Course Content
Section Name Topic Name 3 Classification of Elements and Periodicity in Properties 3.1 Why do we Need to Classify Elements ? 3.2 Genesis of Periodic Classification 3.3 Modern Periodic Law and the present form of the Periodic Table 3.4 Nomenclature of Elements with Atomic Numbers > 100 3.5 Electronic Configurations of Elements and the Periodic Table 3.6 Electronic Configurations and Types of Elements: s-, p-, d-, f – Blocks 3.7 Periodic Trends in Properties of Elements
Section Name Topic Name 7 Equilibrium 7.1 Equilibrium in Physical Processes 7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium 7.3 Law of Chemical Equilibrium and Equilibrium Constant 7.4 Homogeneous Equilibria 7.5 Heterogeneous Equilibria 7.6 Applications of Equilibrium Constants 7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G 7.8 Factors Affecting Equilibria 7.9 Ionic Equilibrium in Solution 7.10 Acids, Bases and Salts 7.11 Ionization of Acids and Bases 7.12 Buffer Solutions 7.13 Solubility Equilibria of Sparingly Soluble Salts
Section Name Topic Name 10 The s-Block Elements 10.1 Group 1 Elements: Alkali Metals 10.2 General Characteristics of the Compounds of the Alkali Metals 10.3 Anomalous Properties of Lithium 10.4 Some Important Compounds of Sodium 10.5 Biological Importance of Sodium and Potassium 10.6 Group 2 Elements : Alkaline Earth Metals 10.7 General Characteristics of Compounds of the Alkaline Earth Metals 10.8 Anomalous Behaviour of Beryllium 10.9 Some Important Compounds of Calcium 10.10 Biological Importance of Magnesium and Calcium
Section Name Topic Name 12 Organic Chemistry – Some Basic Principles and Techniques 12.1 General Introduction 12.2 Tetravalence of Carbon: Shapes of Organic Compounds 12.3 Structural Representations of Organic Compounds 12.4 Classification of Organic Compounds 12.5 Nomenclature of Organic Compounds 12.6 Isomerism 12.7 Fundamental Concepts in Organic Reaction Mechanism 12.8 Methods of Purification of Organic Compounds 12.9 Qualitative Analysis of Organic Compounds 12.10 Quantitative Analysis
Class 11th Chemistry Online Class: Elevate Your CBSE Board Success
About Lesson

Reactions in Solution

  • A majority of reactions in the laboratories are carried out in solutions.
  • The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways:-
  1. Mass per cent or weight percent (w/w% )
  2. Mole fraction
  3. Molarity
  4. Molality

Mass per cent or weight percent

  • Mass percentage is one way of representing the concentration of an element in a compound or a component in a mixture.
  • Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%.
    • For example: – Consider a compound of H2O which contains 2 moles of H2 and 1 mole of O2.
      • Or it can be also stated as 2 x 1gram of H and 16grams of O.
      • Therefore Mass % of H2=(2/18) x 100% and Mass% of O2=(16/18)x100%


A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.


Mass of percent of A = (Mass of A)/ (Mass of solution) x100

= (2g)/ (2g of A + 18g of water) x100

= (2g/20g)/100

= 10 %

Mole Fraction

  • It is the ratio of number of moles of a particular component to the total number of moles of the solution.
    • For example: – Consider a substance ‘A’ dissolves in substance ‘B’ and their moles are nA and nB
    • Mole fraction of A
      • = (No. of moles of A)/(No. of moles of solution)
      • (nA )/( nA + nB)
    • Mole fraction of B
      • =(No. of moles of B)/(No. of moles of solution)
      • (nB )/( nA + nB)


  • Molarity is defined as the number of moles of the solute in 1 litre of the solution.
  • It is widely used unit and is denoted by ‘M’.
  • Molarity (M) = (No. of moles in solute)/(Volume of solution in litres)


A 4 g sugar cube (sucrose: C12H22O11) is dissolved in a 350 ml teacup filled with hot water. What is the molarity of the sugar solution?


Equation of molarity:-

M = (m/V)

Where M is molarity (mol/L)

m = number of moles of solute

V = volume of solvent (Litres)

For each of the atoms to get the total grams per mole:

C12H22O11 = (12) (12) + (1) (22) + (16) (11)

C12H22O11 = 144 + 22+ 176

C12H22O11 = 342 g/mol

To get the number of moles in a specific mass, divide the number of grams per mole into the size of the sample:

(4 g)/ (342 g/mol) = 0.0117 mol


  • It is defined as the number of moles of solute present in 1 kg of solvent.
  • It is denoted by m.
    • Thus Molality(m) =
    • (No. of moles of solute)/(Mass of solvent in kg)


The density of 3 M solution of NaCl is 1.25 g mL–1. Calculate molality of the solution.


M = 3 mol L–1

Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g

Mass of 1L solution = 1000 × 1.25 = 1250 g

(Since density = 1.25 g mL–1)

Mass of water in solution = 1250 –175.5

= 1074.5g

Molality= (No. of moles of solute)/ (Mass of solvent in kg)

= (3 mol)/ (1.0745kg)

= 2.79 m.


How much copper can be obtained from 100 g of copper sulphate (CuSO4)?


1 mole of CuSO4 contains 1 mole of copper.

Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)

= 63.5 + 32.00 + 64.00

= 159.5 g

159.5 g of CuSO4 contains 63.5 g of copper.

⇒ 100 g of CuSO4 will contain (63.5 x 100g)/ (159.5) of copper.

Therefore Amount of copper that can be obtained from 100 g CuSO4

= (63.5 x 100)/ (159.5) =39.81 g


Wisdom TechSavvy Academy