Course Content
CHAPTER 3: CLASSIFICATION OF ELEMENTS
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
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CHAPTER 7: EQUILIBRIUM
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
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CHAPTER 10: S-BLOCK ELEMENTS
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
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CHAPTER 12: CHARACTERIZATION OF ORGANIC COMPOUND
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
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Class 11th Chemistry Online Class For 100% Result
About Lesson

Dalton’s Atomic Theory

  • Matter consists of indivisible atoms.
  • All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
  • Compounds are formed when atoms of different elements combine in a fixed ratio.
  • Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction.

John Dalton

Class_11_Concepts_Of_Chemistry_John_Dalton
Concept of Chemistry

Atomic Mass

  • Atomic mass is the mass of the atom.
  • Nowadays sophisticated techniques e.g. mass spectrometry is used to determine the atomic mass of the atom.
  • But in 19th century atomic mass was determined by calculating the mass of the atom relative to hydrogen by using stoichiometry.

History of Atomic Mass

  • In 1803 John Dalton determined the atomic mass or atomic weight.
  • Atomic weight was originally defined as relative to that of lightest element hydrogen taken as 1.
  • Oxygen was taken as base as its atomic mass was 16(whole number).All the observations were based on stoichiometry.
  • Chemists picked naturally occurring Oxygen, which is a mixture of isotopes Oxygen-16, Oxygen-17 and Oxygen-18 (Atomic mass 16.008).
  • Physicists picked up carbon 12 as base  based  on mass  spectrometry & mass of one carbon -12 atom is a whole number.
  • Carbon-12 was agreed to be taken as standard in 1961 because the mass of one carbon -12 atom is a whole number.
  • The masses of all other atoms are taken relative to mass of 12C.

Average Atomic Mass

  • Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (per cent occurrence).
Class_11_Concepts_Of_Chemistry_Average_Atomic_mass
Concept of Chemistry
  • From the above data, the average atomic mass of carbon will come out to be :
    • (0.98892) (12 u) + ( 0.01108) (13.00335 u) + (2 × 10–12) (14.00317 u)
    • = 12.011 u

Problem:-

Calculate the atomic mass (average) of chlorine using the following data:-

Isotopes of Chlorine        % Natural Abundance                Molar Mass

35Cl                         75.77                                      34.9689

37Cl                       24.23                              36.9659

 Answer:-

Atomic mass of first isotope = 34.9689

Natural abundance of first isotope = 75.77% or 0.757

 Atomic mass of second isotope= 36.9659

 Natural abundance of second isotope= 24.23% or 0.242

Now average atomic mass of chlorine

= [34.9689x 0.757 + 36.9659x 0.242]/ (0.757+0.242) =35.4521

So, the average atomic mass of chlorine = 35.4527 u

Problem:-

Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope        Isotopic molar mass         Abundance

36Ar              35.96755 gmol–1               0.337%

40Ar            39.9624 gmol–1                  99.600%

Answer:-

Molar mass of argon =

Atomic mass of 36Ar = 35.96755 & abundance = 0.337

Or

Total atomic mass of 36Ar = 35.96755 * 0.337= 0.121g/mol

Atomic mass of 38Ar = 37.96272 & abundance = 0.063

Or

Total atomic mass of 38Ar = 37.96272 * 0.063= 0.024g/mol

Atomic mass of 40Ar =39.9624 & abundance = 99.600

Or

Total atomic mass of 40Ar = 39.9624 * 99.600= 39.802g/mol

 Therefore molar mass of argon = total mass of 36Ar+ total mass of 38Ar + atomic mass of 40Ar

= 0.121 + 0.024 + 39.802= 39.947 g/mol

Molecular Mass

  • Molecular mass is the sum of atomic masses of the elements present in a molecule.
  • For example:-Molecular mass of methane CH4 = (12.011 u) + 4 (1.008 u) = 16.043 u.

Formula Mass

  • Some substances such as sodium chloride do not contain discrete molecules as their constituent units. E.g. NaCl.
  • In such compounds, positive (sodium) and negative (chloride) entities are arranged in a three-dimensional structure as shown in the figure.
    • For example: – Formula mass of sodium chloride (NaCl) = atomic mass of sodium (Na) + atomic mass of chlorine (Cl) = 23.0 u + 35.5 u = 58.5 u.

(Packing of Na+ and Cl ions in sodium chloride NaCl)

Class_11_Concepts_Of_Chemistry_NaCl_Molecules
Concept of Chemistry

Problem:-

Calculate the molecular mass of the following:

(i) H2O (ii) CO2 (iii) CH4

Answer:-

(i) H2O:

The molecular mass of water, H2O

= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)

= [2(1.0084) + 1(16.00 u)]

= 2.016 u + 16.00 u

= 18.016

= 18.02 u

(ii) CO2:

The molecular mass of carbon dioxide, CO2

= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)

= [1(12.011 u) + 2 (16.00 u)]

= 12.011 u + 32.00 u

= 44.01 u

(iii) CH4:

The molecular mass of methane, CH4

= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)

= [1(12.011 u) + 4 (1.008 u)]

= 12.011 u + 4.032 u

= 16.043 u

Question 1.
Calculate the molecular mass of the following :
(i) H2o
(ii) CO2
(iii) CH4
Solution:
(i) Molecular mass of H2O :2 × 1 + 1 × 16 = 18u
(ii) Molecular mass of CO2 :1 × 12 + 2× 16 = 44 u
(iii) Molecular mass of CH4 : 12 + 4 × 1 = 16 u

 

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