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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About Lesson

Law of chemical combination

  • The combination of elements to form compounds is governed by the following five basic laws:-
  1. Law of Conservation of Mass.
  2. Law of Definite Proportions.
  3. Law of Multiple Proportions.
  4. Gay Lussac’s Law of Gaseous Volumes.
  5. Avogadro Law.

Law of Conservation of Mass

  • Law of conservation of mass states that the matter cannot be created nor be destroyed.
  • This law was put forth by Antoine Lavoisier in 1789.
  • He performed careful experimental studies for combustion reactions for reaching to the above conclusion.

Antoine Lavoisier

Class_11_Concepts_Of_Chemistry_Antoine_Lavoisier

Law of Multiple Proportions

  • According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.
    • For example: – Hydrogen combines with oxygen to form 2 compounds, water and hydrogen peroxide.
    • H2(2g) + (1/2)O2 (16g) à H2O (18g)
    • H2(2g) + O2 (32g) àH2O2 (34g)
    • The masses of oxygen O (16g and 32g) combine with the fixed mass of (2g) hydrogen H. Therefore the simple ratio is 16:32 or 1:2.
  • This law was given by Dalton in 1803.

John Dalton

Class_11_Concepts_Of_Chemistry_John_Dalton

Law of Definite Proportions

  • According to this law, a given compound always contains exactly the same proportion of elements by weight.
  • This law was given by French chemist, Joseph Proust in 1806.
  • He observed 2 samples of cupric carbonate.
    • One was of natural origin and another was of synthetic origin.
  • He found that the composition of elements present in it was same for both the samples as shown below:-
Class_11_Concepts_Of_Chemistry_Composition_Of_Elements
  • Note: – It is sometimes also referred to as law of definite composition.

Gay Lussac’s Law of Gaseous Volumes

  • Gay Lussac’s law was given by Gay Lussac in 1808.
  • He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.
    • For example: – H (Hydrogen) (100mL) + O (Oxygen) (50mL)à Water (100mL).
    • The volumes of hydrogen (H) and oxygen (O) which combine together (i.e. 100mL and 50mL) bear a simple ratio of 2:1.

Gay Lussac

Class_11_Concepts_Of_Chemistry_Gay_Lussac

Joseph Proust

Class_11_Concepts_Of_Chemistry_Joseph

Avogadro Law

  • In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules.
  • He made distinction between atoms and molecules.
Class_11_Concepts_Of_Chemistry_Avogadro

Problem:-

The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

Mass of dinitrogen Mass of dioxygen

(i) 14 g                            16 g

(ii) 14 g                           32 g

(iii) 28 g                          32 g

(iv) 28 g                          80 g

(a)Which law of chemical combination is obeyed by the above experimental data? Give its statement.

(ii) 1 mg = …………………. kg = …………………. ng

(iii) 1 mL =…………………. L = …………………. dm3

Answer:-

(a)

If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g, 32 g, and 80 g.

The masses of dioxygen bear a whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the law of multiple proportions. The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.

(b)

(i) 1 km = 1 km × (1000m/1km) x (100cm/1m) x (10mm/1cm)

Therefore 1 km = 106 mm

1 km = 1 km × (1000m/1km) x (1pm/10-12m)

Therefore 1 km =1015 pm

Hence, 1 km = 106 mm = 1015 pm

(ii) 1 mg = 1 mg × (1 g/1000mg) x (1kg/1000g)

⇒ 1 mg = 106 kg

1 mg = 1 mg × (1 g /1000mg) x (1 ng /10-9g)

⇒ 1 mg = 106 ng

1 mg = 10–6 kg = 10ng

(iii) 1 mL = 1 mL × (1L/1000mL)

⇒ 1 mL = 103 L

1 mL = 1 cm3 = 1 cm3 = 1 x ((1 dm x 1dm x 1dm) / (10cm x 10cm x10 cm)) cm3

⇒ 1 mL = 103 dm3

1 mL = 10–3 L = 10–3 dm3

The laws of chemical combinations (Chemistry) multiple choice questions and answers.

Ques. Which of the following pairs of substances illustrate the law of multiple proportions
(a) CO and CO2
(b) H2O and D2O
(c) NaCl and NaBr
(d) MgO and Mg(OH)2
Ans. (a)

Ques. 1.0 g of an oxide of A contained 0.5 g of A. 4.0 g of another oxide of A contained 1.6 g of A. The data indicate the law of
(a) Reciprocal proportions
(b) Constant proportions
(c) Conservation of energy
(d) Multiple proportions
Ans. (d)

Ques. Among the following pairs of compounds, the one that illustrates the law of multiple proportions is
(a) NH3 and NCl3
(b) H2S and SO2
(c) CuO and Cu2O
(d) CS2 and FeSO4
Ans. (c)

Ques. Chemical equation is balanced according to the law of
(a) Multiple proportion
(b) Reciprocal proportion
(c) Conservation of mass
(d) Definite proportions
Ans. (c)

Ques. Different propartions of oxygen in the various oxides of nitrogen prove the
(a) Equivalent proportion
(b) Multiple proportion
(c) Constant proportion
(d) Conservation of matter
Ans. (b)

Ques. Carbon and oxygen combine to form two oxides, carbon monoxide and carbon dioxide in which the ratio of the weights of carbon and oxygen is respectively 12 : 16 and 12 : 32. These figures illustrate the
(a) Law of multiple proportions
(b) Law of reciprocal proportions
(c) Law of conservation of mass
(d) Law of constant proportions
Ans. (a)

Ques. n g of substance X reacts with m g of substance Y to form p g of substance R and q g of substance S. This reaction can be represented as, X + Y = R + S. The relation which can be established in the amounts of the reactants and the products will be
(a) n – m = p – q
(b) n + m = p + q
(c) n = m
(d) p = q
Ans. (b)

Ques. In compound A, 1.00 nitrogen unites with 0.57 g oxygen. In compound B, 2.00 g nitrogen combines with 2.24 g oxygen. In compound C, 3.00 g nitrogen combines with 5.11 g oxygen. These results obey the following law
(a) Law of constant proportion
(b) Law of multiple proportion
(c) Law of reciprocal proportion
(d) Dalton’s law of partial pressure
Ans. (b)

Ques. An element forms two oxides containing respectively 53.33 and 36.36 percent of oxygen. These figures illustrate the law of
(a) Conservation of mass
(b) Constant proportions
(c) Reciprocal proportions
(d) Multiple proportions
Ans. (d)

Ques. The law of definite proportions is not applicable to nitrogen oxide because
(a) Nitrogen atomic weight is not constant
(b) Nitrogen molecular weight is variable
(c) Nitrogen equivalent weight is variable
(d) Oxygen atomic weight is variable
Ans. (c)

Ques. The law of multiple proportions is illustrated by the two compounds
(a) Sodium chloride and sodium bromide
(b) Ordinary water and heavy water
(c) Caustic soda and caustic potash
(d) Sulphur dioxide and sulphur trioxide
Ans. (d)

Ques. A sample of pure carbon dioxide, irrespective of its source contains 27.27% carbon and 72.73% oxygen. The data support
(a) Law of constant composition
(b) Law of conservation of mass
(c) Law of reciprocal proportions
(d) Law of multiple proportions
Ans. (c)

Ques. The percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This illustrates the law of
(a) Constant proportions
(b) Conservation of mass
(c) Multiple proportions
(d) Reciprocal proportions
Ans. (a)

Ques. Hydrogen combines with oxygen to form H2O in which  16 g of oxygen combine with 2 g of hydrogen. Hydrogen also combines with carbon to form CH4 in which 2 g of hydrogen combine with 6 g of carbon. If carbon and oxygen combine together then they will do show in the ratio of
(a) 6 : 16 or 12 : 32
(b) 6 : 18
(c) 1 : 2
(d) 12 : 24
Ans. (a)

Ques. Which of the following is the best example of law of conservation of mass?
(a) 12 g of carbon combines with 32 g of oxygen to form 44 g of CO2
(b) When 12 g of carbon is heated in a vacuum there is no change in mass
(c) A sample of air increases in volume when heated at constant pressure but its mass remains unaltered
(d) The weight of a piece of platinum is the same before and after heating in air
Ans. (a)

Avogadro number is
(a) Number of atoms in one gram of element
(b) Number of millilitres which one mole of a gaseous substances occupies at NTP
(c) Number of molecules present in one gram molecular mass of a substance
(d) All of these
Ans. (c)

Two elements X and Y have atomic weights of 14 and 16. They form a series of compounds ABCD and E in which the same amount of element XY is present in the ratio 1 : 2 : 3 : 4 : 5. If the compound A has 28 parts by weight of X and 16 parts by weight of Y, then the compound of C will have 28 parts weight of X and
(a) 32 parts by weight of Y
(b) 48 parts by weight of Y
(c) 64 parts by weight of Y
(d) 80 parts by weight of Y
Ans. (b)

 
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