JEE Advanced 2025 Revision Notes for Chemistry Solutions

A solution is homogeneous of two or more non-reacting components. The formation of the solution is a physical process.

• Solution of a solid in another solid is known as a solid solution. Many alloys are solid solutions. An alloy of a metal with mercury is called an amalgam.

• Formation of solution from solvent and solute is a reversible process.

$Solute + Solvent \rightleftharpoons Solution$

Types of Solution and Common Examples:

• When an undissolved solute is in dynamic equilibrium with solute in the solution, that state of solution is called a saturated solution. This solution cannot dissolve any more solute.

• The solution which contains an excess of solute than required for saturation at a given temperature is called a supersaturated solution. It is a meta-stable state. At this stage, by a slight disturbance, the extra solute is separated and the super saturated solution is converted into a saturated solution.

• When a solute crystal is added to an unsaturated solution, it dissolves.

• When a solute crystal is added to a saturated solution, there is a definite change in the shape but the weight in the dissolved form remains constant, because of the dynamic equilibrium that exists in a saturated solution.

• When a solute crystal is added to a super saturated solution, the size is increased because of the crystallization of extra solute on the crystal. It is called the seeding process.

Methods of Expressing the Concentration of a Solution:

The concentration of a solution can be expressed quantitatively in serval ways.

• Weight percent (or mass percent): $\dfrac{{{\text{Weight of solute}}}}{{{\text{Weight of solution}}}}{\text{} \times 100}$

• Weight volume percent of solute=$\dfrac{{{\text{Weight of solute}}}}{{{\text{Volume of solution}}}}{\text{} \times 100}$

• Volume percent of solute=$\dfrac{{{\text{Volume of solute}}}}{{{\text{Volume of solution}}}}{\text{ }\times 100}$

Molarity:

Molarity indicates the number of moles of the solute dissolved in one liter of the solution. it is also indicating the number of millimoles of the solute dissolved in one milliliter of the solution.

Molarity (M)=$\dfrac{{{\text{Number of moles of solute(n)}}}}{{{\text{volume in liters of solution(V)}}}}$

or

${\text{M = }}\dfrac{{\text{w}}}{{{\text{Molar mass}}}}{\text{ }\times }\dfrac{{{\text{1000}}}}{{\text{V}}}\because {\text{n = }}\dfrac{{\text{w}}}{{{\text{Molar mass}}}}$

Normality

The number of gram equivalents of the solute present in one liter of a solution. As temperature increases, the volume increases and normality decreases.

${\text{N = }}\dfrac{{{\text{Number of equivalent of solute}}}}{{{\text{Volume of solution in litres}}}}$

(or)

${\text{N = }}\dfrac{{{\text{wt of solute}}}}{{{\text{Gram equivalent weight}}}}{\text{ }\times }\dfrac{{{\text{1000}}}}{{{\text{Volume of solution in mL}}}}$

Equivalent Weight Concept:

1. Equivalent weight of acid:

${{\text{E}}_{{\text{Acid}}}}{\text{ = }}\dfrac{{{\text{Formula weight of acid}}}}{{{\text{Basicity of acid}}}}$

2. Equivalent Weight of Base:

${{\text{E}}_{{\text{Base}}}}{\text{ = }}\dfrac{{{\text{Formula weight of base}}}}{{{\text{Acidity of base}}}}$

3. Equivalent Weight of Salt:

${{\text{E}}_{{\text{Salt}}}}{\text{ = }}\dfrac{{{\text{Formula weight of the salt}}}}{{{\text{Total charge of the cation or anion of the salt}}}}$

4. Equivalent weight of an oxidizing agent (or) oxidant:

${{\text{E}}_{{\text{oxidant}}}}{\text{ = }}\dfrac{{{\text{Formula weight of the oxidant}}}}{{{\text{Total number of electrons gained in a chemical reaction}}}}$

Normality of the mixture when two solutions of the same solute are mixed.

$N = \dfrac{{{N_1}{V_1} + {N_2}{V_2}}}{{{V_1} + {V_2}}}$

Normality, Molarity Interrelation:

1. For Acids:

Normality = Molarity × Basicity of acid

2. For Bases:

Normality = Molarity × Acidity of acid

3. For Salts:

Normality = Molarity × total +ve or -ve charge of the salt.

4. For Oxidizing (or) Reducing Agents:

Normality = Molarity × Total change in oxidation state per mole of oxidant or reductant.

For Exact Neutralization of an Acid With Base:

$\dfrac{{{\text{wt. of base}}}}{{{\text{GEW(gram equivalent weight) of base}}}}{\text{ = }}\dfrac{{{{\text{N}}_{\text{a}}}{{\text{V}}_{\text{a}}}\left( {{\text{mL}}} \right)}}{{{\text{1000}}}}$

$\dfrac{{{\text{wt. of acid}}}}{{{\text{GEW of acid}}}}{\text{ = }}\dfrac{{{{\text{N}}_{\text{b}}}{{\text{V}}_{\text{b}}}{\text{(mL)}}}}{{{\text{1000}}}}$

Molality:

The number of gram moles of the solute present in one kilogram of solvent, It is denoted by ‘m’.

Molality (m) is given as

${\text{m = }}\dfrac{{{\text{Number of moles of solute(n)}}}}{{{\text{Weight of solvent(w)}}}}$ 

(or)

$m = \dfrac{w}{{Molar~mass}} \times \dfrac{{1000}}{W}$

where, w and W are masses of solute and solvent respectively in grams.

Parts Per Million:

Trace quantities of solute in a solution are conveniently expressed in parts per million (ppm).

${\text{ppm = }}\dfrac{{{\text{Number of parts of component  }\times { 1}}{{\text{0}}^{\text{6}}}}}{{{\text{Total number of parts of solution}}}}$

Mole Fraction:

• The ratio of the number of moles of a component to the total number of moles of the solution. It is denoted by the symbol ‘x’.

• If the number of moles of component substance A and B in a solution is respectively ${n_A}$and ${n_B}$.

Mole fraction of component A=${x_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}}$

Mole fraction of component B=${x_B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}}$

Effect of Temperature on the Solubility of a Solid in a Liquid

• In a nearly saturated solution, if the dissolution process is endothermic, the solubility of a substance should increase with rise in temperature.

• For salts whose lattice energy is greater than hydration energy, solubility increase with the increase in temperature.

• In a nearly saturated solution, if the dissolution process is exothermic, The solubility of a substance should decrease with rising temperature.

Vapour Pressure:

• The pressure exerted by the vapour molecules of a liquid when they are in equilibrium with the liquid at a given temperature is called the vapour pressure of a liquid at that temperature.

• The vapour pressure of a liquid is more if the rate of evaporation is more.

• vapour pressure of a liquid depends on the nature of the liquid, Surface area of the liquid, temperature, and flow of air current over the surface.

• Rapid evaporation results in a decrease in temperature leading to intense cooling.

• Variation of vapour pressure with temperature is given by the Clausius Clapeyron equation

Vapour pressure of a solution of a non-volatile solute is less than the vapour pressure of the pure solvent at the same temperature.

• If ${P_0}$is the vapour pressure of the pure solvent and ${P_s}$ is the vapour pressure of the solution at the same the temperature then, lowering of vapour pressure is ${P_0} - {P_s}$

• The vapour pressure of a solution of a non-volatile solute, is directly proportional to the mole fraction of the solvent in the solution. ${P_s} = {P_0}{x_{solvent}}$

• Lowering of vapour pressure is directly proportional to the mole fraction of solute.

• Variation of vapour pressure with temperature is given by the clausius-clapeyron equation.

$\log \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{ \Delta {H_{vap}}}}{{2.303RT}}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$.

• Different solutions having the same vapour pressure are called Isopiestic solutions.

Raoult’s Law

states that the relative lowering of the vapour pressure of a solution is equal to the mole fraction of the solute in the solution.

Let a solution contains, $n_{A}$ moles of solvent and $n_{B}$ moles of solute, then according to Raoult's law .

$P_{A} \propto X_{A^{\prime}} P_{A}=k X_{A}$

Where pure solvent is taken, i.e, $X_{A}=1$, then $k_{=} P_{A}^{0}=$ vapour pressure of pure solvent.

Substitute $k=P_{A} X_{A} ; \dfrac{P_{A}}{P_{A}}=X_{A}$

Subtract the above two quantities from 1, we get

$1-\dfrac{P_{A}}{P_{A}{ }^{0}}=1-X_{A}$ but $X_{A}+X_{B}=1$

$\mathrm{P}_{\mathrm{A}}^{0}-P_{A} / \mathrm{P}_{\mathrm{A}}^{0}=X_{B}=$ moles fraction of solute.

$\mathrm{P}_{\mathrm{A}}^{0}-P_{A} / \mathrm{P}_{\mathrm{A}}^{0}=n_{B} / n_{A}+n_{B}$

For dilute solutions $n_{A} \gg n_{B}$

$\therefore n_{A}+n_{B}=n_{A}$

$P_{A}^{0}-P_{A} / P_{A}^{0}=n_{B} / n_{A}$

$P_{A}^{0}-P_{A} / P_{A}^{0}=W_{B} / m_{B} \times M_{A} / W_{A}$

It is called simplified Raoult's law. where, $w_{B^{\prime}} W_{A}$ are the weight of soliute and solvent. $m_{B^{\prime}} M_{A}$ are the molecular weight of the solute and solvent.

• Raoult's law is applicable to dilute solutions only.

• When the solute is non-volatile, if the solute does not undergo either ionization (or) association, and if the solution behaves as an ideal solution Raoult's law is good to apply.

Relative lowering of vapour pressure is determined by Ostwald and walker method.

Colligative Properties

The properties which depend on the number of particles of the solute but not on the nature of the solute are called colligative properties.

• All the colligative properties can be used to determine the molecular weight of non-volatile solute. 

• But the best method is by using Osmotic pressure. When a non-volatile solute is dissolved in the pure solvent its boiling point increases. 

Example: Seawater boils at greater than 100℃

Relationship of Osmotic Pressure With Other Colligative Properties:

1. $\pi ~and ~\dfrac{{\Delta P}}{{P_A^0}};\pi  = \dfrac{{\Delta P}}{{P_A^0}} \times \dfrac{{dRT}}{{{M_A}}}$

2. $\pi ~and~\Delta {T_b};\pi  = \Delta {T_b} \times \dfrac{{dRT}}{{1000 \times {K_b}}}$

3. $\pi ~and~\Delta {T_f};\pi  = \Delta {T_f} \times \dfrac{{dRT}}{{1000 \times {K_f}}}$

In the above relations:

𝛑= Osmotic pressure, d= density of solution at temperature T

R= universal gas constant, ${M_A}$= Molar mass of solvent, ${K_b}$= molar elevation constant of the solvent, ${K_f}$=molar depression constant of the solvent.

• Relation between depressing in freezing point and relative lowering of vapour pressure.$\Delta {T_f} = \dfrac{{\Delta P}}{{P_A^0}} \times \dfrac{{1000 \times {K_f}}}{{{M_A}}}$

• Relation between elevation in boiling point and relative lowering of vapour pressure.

$\Delta {T_b} = \dfrac{{\Delta P}}{{P_A^0}} \times \dfrac{{1000 \times {K_b}}}{{{M_A}}}$

• Relation between elevation in boiling point and depression in freezing point

$\Delta {T_b} = \dfrac{{{K_b}}}{{{K_f}}} \times \Delta {T_f}$

Van’t Hoff Factor:

$\left( i \right) = \dfrac{{{\text{Normal molar mass}}}}{{{\text{Observed molar mass}}}}$

Theoretically,

${\text{i = }}\dfrac{{{\text{Observed colligative property}}}}{{{\text{Normal colligative property}}}}$

• For ideal solutions with no association or dissociation of solute,  the Van't Hoff factor i = 1. For solutes showing association, I < 1 and showing dissociation, i > I.

If a molecule of solute on dissociation gives 'n' ions and α is the degree of dissociation, $i = 1 + \alpha \left( {n - 1} \right)$

Degree of dissociation $(\alpha ) = \dfrac{{i - 1}}{{n - 1}}$

If a solute from associated molecules ${\left( A \right)_n}$and $\alpha $ is the degree of association. $i = 1 - \alpha \left[ {1 - \dfrac{1}{n}} \right]$

Degree of association $(\alpha ) = \dfrac{{(1 - i)n}}{{n - 1}}$

In Terms of Van’t Hoff Factor (i)

• Elevation of boiling point. $\Delta {T_b} = i{K_b}m$

• Depression of freezing point. $\Delta {T_f} = i{K_f}m$

• Osmotic pressure freezing point $\pi=C R T=\dfrac{w}{m} \times \dfrac{R T}{V}$, where, $M$ is molecular weight of solute and $w$ is weight of solute.

Solved Examples:

Question 1: 100mL of ethyl alcohol is made up of a liter of distilled water. If the density of the ethyl alcohol is 0.46g/mL. what is the molality of the solution?

Answer:

Density of ethyl alcohol =0.49 g/mL

Volume of ethyl alcohol = 100 mL

Wt. of ethyl alcohol = d×v=0.46×100=46 g.

Wt. Of the water in the solution= 900×1 =900 g 

(density of ${H_2}O = 1 g/mL$)

Molality =$\left[ {\dfrac{w}{{Molar mass}}} \right] \times \dfrac{{1000}}{{{\text{wt of solvent in gms}}}} = \dfrac{{46}}{{46}} \times \dfrac{{1000}}{{900}} = 1.11~m$

Question 2: Calculate the mole fraction of ${H_2}S{O_4}$ in a solution containing 98% ${H_2}S{O_4}$ by weight

Answer:

98% ${H_2}S{O_4}$ by weight mean 98 g of ${H_2}S{O_4}$ in 2g of ${H_2}O$

Mole fraction of ${H_2}S{O_4}$=

$\dfrac{{{n_{{H_2}S{O_4}}}}}{{{n_{{H_2}S{O_4}}} + {n_{{H_2}O}}}} = \dfrac{{\left( {\dfrac{{98}}{{98}}} \right)}}{{\left( {\dfrac{{98}}{{98}} + \dfrac{2}{{18}}} \right)}}$

$= \dfrac{1}{{1 + \dfrac{1}{9}}} = \dfrac{9}{{10}} = 0.9$

Importance of Physical Chemistry Solutions

This chapter will introduce the concepts of different types of solutions and how they attain equilibrium. You will learn how the constituents of a solution behave in ideal conditions. The factors controlling the features of a solution will also be explained.

The laws and scientific principles related to this chapter will all be explained using mathematical derivations. Every term used in such derivations will be explained elaborately. You will learn how these terms are used in deriving a formula. It will explain and correlate vapour pressure and osmotic pressure with the gas laws.

The chapter will also teach how to execute problems related to solutions and solubility using these scientific formulas derived from the laws and principles. It will also explain the limitation of the laws we use to derive the formulas. It will also teach the effect of changing various factors such as temperature, vapour pressure, etc.

Learn what an ideal solution is and whether it obeys the laws. Find out the implications of molecular attractions in an ideal solution using the fundamental concepts described in this chapter. The importance of this chapter lies in the description of a mixture called solution and its different aspects.

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