Course Content
CHAPTER 1: SOME BASIC CONCEPTS OF CHEMISTRY
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Introduction of Chemistry
14:32
Matter || Pure Substance || Properties of Matter
Quantitative property || The International System of Units (SI) || Prefixes used in SI System || Scientific Notation
Precision & Accuracy || Significant figures || Dimensional Analysis
Law of Chemical Combination
Dalton’s Atomic Theory || Atomic Mass || Average Atomic Mass || Molecular Mass
Mole Concept and Molar Masses || Percentage Composition || Empirical Formula for Molecular formula
Stoichiometry || Limiting Reagent
Reactions in Solution – Mass per cent or weight percent || Mole fraction || Molarity || Molality
NCERT Solutions for Class 11 Chemistry Chapter 1 Some basic Concepts of Chemistry
CBSE Class 11 Chemistry Chapter-1 Important Questions
Chemistry Question Bank || NEET || JEE || AIIMS
Class 11th Chemistry Chapter 1 Quiz Test For Students
01:30:00
CHAPTER 2: STRUCTURE OF ATOM
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INTRODUCTION || DISCOVERY OF ELECTRON
30:35
DISCOVERY OF PROTON || DISCOVERY OF NEUTRON || ATOMIC MODELS OF ATOM
ISOTOPES || ISOBARS || ISOTONES || ISOELECTRONIC
DUAL NATURE OF LIGHT: DEVELOPMENT LEADING TO BOHR’S MODEL || Black body radiation and Photoelectric effect
Planck’s quantum theory || BOHR’S THEORY
EXPLANATION OF HYDROGEN SPECTRA || LIMITATIONS OF BOHR MODEL OF AN ATOM || DUAL NATURE OF PARTICLE || QUANTUM MECHANICS || Pauli’s exclusion Principle
NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom
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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Introduction || Doberneiner’s Triads
26:18
Newland’s Law of Octave || Limitations of Newlands’ law of octaves
Mendeléev’s periodic table || Achievements of Mendeléev’s periodic table || Limitation of Mendeleev’s classification
The Modern Periodic Table || Position of Element in Modern Periodic Table || Trends in the modern periodic table
NCERT Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties
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CHAPTER 4: CHEMICAL BONDING AND MOLECULAR STRUCTURE
4 Chemical Bonding and Molecular Structure 4.1 Kössel-Lewis Approach to Chemical Bonding 4.2 Ionic or Electrovalent Bond 4.3 Bond Parameters 4.4 The Valence Shell Election Pair Repulsion (VSEPR) Theory 4.5 Valence Bond Theory 4.6 Hybridisation 4.7 Molecular Orbital Theory 4.8 Bonding in Some Homonuclear Diatomic Molecules 4.9 Hydrogen Bonding
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Introduction || Kossel and Lewis Approach
15:00
Types of chemical bond || Factors affecting the formation of ionic bond || General properties of ionic compounds
COVALENT BOND || COORDINATE BOND
Bond parameters : Bond Length , Bond Angle, Bond Enthalpy, Bond Order || Resonance
Polarity & Dipole moment || Ionic Percentage || Polarization
VSEPR THEORY || Hybridization || Valence bond theory || Overlapping: sigma and pi bond || Molecular orbital theory
Hydrogen bonding || Hydrogen bonding in ice and water
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CHAPTER 5: STATE OF MATTER
Section Name Topic Name 5 States of Matter 5.1 Intermolecular Forces 5.2 Thermal Energy 5.3 Intermolecular Forces vs Thermal Interactions 5.4 The Gaseous State 5.5 The Gas Laws 5.6 Ideal Gas Equation 5.7 Kinetic Energy and Molecular Speeds 5.8 Kinetic Molecular Theory of Gases 5.9 Behaviour of Real Gases: Deviation from Ideal Gas Behaviour 5.10 Liquefaction of Gases 5.11 Liquid State
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Introduction || State of Matter
27:52
Intermolecular Forces || Dispersion Forces or London Forces
Dipole – Dipole Forces || Dipole – Induced Dipole Force || Hydrogen Bond || Thermal Energy
Intermolecular Forces Vs Thermal Interations || Boyle’s Law || Charles’ Law || Gay Lussac’s Law
Avogadro Law || Ideal Gas Equation || Density and Molar Mass of a Gaseous Substance
Dalton’s Law of Partial Pressure || Partial pressure in terms of mole fraction || Kinetic molecular theory of gases || Behaviour of real gases: deviation from ideal gas behaviour || Surface Tension
Viscosity
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter
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State of Matter Question For NEET || JEE || AIIMS
CHAPTER 6: THERMODYNAMICS
Section Name Topic Name 6 Thermodynamics 6.1 Thermodynamic Terms 6.2 Applications 6.3 Measurement of ∆U and ∆H: Calorimetry 6.4 Enthalpy Change, ∆rH of a Reaction – Reaction Enthalpy 6.5 Enthalpies for Different Types of Reactions 6.6 Spontaneity 6.7 Gibbs Energy Change and Equilibrium
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Thermodynamics of Thermodynamics || Universe || System || Surrounding || types of systems || Internal Energy
24:40
1st Law of Thermodynamics || Enthalpy || Heat capacity || Types of molar heat capacities || Relation between Cp &Cv || Calorimeter
Enthalpies of Reaction || Enthalpy of Combustion || Calorific value || Hess’s law of heat summation || Other type of enthalpies || Born- Haber cycle || Enthalpies of phase transition || Bond enthalpy || Limitation of 1st law of thermodynamics
Entropy || How s & q related? || How S&T related? || Second law of thermodynamics || Standard molar entropy || Gibbs free energy || Third law of thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics
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NCERT Solutions for Class 11 Chemistry Chapter 6 Short Answer Type Questions
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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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Introduction || Types of equilibrium || Boiling point || Equilibrium involving dissolution of solids || General characteristics of physical equilibrium || Equilibrium Involving in Chemical Process
22:43
Law of mass action || Characteristics of Equilibrium constant (k) || Characteristics of chemical equilibrium || Effect of temperature on K
Chemical equilibrium || Characteristics of K || Application of Equilibrium constant || Factors affecting state of equilibrium “Le Chartlier’s principle || Substances: electrolytes and non electrolytes
Acids, bases and salts || 2nd Concept: Bronsted Lowry concept || Conjugate acid –base concept || 3rd concept: Lewis concept || Expressing Hydrogen ion concentration: pH scale || Calculation of PH Value
Protic and non- Protic acids || Hydrolysis of salt || Different kind of salts || Buffer solution || Solubility product ||
NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium
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CHAPTER 8: REDOX REACTIONS
Section Name Topic Name 8 Redox Reactions 8.1 Classical Idea of Redox Reactions-Oxidation and Reduction Reactions 8.2 Redox Reactions in Terms of Electron Transfer Reactions 8.3 Oxidation Number 8.4 Redox Reactions and Electrode Processes
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Introduction || Oxidation and Reduction || oxidation number || Redox reaction in terms of oxidation number
Balancing of Redox reactions || Types of Redox reactions
Redox couple || Electrochemical cell || Functions of salt bridge || Representation of the cell || Electrode potential || Electromotive force || Electromotive force || Electrochemical series || Redox reaction as the basis for Titrations
NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions
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NCERT Solutions for Class 11 Chemistry Chapter 8 Short Answer Type Questions
CHAPTER 9: HYDROGEN
Section Name Topic Name 9 Hydrogen 9.1 Position of Hydrogen in the Periodic Table 9.2 Dihydrogen, H2 9.3 Preparation of Dihydrogen, H2 9.4 Properties of Dihydrogen 9.5 Hydrides 9.6 Water 9.7 Hydrogen Peroxide (H2 O2) 9.8 Heavy Water,D2O 9.9 Dihydrogen as a Fuel
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Introduction || Characteristics of Hydrogen || Difference with alkali metals || Similarities with Halogens || Differences with Halogen || Dihydrogen
15:47
Physical properties of H2 || Chemical Properties || Uses of H2
Hydrides || Covalent or molecular hydrides || Metallic  hydrides
Water || Physical properties || Chemical properties || Hardness of water  || Methods to remove hardness
Hydrogen peroxide || Physical Properties || Chemical Properties || Uses of Hydrogen peroxide
Heavy water || Physical properties || Chemical Properties
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen
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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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Introduction || Diagonal Relationship || Atomic and Ionic Radii of alkali metals
25:32
Ionization Enthalpy and Hydration Enthalpy of alkali metals || Physical properties of alkali metals || Chemical Properties of alkali metals || Anomalous properties of lithium || Washing Soda
Preparation of washing soda by solvay’s process || Sodium Chloride || Caustic Soda || Baking Soda || Electronic Configuration of alkaline earth metal || Atomic and Ionic Radii of alkaline earth metal
Ionization Enthalpy of alkaline earth metal || Hydration Enthalpy of alkaline earth metal || Physical Properties of alkaline earth metal || Chemical Properties of alkaline earth metal
Quick Lime || Slaked Lime || Calcium Carbonate (CaCO3) || Plaster of Paris || Cement || Uses of s-block elements
NCERT Solutions for Class 11 Chemistry Chapter 10 The s-Block Elements
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CHAPTER 11: P-BLOCK ELEMENTS
Section Name Topic Name 11 The p-Block Elements 11.1 Group 13 Elements: The Boron Family 11.2 Important Trends and Anomalous Properties of Boron 11.3 Some Important Compounds of Boron 11.4 Uses of Boron and Aluminium and their Compounds 11.5 Group 14 Elements: The Carbon Family 11.6 Important Trends and Anomalous Behaviour of Carbon 11.7 Allotropes of Carbon 11.8 Some Important Compounds of Carbon and Silicon
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Introduction of p- block || Inert pair effect || Group 13 Elements
Chemical properties of group 13 || Important trends || Uses of Boron || Uses of Aluminum || Borax || PROPERTIES OF BORAX || Uses of Borax
Introduction of Group – 14 || Physical Properties of group 14 || Chemical properties of group 14 || Structure of CCl4 || Allotropes of carbon || Properties and Uses of diamond and graphite || Types of Charcoal || Uses of Carbon, Germanium, Lead, & Tin.
Carbon monoxide (CO) || Carbon dioxide (CO2) || Silicon dioxide (SiO2) || Crystalline form of silica || Structure of Silicon dioxide || Silicones || Silicates || Zeolites
NCERT Solutions for Class 11 Chemistry Chapter 11 The p-Block Elements
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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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Introduction || COMPOUNDS || Friedrich Wohler || IMPORTANT FEATURE OF π BONDS  || Branches of Organic Compounds || Functional Group
19:51
Homologous series || IUPAC system || Different Type of Organic Compounds ||  Alkyl halides
IUPAC and Common names of compounds || RULES FOR IUPAC || In case of double & triple bond
RULES FOR NAMING ORGANIC COMPOUND WITH FUNCTIONAL GROUP || PREFIX USED FOR SECONDARY FUNCTIONAL GROUP || NOMANCLATURE OF AROMATIC COMPOUNDS
ISOMERISM || The structural isomerism type || The different types of stereoisomerism || CONFORMATIONAL ISOMERISM
ELECTRON DISPLACEMENT IN COVALENT BONDS || HYPER CONJUGATION || Fundamental concepts of organic reactions || REACTION INTERMEDIATES
Types of Carbocations || Carboanion || Types of Carboanions ||  Attacking agents
Types of organic reactions || Addition Reaction || Elimination reaction
Methods of purification of organic compounds || Crystallisation || Distillation || Fractional distillation || Distillation under reduced pressure || Steam distillation
Differential extraction || Chromatography || Adsorption chromatography || Column chromatography || Thin layer chromatography || Position chromatography
Detection of carbon and hydrogen || Detection of other elements || Test for Nitrogen || Test for Sulphur || Test for halogens ||  Test for phosphorous || Quantitative analysis || Dumas method || Kjeldahl’s method
NCERT Solutions for Class 11th Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques
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CHAPTER 13: HYDROCARBON
Section Name Topic Name 13 Hydrocarbons 13.1 Classification 13.2 Alkanes 13.3 Alkenes 13.4 Alkynes 13.5 Aromatic Hydrocarbon 13.6 Carcinogenicity and Toxicity
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Introduction || Classification of hydrocarbons || Aromatic || Alkanes || Introduction of alkanes || Isomerism is alkanes
20:34
Classification of Alkanes || Preparation of alkanes
Physical Properties of alkanes || Chemical properties of alkanes
Conformational Isomerism || Torsional strain
Alkenes || ISOMERISM || Stereoisomerism || Stability of Alkenes || Preparation  of Alkenes ||  Physical Properties Of alkenes
Chemical properties of alkenes
Alkynes ||  Structure of ethyne || Isomerism || Preparation of Alkynes || Physical properties of alkynes || Chemical properties of alkynes
Aromatic Compounds || Nomenclature || Isomerism || Stability and structure of benzene || Huckle rule of Aromatic compounds || Preparation of Benzene || Physical properties of benzene || Chemical properties of benzene
NCERT Solutions for Class 11 Chemistry Chapter 13 Hydrocarbons
NCERT Solutions for Class 11 Chemistry Chapter 13 Very Short , Short and Long Answer Type Questions
CHAPTER 14: ENVIRONMENTAL CHEMISTRY
Section Name Topic Name 14 Environmental Chemistry 14.1 Environmental Pollution 14.2 Atmospheric Pollution 14.3 Water Pollution 14.4 Soil Pollution 14.5 Industrial Waste 14.6 Strategies to control Environmental Pollution 14.7 Green Chemistry
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Introduction || Atmosphere || Environmental chemistry || Environmental pollution || Atmospheric pollution || causes of pollution
Types of pollutants || Primary pollutant || Secondary pollutant || Biodegradable || Non-biodegradable
Types of pollution || Air pollution || Water pollution || Soil pollution
Green House Effect || Global warming || Acid rain || Hydrocarbons || Chlorofluorocarbons
Ozone || Ozone hole || Montreal protocol || Particulate pollutants
Smog and Photochemical smog
Water pollution
Soil pollution || Management of industrial waste
Green chemistry
NCERT Solutions for Class 11 Chemistry Chapter 14 Environmental Chemistry
NCERT Solutions for Class 11 Chemistry Chapter 14 Very Short, Short & Long Answer Type Questions
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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 = 10–6 kg

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

⇒ 1 mg = 106 ng

1 mg = 10–6 kg = 106 ng

(iii) 1 mL = 1 mL × (1L/1000mL)

⇒ 1 mL = 10–3 L

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

⇒ 1 mL = 10–3 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 g 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 A, B, C, D and E in which the same amount of element X, Y 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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