]

Let initial particles \[({{\text{n}}_{\text{i}}})=1\]

Final number \[({{\text{n}}_{_{\text{f}}}})=1-\alpha +\alpha /\text{n}\]

van’t Hoff Factor \[\text{i= }{{\text{n}}_{\text{f}}}/{{\text{n}}_{\text{i}}}=\text{ 1-}\alpha \text{+}\alpha \text{/n}\]

Dissociation: Due to dissociation, the number of particles will always rise, hence \[\text{i1}\]

\[{{\text{A}}_{\text{n}}}\to \text{nA}\]

Initial particle = 1

Final particles = \[1-\alpha +\text{n}\alpha \]

\[\text{i=1-}\alpha \text{+n}\alpha \]

Modified Expressions: Relatively lower solvent vapour pressure

\[\dfrac{{{\text{p}}_{\text{1}}}^{0}-{{\text{p}}_{1}}}{{{\text{p}}_{\text{1}}}^{0}}=\text{i}\text{.}\dfrac{{{\text{n}}_{2}}}{{{\text{n}}_{1}}}\]

Elevation of boiling point \[\Delta {{\text{T}}_{\text{b}}}=\text{i}{{\text{K}}_{\text{b}}}\text{m}\]

Depression of Freezing point \[\Delta {{\text{T}}_{\text{f}}}=\text{i}{{\text{K}}_{\text{f}}}\text{m}\]

Osmotic pressure of solution, \[\pi =\text{ i }{{\text{n}}_{2}}\text{ RT/V}\]

5. Vapour Pressure

When the temperature of a liquid is raised, the molecules' desire to escape increases, and the vapour pressure rises.

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In a liquid, the distribution of molecule kinetic energy.

Claussius - Clapeyron Equation

\[{{\text{p}}_{2}}/{{\text{p}}_{1}}=(\Delta {{\text{H}}_{\text{VAP}}}/\text{R)(1/}{{\text{T}}_{1}}-1/{{T}_{2}})\]

Where \[\Delta {{\text{H}}_{\text{VAP}}}\] represents the liquid's vaporisation enthalpy.

6. Thermodynamics of Dissolution

The process is exothermic if the contacts increase stronger during solution formation, and it is endothermic if they become weaker. In the dissolving process, \[\Delta S\] is often positive. \[\Delta \text{G}\] must be negative if the mixing process is spontaneous/natural.

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6.1. Boiling Point Elevation

$ \Delta {{\text{G}}_{\text{VAP}}}=0$ 

${{\text{T}}_{\text{b}}}=\Delta {{\text{H}}_{\text{VAP}}}/\Delta {{\text{S}}_{\text{VAP}}}$ 

The non-volatile solute improves the unpredictability of the solution phase while maintaining the same entropy of the vapours. As a result, \[\Delta {{\text{S}}_{\text{VAP}}}\] lowers, causing the boiling point to rise.

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The difference in their entropies of vapourization, \[\Delta {{\text{S}}_{\text{VAP}}}\], causes a solution's boiling point to be greater than that of a pure solvent.

6.2. Freezing Point Depression

$\Delta {{\text{G}}_{\text{FUS}}}=0$

${{\text{T}}_{\text{f}}}=\Delta {{\text{H}}_{_{\text{FUS}}}}/\Delta {{\text{S}}_{_{\text{FUS}}}}$

Due to the increase in the entropy of the solution, the entropy difference will increase in this scenario. According to the above relationship, an increase in entropy causes a drop in the freezing point.

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A difference in their entropies of fusion, \[\Delta {{\text{S}}_{_{\text{FUSION}}}}\], causes a solution to have a lower freezing point than a pure solvent.

7. Ostwald Walker Method

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This is a common method for determining a solution's relative decrease in vapour pressure. Dry air is carried through three systems in order: solution, pure solvent, and finally a drying agent.

The decrease in weight of the vessels is represented by \[{{\text{w}}_{1}}\] and \[{{\text{w}}_{2}}\], whereas the rise in weight of the third vessel is represented by \[{{\text{w}}_{3}}\].

\[{{\text{w}}_{1}}\propto {{\text{P}}_{\text{SOLUTION}}}\text{ }{{\text{w}}_{2}}\propto {{\text{P}}_{\text{SOLVENT}}}\text{- }{{\text{P}}_{\text{SOLUTION}}}\]

(since the air was already saturated)

\[{{\text{w}}_{3}}\propto {{\text{P}}_{\text{SOLVENT}}}\]

The relative decrease in vapour pressure can be computed using the relationships.

A denotes the solvent and B represents the solute in the formulas below.

\[{{\text{M}}_{\text{A}}}=\]Molar mass of solvent                                         \[{{\text{M}}_{\text{B}}}=\] Molar mass of solute

\[{{\text{W}}_{\text{A}}}=\] Mass of solvent                                                    \[{{\text{V}}_{\text{B}}}=\] volume of solute

\[\text{V}=\]volume of solution                                                   \[\text{d}=\]density of solution

Some Important Formulas

1. Mass of percentage (w/w) = 

\[\dfrac{{{\text{W}}_{\text{B}}}}{{{\text{W}}_{\text{A}}}+{{\text{W}}_{\text{B}}}}\times 100\]

Volume percentage(v/v) = \[\dfrac{{{\text{V}}_{\text{B}}}}{{{\text{V}}_{\text{A}}}+{{\text{V}}_{\text{B}}}}\times 100\]

Mass by volume percentage \[\left[ \dfrac{\text{w}}{v} \right]=\dfrac{{{\text{W}}_{\text{B}}}\times 100}{\text{V(mL)}}\]

Parts per million (ppm) =\[\dfrac{{{\text{W}}_{\text{B}}}}{{{\text{W}}_{\text{A}}}+{{\text{W}}_{\text{B}}}}\times {{10}^{6}}\]

2. Mole Fraction of \[\text{A, }{{\text{x}}_{\text{A}}}=\text{ }\dfrac{{{\text{n}}_{\text{A}}}}{{{\text{n}}_{\text{A}}}+{{\text{n}}_{\text{B}}}}\]

Mole fraction of \[\text{B, }{{\text{x}}_{\text{B}}}=\text{ }\dfrac{{{\text{n}}_{\text{B}}}}{{{\text{n}}_{\text{A}}}+{{\text{n}}_{\text{B}}}}\]

\[{{\text{x}}_{\text{A}}}+{{\text{x}}_{\text{B}}}\]

3. Molarity (M)

\[\text{Molarity(M)= }\dfrac{\text{Moles of solute}}{\text{Volume of solution in litre}}=\dfrac{{{\text{n}}_{\text{B}}}}{{{\text{V}}_{(\text{L)}}}}=\dfrac{{{\text{W}}_{\text{B}}}}{{{\text{M}}_{\text{B}}}\times {{\text{V}}_{(\text{L)}}}}\]

4. Molality (M)

\[\text{Molality(M)= }\dfrac{\text{Moles of solute}}{\text{Volume of solvent in kg}}=\dfrac{{{\text{n}}_{\text{B}}}}{{{\text{W}}_{A}}}\times 1000=\dfrac{{{\text{W}}_{\text{B}}}\times 1000}{{{\text{M}}_{\text{B}}}\times {{\text{W}}_{A}}}\]

5. Normality(N)

\[\text{Normality(N)= }\dfrac{\text{Gram Equivalents of solute}}{\text{Volume of solution in litre}}=\dfrac{{{\text{W}}_{\text{B}}}}{\text{Gem of solute}\times {{\text{V}}_{(L)}}}\]

GEM= Gram Equivalent Mass

6. Relationship Between Molarity and Normality

A solution's normality (N) and molarity (M) are connected as follows:

\[\text{Normality}\times \text{Equivalent}\text{.mass(solute)=Molarity}\times \text{Molar mass(solute)}\]

7. Relationship Between Molarity and Mass percentage (p)

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

\[\text{Molarity=}\dfrac{\text{p}\times \text{d}\times \text{10}}{\text{Mol}\text{. mass(solute)}}\]

\[\text{Normality=}\dfrac{\text{p}\times \text{d}\times \text{10}}{\text{Eq}\text{. mass(solute)}}\]

8. Dilution Formula

When a substance's solution is diluted by adding solvent from volume \[{{\text{V}}_{1}}\] to volume \[{{\text{V}}_{2}}\],

\[{{\text{M}}_{1}}{{\text{V}}_{1}}={{\text{M}}_{2}}{{\text{V}}_{2}}\], similarly, \[{{\text{N}}_{1}}{{\text{V}}_{1}}={{\text{N}}_{2}}{{\text{V}}_{2}}\]

9. Molarity of a Mixture

When \[{{\text{V}}_{1}}\]mL of a molarity \[{{\text{M}}_{1}}\] solution is mixed with another solution of the same substance with volume \[{{\text{V}}_{2}}\] and molarity\[{{\text{M}}_{2}}\], the molarity of the resulting mixture of solution (M) can be calculated as follows:

\[\text{M=}\dfrac{{{\text{M}}_{1}}{{\text{V}}_{1}}={{\text{M}}_{2}}{{\text{V}}_{2}}}{{{\text{V}}_{1}}+{{\text{V}}_{2}}}\]

10. Raoult’s Law for Volatile Solute.

\[{{\text{p}}_{\text{A}}}={{\text{p}}_{\text{A}}}^{0}{{\text{x}}_{\text{A}}}\text{ and }{{\text{p}}_{\text{B}}}={{\text{p}}_{\text{B}}}^{0}{{\text{x}}_{\text{B}}}\]

The partial vapour pressures of component ‘\[\text{A}\]' and component ‘\[\text{B}\]' in the solution are \[{{\text{p}}_{\text{A}}}\] and \[{{\text{p}}_{\text{B}}}\], respectively. The vapour pressures of pure components ‘\[\text{A}\]' and ‘\[\text{B}\]' are \[{{\text{p}}_{\text{A}}}^{0}\] and \[{{\text{p}}_{\text{B}}}^{0}\], respectively.

Total vapour pressure = \[\text{p = }{{\text{p}}_{\text{A}}}\text{+ }{{\text{p}}_{\text{B}}}={{\text{p}}_{\text{A}}}^{0}{{\text{x}}_{\text{A}}}\text{+}{{\text{p}}_{\text{B}}}^{0}{{\text{x}}_{\text{B}}}\]

11. Raoult’s Law for Non-Volatile Solute

\[\dfrac{{{\text{p}}_{\text{A}}}^{0}-\text{p}}{{{\text{p}}_{\text{A}}}^{0}}={{\text{x}}_{\text{B}}}=\dfrac{{{\text{n}}_{\text{B}}}}{{{\text{n}}_{\text{A}}}+{{\text{n}}_{\text{B}}}}=\dfrac{{{\text{n}}_{\text{B}}}}{{{\text{n}}_{\text{A}}}}=\dfrac{{{\text{W}}_{\text{B}}}}{{{\text{M}}_{\text{B}}}}\times \dfrac{{{\text{M}}_{\text{A}}}}{{{\text{W}}_{\text{A}}}}\]

where \[{{\text{x}}_{\text{B}}}\] is the mole-fraction of the solute and \[\dfrac{{{\text{p}}_{\text{A}}}^{0}-\text{p}}{{{\text{p}}_{\text{A}}}^{0}}\]is the relative reduction in vapour pressure.

12. Elevation in Boiling Point. \[\Delta {{\text{T}}_{\text{b}}}={{\text{K}}_{\text{b}}}\times \text{m}\]

\[{{\text{M}}_{\text{B}}}=\dfrac{{{\text{K}}_{\text{b}}}{{\text{W}}_{\text{B}}}}{\Delta {{\text{T}}_{\text{b}}}\times {{\text{W}}_{\text{A}}}}\]

Where \[\Delta {{\text{T}}_{\text{f}}}={{\text{T}}_{\text{f}}}^{0}-{{\text{T}}_{\text{f}}}\]

14. Osmotic Pressure \[(\pi )\]

\[\pi =\text{cRT}\]

where ‘c’ is molarity

15. Van’t Hoff Factor.

\[i\text{=}\dfrac{\text{Normal molar mass}}{\text{Abnormal molar mass}}\]

Or

\[i\text{=}\dfrac{\text{Observed colligative property}}{\text{Calculated colligative property}}\]

Or

\[i=\dfrac{\text{Total no}\text{. of moles of particles after association(dissociation)}}{\text{No}\text{. of moles of particles before association(dissociation)}}\]

Modified Forms of Colligative Properties:

(a) \[\dfrac{{{\text{p}}_{\text{A}}}^{0}-{{\text{p}}_{\text{A}}}}{{{\text{p}}_{\text{A}}}^{0}}=\text{i}{{\text{x}}_{\text{B}}}\]

(b) \[\Delta {{\text{T}}_{\text{b}}}=\text{i }{{\text{K}}_{\text{b}}}\text{m}\]

(c) \[\Delta {{\text{T}}_{\text{f}}}=\text{i}{{\text{K}}_{\text{f}}}\text{m}\]

(d) \[\pi \text{V=i}{{\text{n}}_{\text{B}}}\text{RT}\]

16. \[\alpha =\dfrac{\text{i-1}}{\text{n-1}}\]where is the degree of dissociation, ‘\[\text{i}\]’ is the van't Hoff factor, and \[\text{n}\] is the number of ions produced per compound formula.

17. \[\alpha =\dfrac{\text{1-i}}{\text{1-}\dfrac{1}{\text{n}}}\] where is the degree of association, 1 is the number of molecules of solute that must associate to produce an associated molecule, and n is the number of molecules of solute that must associate to form an associated molecule where \[\dfrac{1}{\text{n}}<1\].

18. If \[\text{i1}\], the solute will dissociate.

\[\text{i1}\], the solute forms a bond with another substance.

There is no linkage or dissociation when \[\text{i=1}\]

The \[\text{i=}\dfrac{\text{1}}{2}\] solute is a dimer.

Solution Class 12 Chemistry Notes - Free PDF Download

In the Class 12 Chemistry Ch 2 notes, students will become familiar with the solution’s basic definition and formation. These notes are available in free PDF that prevents students from purchasing expensive study material and books. 

Why buy something when you are availed with free solutions? These Class 12 chemistry chapter 2 notes PDF are created by qualified chemistry teachers with many years of experience. They work according to the new terms and conditions, as stated by CBSE every year and form questions according to the latest edition of CBSE. The Solution chapter covers many formulas that must be on tips to solve any numerical problem without wasting time. Solution chapter notes PDF is a separate section where all the formulas to be covered in the chapter are mentioned in one place. Thus this free PDF for notes will also help students to score well in their exams. 

Topics to Cover Under NCERT Solutions of Chemistry Chapter Solutions Class 12 Notes

The Solution Class 12 notes covers in-depth knowledge of solutions, solvents, and solute. There will be questions based upon the concentration of solvent or solute and the entire solution. You need to set your mind according to the tricky questions asked in the paper. Class 12 chemistry Chapter 2 notes PDF cover all basic details about the chapter along with numerical solutions. Also, there are different experiments to perform accordingly. 

Below Are the Mentioned Topics Covered Within Class 12 Chemistry Ch 2 Notes

• What is the solution?

• Solvents and solutes

• Different types of solutions

• Molarity, Molality, moles

• Mole fraction and percentages

• Henry’s Law

• Ideal Solutions

• Raoult’s Law

• The concentration of the solution

• Normality

• Solubility

• Saturated and Unsaturated Solutions

There is a detailed material provided in the Solutions class 12 notes PDF for each concept to clarify your concepts efficiently. These Class 12 Chemistry Ch 2 notes are readily available for free so that students can put in their maximum practice for clarifying their problems. 

Mastering Chemistry Solutions Class 12 Notes: Solutions - Quick Review of Notes

You will get an answer to each problem that you face during preparation for exams through Solutions class 12 chemistry notes. These notes are to make sure you prepare well to score the best. Also, it depends upon the self-study that lets you prepare well and thoroughly you revise it. Below is the detailed description of the topics covered in solution chemistry notes PDF:

Solution

• The solution is the homogeneous mixture of two non-reactive substances. It is a phase with two or more components. 

• The solution is made with two main components- solute and solvent. The solute is present in less amount while the solvent is present in a larger amount.  

Types of Solutions

There are three different types of solutions:

• Solid

• Liquid

• Gaseous

Saturated Solutions

Solutions in which solute is present in a higher amount that no more solutes can be added to the solvent are saturated. 

Other Terms

Molality: The symbol to define molality is ‘m.’ It is defined by moles of solute in a one-kilogram solvent. 

Molarity: The symbol to define molarity is ‘M’ which is moles of solute in a litre of solvent. 

Moles: Moles have different formulas to state. However, in terms of mass, we define moles as a mass of solute or solvent in the solution divided by their atomic masses. 

Mole Fraction: It is given by the number of molecules of a substance divided by total molecules. 

Normality: It is given by gram equivalent of solute in one litre of the total solution.

Solubility: It is the maximum solute that can be dissolved in the solvent at a given temperature. 

Henry’s Law

If the temperature remains constant, a gas’s solubility in a liquid substance will be proportional to the pressure exerted. 

Raoult’s Law

In volatile liquids, the partial vapour pressure of each substance in a solution will be directly proportional to the mole fraction of that substance.

Ideal Solutions

Solutions that will follow Raoult’s Law throughout the defined time are ideal. The attractive forces between two different substances in a solution will be the same as the attractive forces within the substance itself.

Important Questions on Solutions Class 12 Chemistry Notes - Chapter 2

Given below are a few important questions from Class 12 Chemistry Ch 2 notes. Try to solve them once you have learnt the chapter well. If you face any problems, go back to our revision notes and take help. Some of these questions have appeared in the CBSE Board exams in the previous years. Hence, it is advisable that Class 12 students should try to solve them while revising Solutions.

Q1. Briefly explain mole fraction.

Q2. What do you understand by molarity?

Q3. Define Henry’s law. 

Q3. Mention a few applications of Henry’s law.

Q4. Briefly discuss molecular interaction in a solution of water and alcohol.

Q5. What is molarity?

Q6. On mixing two liquids A and B, a warm solution is produced. State which type of deviation from Raoult’s law can be observed.

Q7. What is meant by an ebullioscopic constant?

Q8. State  Raoult’s law. What is an ideal solution?

Q9. Mention a few points of similarities between Raoult’s law and Henry’s law.

Q10. By positive deviation from Raoult’s law, what type of azeotrope will be formed? Give an example.

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Conclusion

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