Modern ABC Chemistry XI

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© Preface & Acknowledgement Modern Publishers. All rights reserved. I feel pleasure in presenting the revised and updated edition of our book \"Modern's abc + of Chemistry\" for Class XI students. The book has been prepared strictly according to the new syllabus proposed by C.B.S.E. New Delhi and Education Board of other Indian States and N.C.E.R.T. Text Book. Salient Features of the Book ✰ Text matter has been planned giving emphasis on fundamental concepts in a simple, clear and systematic method. ✰ The text has been presented in an interesting style with a large number of Illustrative examples and Numerical problems. ✰ Additional information about the topic is given in Key Note, Key Facts, Facts File, Learning Plus, In focus, Competition Plus to provide stimulus to find more about the subject. ✰ A number of Problems and Short answer questions have been given under the heading Practice Problems promoting problem solving skills in students. ✰ Alarge number of Conceptual Questions (Solved) are given in each chapter. ✰ Advanced Level Problems with solutions are added to accelerate the potential of the students for SOLVING NUMERICAL PROBLEMS for JEE (Advanced). ✰ All the numerical problems of Practice Problems are completely solved in Solution File at the end of each chapter. ✰ Large number of Numerical Problems from I.I.T., Roorkee, M.L.N.R. and other Competitive Examinations have been given in Competition File. ✰ A variety of Multiple Choice Questions from different competitive examinations have been added to make the book useful for the preparation of competitive examinations in Competition File. ✰ MCQs with more than one correct answers and MCQs based on given comprehension/passage, Matrix Match Type Questions and Integer Type or Numerical Value Type Questions according to latest IIT pattern are also given. ✰ In NCERT File, all the Textbook Exercises and Exemplar Problems are fully solved. ✰ At the end of each chapter, Revision Exercises are given according to Latest Examination Pattern covering Objective and Subjective Questions. ✰ A number of Higher Order Thinking Skills (HOTS) and Advanced Level Questions with answers are included at the end of each chapter. ✰ The definition of Key Terms &Laws and QUICK CHAPTER ROUND UP are given at the end under the heading Chapter Summary. ✰ UNIT PRACTICE TESTS with Hints & Solutions are given at the end of each chapter to help the students to check their performance after covering the chapter. A Mock Test with solutions according to CBSE pattern is given at the end of the book. I am extremely thankful to many teachers and students who have been sending their valuable suggestions and comments for the improvement of the book. I am greatly indebted to them. I wish to acknowledge my sincere thanks to Dr. (Mrs.) Sheenu Jauhar for her untiring efforts and valuable contribution in critical revision of the book and updating Objective Questions. I also feel pleasure to express my thanks to Mr. Arun Kamboj (G.S.S. School, Tohana), Dr. G. J. P. Singh (Chemistry Deptt., P.U., Chandigarh), Dr. Sajeev Soni (S.D. College, Chandigarh), Dr. A.N. Sharma (Govt. College, Hamirpur), Ms. Parveen (Faridkot), Mrs. Jagmohan Kaur (Govt. Model S.S. School Sec. 35 Chandigarh), Mrs. Anju Goel (Govt. S. S. School Sec. 16, Chandigarh), Dr. G.S. Arora, Mrs. Sunita Saroha (D.A.V. S.S. School, Sec. 15, Chandigarh), Mr. Gaurav Chakraborty, Mrs. Amita Sharma (Saraswati Institute, Chandigarh), Mrs. Seema (Seema Chemistry classes), Mr. Sudhanshu Jaitley (Chandigarh), Mr. Deepak Mishra (H.O.D. Shri Gauri Shankar Inter College, Ferozabad), Mr. Praveen Kumar (S.V.M. Senior Sec. School, Kosi Kalan, Mathura), Mr. Vikas Chander (Saint Mary S.S. School Gurdaspur), Shri Rakesh Jassotia, (Sr. Lecturer, Govt. G.H.S.S. Nagari PAROLE, Kathua) and Mr. Vaibhav Yavlekar (Ujjain. I am also highly thankful to Mr. Vinod Kumar Jangra (G.S.S.S. Akbarpur Baroto, Sonepat), Mr. Chander Mohan Kumar (Retd. P.G.T., Chemistry, Sacred Heart Convent School, Sarabha Nagar, Ludhiana) and Dr. Anjul Rajput (Dhampur) for thoroughly checking of the book and providing valuable suggestions. I am also thankful to Simran Kaur for her sincere and dedicated editorial work. Finally, I am happy to express my sincerest thanks and indebtness to our dynamic and versatile publisher and his efficient staff for making the project successful. I am also thankful to Mr. Manik Juneja, Director — Content and Production, Mr. B.S. Rawat, Mr. S.K. Sikka, Mr. Ravinder Pathania and L.B. Mishra who have taken great pains in bringing up the book. I would also like to gratefully appreciate the tremendous and valuable efforts of Mr. Manu Jauhar for excellent thoughtful editing of the book. I hope that the present book will be warmly received by the students and the teachers. Suggestions for the further improvement of the book will be gratefully acknowledged. – Dr. S.P. Jauhar

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Atomic Masses (C12 = 12.00) and Electronic Configurations of Elements © Element Symbol Atomic Atomic Electronic Configuration Modern Publishers. All rights reserved. Number Mass Actinium Ac [Rn]86 6dl 7s2 Aluminium Al 89 227 [Ne]10 3s2 3pl Americium Am 13 26.9 [Rn]86 5f 7 7s2 Antimony Sb 95 243 [Kr]36 4d10 5s2 5p3 Argon Ar 51 121.75 [Ne]10 3s2 3p6 Arsenic As 18 39.94 [Ar]18 3d10 4s2 4p3 Astatine At 33 74.92 [Xe]54 4f14 5d10 6s2 6p5 Barium Ba 85 210 [Xe]54 6s2 Berkelium Bk 56 137.3 [Rn]86 5f9 7s2 Beryllium Be 97 247 [He]2 2s2 Bismuth Bi 4 9.01 [Xe]54 4f14 5d10 6s2 6p3 Boron B 83 208.98 [He]54 2s2 2p1 Bromine Br 5 10.81 [Ar]18 3d10 4s2 4p5 Cadmium Cd 35 79.90 [Kr]36 4d10 5s2 Calcium Ca 48 112.40 [Ar]18 4s2 Californium Cf 20 40.02 [Rn]86 5f10 7s2 Carbon C 98 251 [He]2 2s2 2p2 Cerium Ce 6 12 [Xe]54 4f1 5d1 6s2 Cesium Cs 58 140.12 [Xe]54 6s1 Chlorine Cl 55 132.90 [Ne]10 3s2 3p5 Chromium Cr 17 35.45 [Ar]18 3d5 4s1 Cobalt Co 24 51.99 [Ar]18 3d7 4s2 Copernicium Cn 27 58.93 [Rn]86 5f14 6d10 7s2 Copper Cu 112 285 [Ar]18 3d10 4s1 Curium Cm 29 63.54 [Rn]86 5f7 6d1 7s2 Dysprosium Dy 96 245 [Xe]54 4f10 6s2 Einsteinium Es 66 162.50 [Rn]86 5f11 7s2 Erbium Er 99 254 [Xe]54 f12 6s2 Europium Eu 68 167.26 [Xe]54 4f7 6s2 Fermium Fm 63 151.96 [Rn]86 5f12 7s2 Flerovium Fl 100 257 [Rn]865f146d107s27p2 Fluorine F 114 289 [He]2 2s2 2p5 Francium Fr 9 18.99 [Rn]86 7s1 Gadolinium Gd 87 223 [Xe]54 4f7 5d1 6s2 Gallium Ga 64 157.25 [Ar]18 3d10 4s2 4pl Germanium Ge 31 69.72 [Ar]18 3d10 4s2 4p2 Gold Au 32 72.59 [Xe]54 4f14 5d10 6s1 Hafnium Hf 79 196.99 [Xe]54 4f14 5d2 6s2 Hanium Ha 72 178.48 [Rn]86 5f14 6d3 7s2 Helium He 105 260 1s2 Holmium Ho 2 4 [Xe]54 4f11 6s2 Hydrogen H 67 164.93 1s1 Indium In 1 1 [Kr]36 4d10 5s2 5pl Iodine I 49 114.82 [Kr]36 4d10 5s2 5p5 Iridium Ir 53 126.90 [Xe]54 4f14 5d7 6s2 Iron Fe 77 192.2 [Ar]18 3d6 4s2 Krypton Kr 26 55.84 [Ar]18 3d10 4s2 4p6 Lanthanum La 36 83.80 [Xe]54 5d1 6s2 Lawrencium Lr 57 138.91 [Rn]86 5f14 6d1 7s2 Lead Pb 103 257 [Xe]54 4f14 5d10 6s2 6p2 Lithium Li 82 207.19 [He]2 1s1 Livermorium Lv 3 6.93 [Rn]86 5f146d107s27p4 Lutetium Lu 116 293 [Xe]54 4f14 5d1 6s2 Magnesium Mg 71 174.97 [Ne]10 3s2 Manganese Mn 12 24.31 [Ar]18 3d5 4s2 25 54.93

© Element Symbol Atomic Atomic Electronic Configuration Modern Publishers. All rights reserved. Number Mass Mendelevium Md [Rn]86 5f13 7s2 Mercury Hg 101 256 [Xe]54 4f14 5d10 6s2 Molybdenum Mo 80 200.50 [Kr]36 4d5 5s1 Moscovium Mc 42 95.94 [Rn]86 5f14 6d10 7s2 7p3 Neodymium Nd 115 289 [Xe]54 4f4 6s2 Neon Ne 60 144.24 [He]2 2s2 2p6 Neptunium Np 10 20.18 [Rn]86 5f4 6d1 7s2 Nickel Ni 93 237 [Ar]18 3d8 4s2 Nihonium Nh 28 58.71 [Rn]86 5f14 6d10 7s2 7p1 Niobium Nb 113 286 [Kr]36 4d4 5s1 Nitrogen N 41 92.90 [He]2 2s2 2p3 Nobelium No 7 14 [Rn]56 5f14 7s2 Oganesson Og 102 254 [Rn]86 5f14 6d10 7s2 7p6 Osmium Os 118 294 [Xe]54 4f14 5d6 6s2 Oxygen O 76 190.2 [He]2 2s2 2p4 Palladium Pd 8 15.99 [Kr]36 4d10 Phosphorus P 46 106.4 [Ne]10 3s2 3p3 Platinum Pt 15 30.97 [Xe]54 4f14 5d9 6s1 Plutonium Pu 78 195.09 [Rn]86 5f6 7s2 Polonium Po 94 244 [Xe]54 4f14 5d10 6s2 6p4 Potassium K 84 210 [Ar]18 4s1 Praseodymium Pr 19 39.102 [Xe]54 4f3 6s2 Promethium Pm 59 140.907 [Xe]54 4f5 6s2 Protactinium Pa 61 145 [Rn]86 5f2 6d1 7s2 Radium Ra 91 231 [Rn]86 7s2 Radon Rn 88 226 [Xe]54 4f14 5d10 6s2 6p6 Rhenium Re 86 222 [Xe]54 4f14 5d5 6s2 Rhodium Rh 75 186.2 [Kr]36 4d8 5s1 Rubidium Rb 45 102.90 [Kr]36 5s1 Ruthenium Ru 37 85.47 [Kr]36 4d7 5s1 Rutherfordium Rf 44 101.07 [Rn]86 5f14 6d2 7s2 Samarium Sm 104 257 [Xe]54 4f6 6s2 Scandium Sc 62 150.35 [Ar]18 3d1 4s2 Selenium Se 21 44.95 [Ar]18 3d10 4s2 4p4 Silicon Si 34 78.96 [Ne]10 3s2 3p2 Silver Ag 14 28.08 [Kr]22 4d10 5s1 Sodium Na 47 107.87 [Ne]10 3s1 Strontium Sr 11 22.98 [Kr]36 5s2 Sulphur S 38 87.62 [Ne]10 3s2 3p4 Tantalum Ta 16 32.06 [Xe]54 4f14 5d 3 6s2 Technetium Tc 73 180.94 [Kr]36 4d5 5s2 Tellurium Te 43 99 [Kr]36 4d10 5s2 5p4 Tennessine Ts 52 127.60 [Rn]86 5f14 6d10 7s2 7p5 Terbium Tb 117 294 [Xe]54 4f9 6s2 Thallium Tl 65 158.92 [Xe]54 4f14 5d10 6s2 6 p1 Thorium Th 81 204.37 [Rn]86 6d2 7s2 Thulium Tm 90 232.03 [Xe]54 4f13 6s2 Tin Sn 69 168.93 [Kr]36 4d10 5s2 5p2 Titanium Ti 50 118.69 [Ar]18 3d2 4s2 Tungsten W 22 47.90 [Xe]54 4f14 5d4 6s2 Uranium U 74 183.85 [Rn]86 5f2 6d1 7s2 Vanadium V 92 238.02 [Ar]18 3d3 4s2 Xenon Xe 23 50.94 [Kr]36 4d10 5s2 5p6 Ytterbium Yb 54 131.04 [Xe]54 4f14 6s2 Yttrium Y 70 173.04 [Kr]36 4d1 5s2 Zinc Zn 39 88.90 [Ar]18 3d10 4s2 Zirconium Zr 30 63.37 [Kr]36 4d2 5s2 40 91.22

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Award© ofModern Publishers. All rights reserved. Excellence First Prize Nytra: a first augmented reality app for K-12 Education in India. To help the students to visualize the concepts more easily and enhance understanding – a digital revolution: Watch chemistry come alive.

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SYLLABUS Total Periods (Theory 160 + Practical 60) CHEMISTRY CLASS XI (THEORY) ©Time : 3 Hours Total Marks : 70 Modern Publishers. All rights reserved. Unit No. Title No. of Periods Marks Unit I Some Basic Concepts of Chemistry 12 11 Unit II Structure of Atom 14 Unit III Classification of Elements and Periodicity in Properties 08 04 Unit IV Chemical Bonding and Molecular Structure 14 Unit V States of Matter : Gases and Liquids 12 Unit VI Chemical Thermodynamics 16 21 Unit VII Equilibrium 14 Unit VIII Redox Reactions 06 16 Unit IX Hydrogen 08 Unit X s-Block Elements 10 18 Unit XI p-Block Elements 14 70 Unit XII Organic Chemistry : Some Basic Principles and Techniques 14 Unit XIII Hydrocarbons 12 Unit XIV Environmental Chemistry 06 Total 160 Unit I : Some Basic Concepts of Chemistry (12 Periods) General Introduction : Importance and scope of chemistry. Nature of matter, laws of chemical combination, Dalton’s atomic theory : concept of elements, atoms and molecules. Atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry. Unit II : Structure of Atom (14 Periods) Bohr’s model and its limitations, concept of shells and subshells, dual nature of matter and light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbitals, quantum numbers, shapes of s, p and d orbitals, rules for filling electrons in orbitals – Aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of atoms, stability of half-filled and completely filled orbitals. Unit III : Classification of Elements and Periodicity in Properties (08 Periods) Modern periodic law and the present form of periodic table, periodic trends in properties of elements – atomic radii, ionic radii, inert gas radii, ionization enthalpy, electron gain enthalpy, electronegativity, valency. Nomenclature of elements with atomic number greater than 100. Unit IV : Chemical Bonding and Molecular Structure (14 Periods) Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules (qualitative idea only), hydrogen bond.

Unit V : States of Matter : Gases and Liquids (12 Periods) Three states of matter, intermolecular interactions, types of bonding, melting and boiling points, role of gas laws in elucidating the concept of the molecule, Boyle’s law, Charles law, Gay Lussac’s law, Avogadro’s law, ideal behaviour, empirical derivation of gas equation, Avogadro’s number, ideal gas equation. Deviation from ideal behaviour, liquefaction of gases, critical temperature, kinetic energy and molecular speeds (elementary idea). © Modern Publishers. All rights reserved.Liquid state : Vapour pressure, viscosity and surface tension (qualitative idea only, no mathematical derivations). Unit VI : Chemical Thermodynamics (16 Periods) Concepts of systems and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics – internal energy and enthalpy, heat capacity and specific heat, measurement of DU and DH, Hess’s law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution. Second law of thermodynamics (brief introduction). Introduction of entropy as a state function, Gibb’s energy change for spontaneous and non- spontaneous processes, criteria for equilibrium. Third law of thermodynamics (brief introduction). Unit VII: Equilibrium (14 Periods) Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium – Le Chatelier’s principle; ionic equilibrium – ionization of acids and bases, strong and weak electrolytes, degree of ionization, ionization of poly basic acids, acid strength, concept of pH. Henderson equation, hydrolysis of salts (elementary idea), buffer solutions, solubility product, common ion effect (with illustrative examples). Unit VIII : Redox Reactions (06 Periods) Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions in terms of loss and gain of electrons and change in oxidation number, applications of redox reactions. Unit IX : Hydrogen (08 Periods) Position of hydrogen in periodic table, occurrence, isotopes, preparation, properties and uses of hydrogen, hydrides-ionic covalent and interstitial; physical and chemical properties of water, heavy water, hydrogen peroxide-preparation, reactions and structure and use; hydrogen as a fuel. Unit X : s-Block Elements (Alkali and Alkaline Earth Metals) (10 Periods) Group 1 and Group 2 elements : General introduction, electronic configuration, occurrence, anomalous properties of the first element of each group, diagonal relationship, trends in the variation of properties (such as ionization enthalpy, atomic and ionic radii), trends in chemical reactivity with oxygen, water, hydrogen and halogens, uses. Preparation and properties of some important compounds : Sodium carbonate, Sodium chloride, Sodium hydroxide and Sodium hydrogencarbonate, Biological importance of Sodium and Potassium. Calcium oxide and Calcium carbonate and their industrial uses, Biological importance of Magnesium and Calcium. Unit XI : p-Block Elements (14 Periods) General Introduction to p-Block Elements Group 13 Elements: General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity, anomalous properties of first element of the group, Boron – physical and chemical properties, some important compounds, Borax, Boric acids, Boron Hydrides, Aluminium: Reactions with acids and alkalies, uses.

Group 14 Elements: General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity, anomalous behaviour of first elements. Carbon-catenation, allotropic forms, physical and chemical properties; uses of some important compounds: oxides. Important compounds of silicon and a few uses: Silicon tetrachloride, Silicones, Silicates and Zeolites their uses. Unit XII : Organic Chemistry – Some Basic Principles and Techniques (14 Periods) © Modern Publishers. All rights reserved.General introduction, methods of purification, qualitative and quantitative analysis, classification and IUPAC nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyper-conjugation. Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions. Unit XIII : Hydrocarbons (12 Periods) Classification of Hydrocarbons Aliphatic Hydrocarbons: Alkanes – Nomenclature, isomerism, conformation (ethane only), physical properties, chemical reactions including free radical mechanism of halogenation, combustion and pyrolysis. Alkenes – Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markownikov’s addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition. Alkynes – Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of hydrogen, halogens, hydrogen halides and water. Aromatic hydrocarbons – Introduction, IUPAC nomenclature, benzene: resonance, aromaticity, chemical properties, mechanism of electrophilic substitution. Nitration, sulphonation, halogenation, Friedel Craft’s alkylation and acylation, directive influence of a functional group in mono-substituted benzene. Carcinogenicity and toxicity. Unit XIV : Environmental Chemistry (06 Periods) Environmental pollution – air, water and soil pollution, chemical reactions in atmosphere, smog, major atmospheric pollutants, acid rain, ozone and its reactions, effects of depletion of ozone layer, greenhouse effect and global warming – pollution due to industrial wastes, green chemistry as an alternative tool for reducing pollution, strategies for control of environmental pollution.

1. Some Basic Concepts of Chemistry© 1/1 – 1/140 2. Structure of AtomModern Publishers. All rights reserved. 2/1 – 2/120 3. Classification of Elements and Periodicity in Properties 4. Chemical Bonding and Molecular Structure 3/1 – 3/70 5. States of Matter : Gases and Liquids 4/1 – 4/114 6. Thermodynamics 5/1 – 5/106 7. Equilibrium 6/1 – 6/132 7/1 – 7/173  Hints & Solution for (Unit Practice Tests)  Additional Questions 1–4 (According to the Latest CBSE Question Paper Designed by CBSE) 1 – 34  Appendices 1–6  Logarithm & Antilogarithm Tables (i) – (v)

UNIT© 22.4 L Modern Publishers. All rights reserved. at STP 1 Volume 1 32g Mole Particles S Mass SOME BASIC Gram atom or 6.022 ´ 1023 CONCEPTS Gram mole OF CHEMISTRY 58.5g NaCl Chemistry is the study of the materials that make up the universe and the changes which these materials undergo. The study deals with the composition, structure and properties of matter. These aspects can be best understood in terms of the constituents of matter: atoms and molecules. In fact, chemistry is called the science of atoms and molecules. Chemistry is defined as: the branch of science which deals with the study of composition, properties and structures of matter and the changes which the matter undergoes under different conditions and the laws which govern these changes. Chemistry plays a central role in science and is often interlinked with other branches of science. It also plays an important role in our daily life. OBJECTIVES Building on..... Assessing..... Preparing for Competition..... Understanding Text 1 REVISION EXERCISES 102 Additional Useful Information 113 Topicwise MCQs 117 Conceptual Questions 32, 45, 76 Answers/Hints for Competitive Examination Qs Advanced Level Problems 47, 77 Revision Exercises 106 SOLUTION FILE HOTS & Advanced Level AIPMT, NEET & Other State Boards’ Hints & Solutions for Practice Questions with Answers109 Medical Entrance 119 Problems 79 JEE (Main) & Other State Boards’ CHAPTER SUMMARY & QUICK UNIT PRACTICE TEST 141 Engineering Entrance 121 CHAPTER ROUND UP 90 JEE (Advance) for IIT Entrance 124 NCERT FILE Hints & Explanations for Difficult Textbook Exercises with Objective Questions 127 Solutions 92 NCERT Exemplar Problems with Answers/Hints 96 1/1

© 1/2 MODERN’S abc + OF CHEMISTRY–XI Modern Publishers. All rights reserved.Fig. 1 Applications of chemistry in different fields.

© SOME BASIC CONCEPTS OF CHEMISTRY 1/3 Modern Publishers. All rights reserved. IMPORTANCE AND SCOPE OF CHEMISTRY tranquilizers (for treatment of stress and mental In chemistry, we study the composition of diseases), antimicrobials (to cure infections), antimalarials (to treat malaria), etc. Antibiotics materials to find out what they are made of. The such as penicillin, amoxycillin, streptomycin and marvellous thing about chemistry is that all the matter broad spectrum antibiotics such as tetracycline, in the universe is made up of tiny smallest particles chloramphenicol, etc., have cured a variety of diseases called atoms or molecules. It is very interesting that due to harmful micro-organisms. Antifertility drugs only about 118 types of atoms make up whole of the have been world-wide used for birth control methods. matter in the universe. We also learn how their composition affects their characteristics and behaviour Chemistry has given us a large number of life so that we can plan to make new materials with saving drugs. properties of our interests. For this, we have to learn how substances undergo changes in composition and Life saving drugs like cisplatin and taxol for cancer properties. These changes in the language of chemistry therapy and AZT (azidothymidine) used for helping are called chemical reactions. Thus, studying AIDS victims are latest contributions of chemistry chemistry is essential for us to understand better the in medicines. world in which we live. Chemistry for our comforts, pleasure and Since there is a large variety of substances in the universe, the scope of chemistry is immense. Whether luxuries we are concerned with living systems in biology and medicine, with materials such as iron, steel and concrete Chemistry has also pioneer contribution towards as in engineering or with manufacture of computer our comforts, pleasure and luxuries. chips, we deal with chemistry. Progress in modern society is completely based on advances in chemistry. ● It has given us building materials, synthetic In today’s technological age, the importance of fibres and variety of articles of domestic use. chemistry is increasing. Chemical principles are playing key role in diverse areas such as weather patterns, ● Chemical industries manufacturing fertilizers, functioning of brain and operation of computers. acids, alkalies, salts, dyes, polymers, drugs, soaps, detergents, metals, alloys and other organic and It is difficult today to imagine our life inorganic chemicals including new materials have without chemistry. improved our national economy. Main Applications of Chemistry Chemistry for giving us new materials Chemistry has helped in agriculture, food, In recent years, chemistry has given us new medicine, warfare, transportation, computers and in materials such as super-conducting ceramics, our everyday life. The important applications of conducting polymers, optical fibres, microalloys, chemistry in different fields are shown in Fig. 1 and carbon fibres, etc. are discussed below: Chemistry in war Chemistry for meeting our basic needs and necessities of life. Chemistry has also increased the striking power of a country in war times. It is responsible for the Chemistry has helped significantly in meeting discovery of highly explosive substances such as human needs for food, health care products and other dynamite, TNT (trinitrotoluene), nitroglycerine, necessities of life. poisonous gases such as mustard gas, phosgene, lewisite and many deadly weapons such as atom bomb ● Chemistry has provided chemical fertilizers, and hydrogen bomb. improved varieties of insecticides, fungicides and pesticides to increase the yield of crops and fruits. Future Goals and Challenges for Chemistry In recent years, chemistry has solved with a fair ● The use of preservatives has helped to keep food materials for longer periods. degree of success some of the challenging aspects of environmental degradation. So far alternatives to ● Chemistry has also helped for better health environmentally hazardous refrigerants like CFCs and sanitation. The epidemics such as cholera, small (chlorofluoro carbons), responsible for ozone depletion pox, plague have now become things of the past. in the stratosphere has been successfully developed. Still many goals are there for the chemists to achieve ● The discovery of anaesthetics has made successfully. Environmental problems, management surgical operations more and more successful. of Green house gases (like methane, carbon dioxide, etc.) understanding bio-chemical processes, use of ● Chemistry has also given us a variety of drugs enzymes for large scale production of chemicals, such as antipyretics (to lower body temperature in synthesis of exotic materials are some main challenges high fever), analgesics (to relieve pain), for the future chemists. A developing country like India looks forward towards intelligent, talented and creative chemists for accepting these challenges.

© 1/4 MODERN’S abc + OF CHEMISTRY–XI Modern Publishers. All rights reserved. NATURE OF MATTER Because of such arrangement of particles, different The entire universe is made up of matter and states of matter exhibit different characteristics. The arrangement of particles in solids, liquids and gaseous energy. state is shown in Fig. 2. Anything that has mass and occupies space is called matter. Fig. 2. Arrangement of particles in solid, liquid and gaseous state. The presence of matter can be felt by one or more of our senses. Everything around us, for example, book, These three states of matter are interconvertible pen, table, sugar cubes, iron rod, water, air are by changing the conditions of temperature and composed of matter because they have mass and they pressure. On heating, a solid changes to a liquid and occupy space. on further heating the liquid changes to the gaseous (or vapour) state. In the reverse process, a gas on Matter can be classified in two ways : cooling liquefies to the liquid and liquid on further A : Physical classification of matter cooling freezes to the solid. B : Chemical classification of matter. A : Physical Classification of Matter SOLID heat LIQUID heat GAS Depending upon the physical state of matter, it can be classified into three states, namely, solid, liquid cool cool and gaseous state. (i) Solid state. A solid has a definite shape and B. Chemical Classification of Matter definite volume. Thus, solids are generally hard and The chemical classification of matter is based upon rigid. For example, wood, table, copper rod, common salt, etc. its composition. Different substances differ from each (ii) Liquid state. A liquid has a definite volume other in their constituents composition. On the basis but not definite shape. A liquid takes the shape of the of chemical composition, matter can be classified as : container in which it is placed. For example, water, milk, oil, etc. A : Pure substances (iii) Gaseous state. A gas neither possesses a B : Mixtures definite volume nor a definite shape. It occupies the A. Pure substances whole of the volume of the vessel in which it is placed. Pure substances consist of single type of For example, air, oxygen, hydrogen, carbon dioxide, etc. particles. All the constituent particles of a pure Particle Nature of Matter substance are same in their chemical nature. For As we know matter is made up of particles. In terms example, copper, silver, gold, water, glucose, sodium of particle concept of matter, in solids, the particles chloride (common salt) are some examples of pure are closely packed and the empty spaces between the substances. As we know, water contains hydrogen and particles are very small. Due to close packing of the oxygen but they are always present in a fixed ratio particles, they can only vibrate about their fixed and therefore, like all other pure substances behave positions i.e. can have only vibratory motion. Due to as single substance having fixed chemical fixed positions of the particles in solids, they have composition. Similarly, glucose contains carbon, highly ordered arrangement and this gives definite hydrogen and oxygen in a fixed ratio and thus behaves regular shape to the crystals. The regular ordered as a pure substance. The constituents of pure arrangement of particles in solids is called lattice. substances cannot be separated by simple physical The attractive forces among the particles called methods (like filtration, evaporation, distillation, interparticle forces are strong in solids. Thus, solids sublimation, mechanical separation, etc.). These can have a definite shape and a definite volume. only be separated by chemical or electrochemical In liquids, the particles are loosely packed and methods. empty spaces between them are relatively large. Due Pure substances can be further classified into to loose packing, the attractive forces between them elements and compounds. are relatively weak. However, these are not weak enough to allow the particles to separate from one another. Thus, liquids have definite volume but donot have fixed shape. They take the shape of the container. On the other hand, in gases, the particles are very loosely packed and empty spaces between them are very large. As a result, the attractive forces between the particles are very very small so that their movement is easy and fast. They can move to each corner of the vessel and therefore gases donot have definite shape and definite volume.

© SOME BASIC CONCEPTS OF CHEMISTRY 1/5 Modern Publishers. All rights reserved. 1. Element The compounds may be classified into two types as : An element is the simplest form of a pure substance. It may be defined as : (a) Inorganic compounds. These are the compounds which are obtained from non-living sources the simplest form of a pure substance which such as rocks, minerals, etc. For example, common can neither be decomposed into nor built salt, marble, washing soda, etc. from simpler substances by ordinary physical or chemical methods. (b) Organic compounds. These are the The common examples of elements are hydrogen, compounds which are obtained from living sources oxygen, nitrogen, sulphur, iron, lead, gold, mercury, such as plants and animals. All these contain carbon. etc. There are about 118 elements known at present. For example, carbohydrates, oils, fats, waxes, proteins, Out of these, 92 have so far been found to occur etc. naturally and the remaining have been prepared in the laboratory. All the elements do not occur in the B. Mixtures crust of earth in equal proportions. About twenty elements make up 99% of the earth’s crust. The most Mixtures contain more than one kind of pure form abundant elements in the earth’s crust are oxygen, of matter, known as substance. It is a simple silicon, aluminium, iron, calcium, sodium, potassium, combination of two or more substances in which the hydrogen, chlorine, carbon, etc. constituent substances retain their identities. The Elements are further classified into following types : composition of the mixture may be varied to any (a) Metals. These elements are generally solids and extent. The substances present in a mixture are called possess characteristics such as bright lustre, hardness and its components. Therefore, a mixture may contain ability to conduct electricity and heat. These are generally two or more substances in any ratio. Thus, a mixture malleable (can be beaten into thin sheets) and ductile (can may be defined as: be drawn into wires). Some common examples of metals are copper, iron, silver, gold, aluminium, etc. About 80% a combination of two or more elements or of the known elements are metals. compounds in any proportion so that the (b) Non-metals. These elements are generally components do not lose their identity. non lustrous, brittle and poor conductors of heat and electricity. The common examples of non-metals are Many of the substances present around us are carbon, hydrogen, oxygen, nitrogen, etc. mixtures. For example, sugar solution in water, air, (c) Metalloids. These are elements which have tea, brass (an alloy of copper and zinc), soft drink, soil, characteristics common to both metals and non-metals. etc., are all mixtures. The common examples of metalloids are silicon, arsenic, bismuth, antimony, etc. Mixtures are of two types: 2. Compound (a) Homogeneous mixture Compounds are pure substances containing more (b) Heterogeneous mixture than one kind of element. A compound may be defined as : (a) Homogeneous mixture. A mixture is said to a pure substance containing two or more be homogeneous if it has a uniform composition than two elements combined together in a throughout. The components of a mixture cannot be seen definite proportion by mass and which can even under a powerful microscope. Homogeneous be decomposed into its constituent elements mixtures are also called solutions. Some examples of by suitable chemical methods. homogeneous mixtures are air, gasoline, sea water, brass, A compound always contains the same elements etc. Air is a homogeneous mixture of a number of gases in definite or fixed proportion by weight. For example, such as oxygen, nitrogen, carbon dioxide, water vapour, water always contains hydrogen and oxygen in the etc. The composition of air is the same everywhere. ratio of 1 : 8 by mass. Similarly, carbon dioxide always contains carbon (b) Heterogeneous mixture. A mixture is said and oxygen in the ratio of 3 : 8. to be heterogeneous if its composition is not uniform throughout. These mixtures consist of two or more The properties of the compounds are totally parts (called phases) which have different different from the elements from which these are compositions. These have visible boundaries of formed. For example, hydrogen is combustible while separation between the different constituents and can oxygen is supporter of combustion but water be easily seen even with naked eye. For example, a (a compound of hydrogen and oxygen) is normally used mixture of iron filings, common salt and sulphur gives for extinguishing fire. a heterogeneous mixture in which the components can be seen lying side by side very easily.

1/6 MODERN’S abc + OF CHEMISTRY–XI Differences between Compound and Mixture Let us sum up important differences between compounds and mixtures. Compounds Mixtures 1. In a compound, two are more elements are 1. In a mixture, two or more elements or compounds combined chemically. just mix together. 2. The compounds contains two or more elements in 2. The components of a mixture may be present in a fixed ratio by mass. Its composition is always fixed. any ratio. Its composition is variable. 3. A compound has a definite formula. 3. A mixture does not have a definite formula. 4. A compound is always homogeneous i.e., has the 4. A mixture may be homogeneous or heterogeneous. same composition throughout. 5. No chemical reaction takes place and therefore, 5. A chemical reaction takes place and therefore, the the formation of mixture is not accompained by any energy change. formation of a compound takes place with absorption or evolution of energy. 6. A mixture shows the properties of its 6. The properties of a compound are entirely different constituents. from those of its constituents. 7. A compound cannot be separated into its 7. A mixture can be separated into its constituents constituents by ordinary physical methods. These by physical methods (like filtration, evaporation, can be separated by chemical or electrochemical distillation, sublimation, mechanical separation reactions. etc.) 8. A compound has a fixed melting point, boilling point, etc. 8. A mixture does not have fixed melting point, boiling point, etc. © The classification of matter has been summed up below :Modern Publishers. All rights reserved. PROPERTIES OF MATTER AND THEIR chemical properties are those properties in MEASUREMENT which a chemical change in the substance takes place. Every substance has unique or characteristic The observation or measurement of chemical properties. These properties may be classified as properties involve chemical reactions. The examples physical properties and chemical properties. of chemical properties are characteristic reactions of different substances such as acidity, basicity, Physical properties are those properties combustibility and reactions with other elements and which can be measured or observed without compounds. changing the identity or composition of the Many physical properties of matter such as mass, substance. length, volume, temperature, pressure, density, etc., For example, colour, odour, melting point, boiling are quantitative in nature. Therefore, these properties point, density, etc. On the other hand, are also called physical quantities.

SOME BASIC CONCEPTS OF CHEMISTRY 1/7 Expressing a Physical Quantity abbreviated as SI units. These units were adopted by The value of a physical quantity is always General Conference of Weights and Measures in 1960. These are abbreviated as SI and the designation comes expressed in two parts : from its French name Systeme Internationale. (i) numerical value and (ii) unit. For example, we may express correctly the weight The SI system has seven base units and are listed in Table 1. These base units pertain to seven of a box as 5700 g or 5.7 kg. The simple numerical figure fundamental scientific quantities and all other units 5700 or 5.7 does not convey any meaningful information. can be derived from these. Thus, it is essential to include units with every© experimental value. A unit is defined asModern Publishers. All rights reserved.Table. 1. Seven Base SI Units. the standard of reference chosen to measure Physical Symbol for Name of Symbol any physical quantity. Quantity Naturally, we must have a convenient system of units quantity Unit for assigning numerical values to the measured or calculated values. Fortunately, it turns out that it is Length l metre m sufficient to define some basic units such as units of Mass m kilogram kg mass, length and time. Since these are independent Time t s units and cannot be derived from any other unit, these Thermodynamic T second K are also called fundamental units. temperature I kelvin The units for other quantities can be derived from Electric current Iv A these units and hence are called derived units. Luminous intensity n ampere cd There are many different systems of units. For Amount of substance candela mol the first time in 1791, a study committee of the French Academy of Science devised a system called “the mole metric system” which became popular in the scientific community throughout the world. The fundamental It may be noted that the unit of temperature units of metric system are the grams for mass, the according to SI system is Kelvin but still celsius scale metre for length and the litre for the volume. In India, (°C) of temperature is commonly used in our daily life. the metric system was adopted in 1957. The metric These two units are related as : system is a decimal system and the different units for Kelvin temperature (K) = °C + 273.15 a physical quantity are related in powers of ten. The different powers are generally indicated by a prefix Similarly, Angstrom (Å) is commonly used as a attached to the unit. unit of length in chemistry. It is almost of the same size as the size of an atom. It is equal to 10–10 m. SI Units In recent years, the scientists have generally 1 Å = 10–10 m However, it must be kept in mind that it is not agreed to use the International System of Units SI unit. In SI units, we may use nanometre (nm) or picometre (pm) : 1 nm = 10–9 m 1 pm = 10–12 m The definitions of the SI base units are given in Table 2. Table 2. Definitions of seven SI base units. Measured Quantity Unit Definition Length metre The metre is the length of the path travelled by light in vacuum during a Mass kilogram time interval of 1/299, 792,458 of a second. The kilogram is the unit of mass and is equal to the mass of the Time second international prototype of the kilogram. It is also defined as the mass of Electric current ampere a platinum block stored at the International Bureau of Weights and Measures in France. Thermodynamic kelvin The second is the duration of 9, 192, 631, 770 periods of the radiation temperature mole corresponding to the transition between the two hyperfine levels of the Amount of substance candela ground state of the cesium- 133 atom. The ampere is that constant current which, if maintained in two straight Luminous intensity or parallel conductors of infinite length, of negligible cross-section, and Luminosity placed one metre apart in vacuum, would produce between these conductors a force equal to 2 × 10–7 newton per metre of length. The kelvin is the unit of thermodynamic temperature and equals exactly 1/273.16 of the thermodynamic temperature of the triple point of water. The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon–12. The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 × 1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.

1/8 MODERN’S abc + OF CHEMISTRY–XI PREFIXES The SI units of some of the physical quantities are either too small or too large. To change the order of magnitude, these are expressed by using prefixes before the name of the base units. The various prefixes are listed in Table 3. Table 3. List of common prefixes used in SI system. © Modern Publishers. All rights reserved.MultiplePrefixSymbolMultiple Prefix Symbol 10−1 deci d 101 deca da 10−2 centi c 102 hecto h 10−3 milli m 103 kilo k 10−6 micro μ 106 mega M 10−9 nano n 109 giga G 10−12 pico p 1012 tera T 10−15 femto f 1015 peta P 10−18 atto a 1018 exa E 10–21 zepto z 1021 zetta Z 10–24 yocto y 1024 yotta Y SOME COMMONLY USED QUANTITIES balance commonly used in laboratories is shown in Let us learn about some common quantities which Fig. 3 (a). These days, many types of electronic balances are available (Fig. 3 b). The analytical balances usually we often come across. weigh to ± 0.0001 g, although some balances are 1. Mass and weight. available which can weigh to ± 0.00001 g. Mass of a substance is the amount of matter The SI unit of mass is kilogram, (kg). Because present in it. kilogram is too large unit for many purposes, its fraction, gram, g, is commonly used and 1 kg = 1000 Weight is the force exerted on an object by the pull g. The smaller units such as milligram, mg (1mg = of gravity. The mass of a substance is constant and is 10–3 g), microgram, μg (1 μg = 10–3 mg or = 10–6 g) are independent of its location. For example, your body also commonly used. has same mass whether you are on Earth or on the moon. However, weight of an object may change from 1 kg = 103 g = 106 mg = 109 μg one place to another due to change in gravity. The mass 1g = 103 mg = 106 μg of a substance can be measured very accurately in the 1 mg = 103 μg laboratory by using an analytical balance.An analytical Rider carrier Rider carrier knob Beam Pan Levelling screw Beam Control Pointer and scale (b) (a) Fig. 3. (a) Analytical balance (b) Electronic balances. 2. Volume volumes are commonly used. Hence, smaller units Volume is the amount of space occupied by such as cubic decimetre, dm3 (1dm3 = 10–3 m3) and cubic centimetre, cm3 (1cm3 = 0.001 dm3 or = 10–6m3) an object. It has the units of (length)3. So in SI system, are commonly used. volume has units cubic metre, m3 and is defined as the amount of space occupied by a cube having However, it may be noted that in normal use, volume each edge of 1 metre. But in laboratories smaller of liquids are commonly expressed in more familiar units

SOME BASIC CONCEPTS OF CHEMISTRY 1/9 such as litre, L and millilitre, mL. Litre is equal in size SI unit of density = SI unit of mass to dm3 and millilitre is equal in size to cm3. SI unit of volume 1L = 1dm3 and 1mL = 1 cm3 = kg or kg m–3 and 1L = 1000 mL = 1000 cm3 = 1dm3 m3 These relationships can be understood from Fig. 4. © 25Modern mL This unit is quite large and it is commonly Publishers. All rights reserved. expressed as g cm–3 for solids and g mL–1 for liquids, where mass is expressed in gram and volume in cm3 or mL. As we know that volume of a substance changes when heated or cooled, therefore, densities depend upon temperature. Hence, while reporting a density, the temperature must also be mentioned. 1 cm 4. Temperature Volume : 1 cm3 Temperature measures the degree of hotness or or 1 mL coldness of an object. There are three common scales to measure temperature 1 cm 10 cm = 1 dm Volume = 1000 cm3 or 1000 mL 1. The SI scale or Kelvin scale measured in = 1 dm3 or 1 L Kelvin (K) Fig. 4. Comparison between the relative sizes of two 2. Celsius scale measured in degree units of volume. Celsius (°C) The volume of liquids or solutions are commonly 3. Fahrenheit scale measured in degrees measured in the laboratory by graduated cylinder, Fahrenheit (°F) burette or pipette. A volumetric flask is used to prepare a known volume of a solution. Volumetric flasks of The Fahrenheit scale has been commonly used in different volumes are available. These devices for United States and is slowly been replaced by Celsius measuring volume are shown in Fig. 5. A precise scale and Kelvin scale. The thermometers with celsius volume of a liquid can also be measured with the help scale are calibrated from 0° to 100° where these two of graduated syringe. temperatures are the freezing point and boiling point of water respectively. 0 The Fahrenheit scale is represented between 32° mL 5 to 212° which represent freezing point and boiling 100 10 point of water respectively. There are 100 degrees 90 15 between these two points on the celsius scale but 180° 80 20 degrees between them in Fahrenheit giving rise to 70 25 exact relationship as 60 30 50 35 180° F = 100°C or 1.8°F = 1°C 40 40 30 45 The temperatures on two scales are related to each 20 50 other by the following relationship : 10 1 litre Graduated Burette Volumetric flask °F = 9 (°C) + 32° cylinder Pipette 5 Fig. 5. Some volume measuring devices. So to convert Celsius to Fahrenheit, we can use the formula 3. Measuring Density Density of a substance is its amount of mass °F = 9 (°C) + 32° 5 per unit volume. So, density is the property which relates the mass of an object to its volume. It is simply and to convert Fahrenheit to Celsius, we can use the mass of an object divided by its volume. The the formula, units of density may be obtained as : °C = 5 [°F – 32°] Density = Mass 9 Volume

1/10 MODERN’S abc + OF CHEMISTRY–XI According to SI system, the unit of temperature is Volume = m × m × m = m3 Kelvin. One kelvin is the same size as one degree celsius. On Kelvin scale, the lowest possible Therefore, the SI units of volume are m3. If temperature is equal to 0 Kelvin or 0 K. It may be noted length, breadth and height of the box are 1.0 m, 2.0 m that kelvin temperature is written without a and 3.0 m respectively, then volume in terms of SI degree sign. On this scale, the freezing point of water units may be expressed as : is – 273.15 K and boiling point of water is 373.15 K. The coldest possible temperature 0 K becomes Volume = (1.0 m) × (2.0 m) × (3.0 m) = 6.0 m3 –273.15°C and it is also sometimes called absolute zero. The other units may also be built up in this way from the definition of the physical quantity. For The Kelvin scale is related to Celsius scale as : example, K = °C + 273.15 (i) Area. The units of area can be easily derived as : or K = °C + 273 (for simplicity) Area = Length × Breadth so that = m × m = m2 Temperature in Kelvin = Temperature in °C + 273.15 (Units of length are m) Temperature in °C = Temperature in K – 273.15 (ii) Density. The density of a substance is mass For example, the human body temperature and per unit volume, i.e. room temperature on these temperature scales are : Human body temperature = 98.6°F = 37°C = 310.15 K Room temperature = 77°F = 25°C = 298.15K The three types of thermometers using different scales are shown in Fig. 6. © Density = Mass of sample = kg Modern Publishers. All rights reserved. Volume of sample m3 = kg m–3 (iii) Velocity. It is the change in distance per unit time, i.e. Celsius Fahrenheit Kelvin Velocity = Distance = m Time s Fig. 6. Thermometers using different temperature scales. = m s–1 It is interesting to note that temperature below (iv) Acceleration. It is change in velocity per 0°C (i.e. –ve values) are possible in Celsius scale but in Kelvin scale, negative values of temperature are unit time, i.e. not possible. Acceleration = Velocity DERIVED UNITS Time The units of different physical quantities can be = (m s−1) derived from the seven base units. These are called (s) derived units because these are derived from the base units. For deriving these units, we can multiply = m s–2 or divide the symbols for units as if they are algebraic (v) Force. It is mass multiplied by acceleration. quantities. For example, to determine the volume of a box, we multiply length, breadth and height as : Force = Mass × Acceleration = (kg) × (m s–2) Volume = Length × Breadth × Height = kg m s–2 If units of length are m, (vi) Pressure. Pressure is force per unit area, i.e. Pressure = Force Area But force has the derived units as kg m s–2. ∴ Units of pressure = kg m s−2 m2 = kg m–1 s–2 Some common derived units are given in Table 4.

SOME BASIC CONCEPTS OF CHEMISTRY 1/11 Table 4. Some common SI derived units. Physical Relation with Unit Symbol and quantity other basic quantities definition Area© Length square — m2 VolumeModern Publishers. All rights reserved.Length cube — m3 Density Mass per unit volume — kg/m3 or kg m–3 Speed Distance travelled per unit time — m/s or m s–1 Acceleration Speed change per unit time — m/s × 1/s or m s–2 Force Mass × acceleration Newton* N = kg m s–2 Pressure Force per unit area Pascal Pa = kg m–1 s–2 Energy, work Force × distance travelled Joule** J = kg m s–2 × m or kg m2 s–2 Power Energy/Time Watt W = kg m2 s–3 or Js–1 Electric charge Current × time Coulomb C = As or potential difference Electric resistance Potential difference/current Ohm Ω = VA–1 Electrical conductance Reciprocal of resistance Ohm–1 or S = AV–1 Siemen Frequency Cycles/sec Hertz Hz = s–1 * Newton is defined as the force which gives an acceleration of 1 m s–2 to a mass of 1 kg; so that f = ma = 1 (1 kg) (1 m s–2) = 1 kg m s–2 = 1 N ** Joule is the work done when a displacement of 1 metre takes place by a force of 1 newton so that w = f × d = (1N) × (1 m) = (1 kg m s–2) × (1m) = 1 kg m2 s–2 = 1 J KEY NOTES • Exponents also operate on prefixes. e.g., 1 cm2 = (10–2 m)2 = 10–4 m2 SOME IMPORTANT NOTES ON THE USE OF (∵ 1 cm = 10–2 m) SI UNITS It is not equal to 10–2 m2 In order to avoid confusion in the SI units, it is Similarly, useful to remember the following points : 1 mm3 = (10–3 m)3 = 10–9 m3 • Unit combinations should be indicated by (∵ 1 mm = 10–3 m) means of either a dot or leaving space in It is not equal to 10–3 m3 between, e.g. metre kelvin may be written It may be noted that some of the units are named as : after the name of the scientist who did pioneering work in the related field. For example, the unit of metre Kelvin = m.K or m K energy is joule abbreviated as ‘J’ and the unit of electric charge is coulomb abbreviated as C. The It should not be written as mK which stands symbols of the units which bear the names of the for milli kelvin. scientists are generally written in capital letters. For example, the unit ‘joule’ is adopted after the name • Words and symbols should not be used in James P. Joule and is denoted by ‘J’ (capital J). mixed forms. Thus, it is not proper to use J per mole. It should be written either as joule MEASUREMENT AND SIGNIFICANT FIGURES per mole or J mol–1. If mathematical Chemistry is largely an experimental science and operations are shown, only symbols should be used e.g., J mol–1 and not joule mol–1. deals with things that can be measured. A number of common devices are used in laboratories to make simple • Symbols of the units do not have a plural measurements. For example, a meter rod measures ending like ‘s’. For example, we should write length, balance measures mass; burette, pipette, 10 cm and not 10 cms. graduated cylinder and volumetric flask measure • A unit written with a prefix and a power is a power for the complete unit e.g., cm3 means (centimetre)3 and not centi (metre)3.

1/12 MODERN’S abc + OF CHEMISTRY–XI volume; thermometer measures temperature, etc. In other words, precision gives the extent of Every scientific measurement is limited by the agreement of the individual values among themselves reliability of the measuring instrument and the skill i.e., between the repeated measurements of the same of the observer. In each measurement, uncertainty quantity. This means that smaller the difference should be reported carefully. If we repeat a particular between the individual values of repeated measurement, we usually do not obtain the same result measurements, the greater is the precision. For because each measurement is subject to experimental example, the analysis of an element gave the following error. Therefore, different measured values vary results by two methods. slightly from one another. To express the results of different measurements two terms ; accuracy and Method A : %Cl = 20.60 ± 0.05 precision are commonly used. Method B : %Cl = 20.46 ± 0.15 Thus, method A gives more precise data because Accuracy and Precision the results differ only by ± 0.05%. The term accuracy denotes the closeness of an The terms accuracy and precision can be illustrated by the following example. Consider the results obtained experimental value or the mean value of a set of by four students for the % of iron in a sample. These are measurements to the true value. Thus, given in Table 5 and are shown graphically in Fig. 7. The accuracy is a measure of the difference between true value is 20.50 mg of iron. the experimental value or the mean value of a set of measurements and the true value. Table 5. Results obtained by four students for the percentage of iron in a sample. If the mean value of different measurements is© close to the true value, the measurement is said to beModern Publishers. All rights reserved.StudentStudentStudent Student accurate. Smaller the difference between the mean A B CD value and the true value, the larger is the accuracy. Suppose a sample of a compound contains 20.30% of 20.50 20.75 20.20 20.20 chloride. The chemical analysis of the sample by one method (say method A) gives mean value of 20.60% 20.48 20.65 20.25 20.40 chloride while another method (say method B) gives mean value of 20.46% chloride. Then, 20.52 20.40 20.15 20.00 Difference between the true value and mean value 20.50 20.28 20.20 20.60 in method A = 20.60 – 20.30 = 0.30% Mean = 20.50 20.52 20.20 20.30 Difference between the true value and mean value in method B = 20.46 – 20.30 = 0.16% Deviation = ± 0.02 ± 0.24 ± 0.05 ± 0.30 Therefore, the method B gives more accurate data The results obtained by student A have good than method A. Thus, accuracy also expresses the precision because the individual measurements are correctness of measurement. close together (± 0.02) and also have good accuracy because the mean result is same as the true value. However, it may be noted that quite often, the true The results of student B have low or poor precision value of a quantity is not known. Therefore, it becomes because the measurements are scattered (± 0.24) but very difficult to calculate the accuracy. In such cases, have good accuracy because the mean value (20.52) is precision of the measurements is calculated. close to the true value. Student C has obtained results of good precision because the measurements are close Precision refers to how closely two or more together (± 0.05) but have poor accuracy because the measurements of the same quantity agree with one mean value (20.20) is far from the true value. The another. Thus, results of student D have poor precision because the precision is expressed as the difference between measurements are scattered and have poor accuracy a measured value and the arithmetic mean value for a series of measurements. Results True value = 20.50 mg Good precision Student A Mean = 20.50 mg Good accuracy Student B Mean = 20.52 mg Low precision Good accuracy Student C Mean = 20.20 Good precision Mean = 20.30 Poor accuracy Student D Poor precision Fig. 7. Illustration of difference between accuracy and precision. Poor accuracy

SOME BASIC CONCEPTS OF CHEMISTRY 1/13 because the mean value (20.30) is far from the true 0.0001 4872 value. ↑ From the above observation, it can be concluded so that exponent n = – 4 and we can write that accurate results are generally precise but precise results need not be accurate. In other words, good 0.00014872 = 1.4872 × 10–4 precision does not assure good accuracy. Thus, in general, number is written in scientific Exponential Number or Scientific Notation. notation as : © In Chemistry, we come across very large and veryModern Publishers. All rights reserved.number with a single number of places decimal small numbers. It is very tedious to write down such non-zero digit point was moved numbers in the ordinary way. For example, it is not ⎯⎯⎯⎯→ convenient to write Avogadro constant as 602, 213, + moved left 700,000,000,000,000,000. These numbers are usually expressed in a simple way known as exponential form N × 10n – moved right or scientific notation. For example, the number 246.38 may be expressed as : Arithmetics Using Scientific Notation, Addition and Subtraction 246.38 = 2.4638 × 10 × 10 or = 2.4638 × 102 where 102 means 10 × 10 and 2 is the power or To add or subtract numbers in scientific notation, exponent to which 10 is raised. the exponent (n) must be the same in both numbers. If the exponent is not same in both numbers, it has to In general, in scientific notation, a number may be made same before adding or subtracting. For be expressed as : example, suppose we want to add 4.236 × 104 and 3.582 × 103. We must first express both numbers so N × 10n that they have the same exponent, n. So we can transform 3.582 × 103 to 0.3582 × 104 and then add where n is an exponent having positive or negative values and N is a single non-zero digit and lies between 4.236 × 104 + 0.3582 × 104 = (4.236 + 0.3582) × 104 1 to 10. N is called digit term and n is called exponent. = 4.5942 × 104 For example, 1487.2 may be written as: 1487.2 = 1.4872 × 10 × 10 × 10 = 1.4872 × 103 Similarly, to subtract these numbers, In a simple way, the decimal point is moved to the 4.236 × 104 – 3.582 × 103 left until there is only one non zero digit before the decimal point. If the decimal point is moved x places, 4.236 × 104 – 0.3582 × 104 = (4.236 – 0.3582) × 104 the exponent n = x. For example, in transforming 1487.2 to scientific notation, the decimal point is moved = 3.8778 × 104 to left three places as : KEY NOTE 1487.2 ↑ Before adding or subtracting scientific numbers, it is important to make both numbers to the same Here exponent, n = 3 and we can write power of 10. 1487.2 = 1.4872 × 103 Multiplication and Division Similarly, Avogadro number, may be expressed as : To multiply two numbers in scientific notation, we 602, 213,700,000,000,000,000,000 = 6.022137 × 1023 make use of the relation : Similarly, 0.00014872 may be written as: (10)x × (10)y = 10(x + y) In other words, we add the exponents. For example, (2.4 × 103) × (5.6 × 105) = (2.4 × 5.6) × 103+5 13.44 × 108 = 1.344 × 109 0.00014872 = 1.4872 = 1.4872 × 10–4 (3.025 × 103) × (6.217 × 10–6) = (3.025 × 6.217) × 103+(–6) 10 × 10 × 10 × 10 = 18.81 × 10–3 = 1.881 × 10–2 Here the exponent n = – 4 means that the number (9.8 × 10–2) × (2.5 × 10–6) = (9.8 × 2.5) × 10–2+(–6) = 24.50 × 10–8 = 2.450 × 10–7 1.4872 has been divided by 10 four times. In a simple To divide two numbers in scientific notation, we way, the decimal point is moved to the right until there make use of the relation : is one non-zero digit before the decimal point. If the 10 x 10 y decimal point is moved y places, the exponent n = – y. = 10x–y For example, in transforming 0.00014872 to scientific notation, the decimal point is moved to right four In other words, we subtract the power of 10 of the places as : number in the denominator from the power of 10 of

1/14 MODERN’S abc + OF CHEMISTRY–XI the number in the numerator. For example, 2.005 has four significant figures 3.0023 has five significant figures F I4.74 × 1012=4.74 × 1012–20 (iii) The zeros preceding to the first non-zero number HG KJ6.82 × 10206.82 (i.e., to the left of the first non-zero number) are not significant. Such zeros indicate the position = 0.695 × 10–8 = 6.95 × 10–9 of decimal point. For example, 0.324 has three significant figures 2.7 × 10−3 ⎛ 2.7 ⎞ 0.0052 has two significant figures 5.5 × 104 ⎝⎜ 5.5 ⎟⎠ 0.0003 has one significant figure © = × 10–3–4 (iv) All zeros placed at the end or to the right of a Modern Publishers. All rights reserved. number are significant provided they are on the = 0.4909 × 10–7 = 4.909 × 10–8 right side of the decimal point. In fact, these represent the accuracy or the precision of the Significant Figures measuring scale. For example, 0.0200 has three significant figures The uncertainty in the experimental or the 243.0 has four significant figures calculated values is mainly due to the skill and 243.00 has five significant figures accuracy of the observer and limitation of the 243.000 has six significant figures measuring instrument as explained below: KEY NOTE (i) Skill and accuracy of the observer. Suppose three students A, B and C measure the volume of the • The results obtained by counting are exact liquid in a cylinder. They report the following values : numbers i.e., numbers without any uncertainty. These exact numbers have infinite Result of A = 23.4 mL ; Result of B = 23.5 mL ; number of significant figures. For example, in 2 balls or 20 eggs, there are infinite number of Result of C = 23.6 mL significant figures because these are exact numbers and may be represented by writing If the correct volume of the liquid is 23.5 mL, it infinite number of zeros after placing a decimal means that the student B has measured the volume i.e. 2 = 2.000000 or 20 = 20.000000. of the liquid correctly while the students A and C have made some error. • Zeros at the end of a number or right of a number are significant provided they are on (ii) Limitations of the measuring instrument. the right side of decimal point. For example, Limitation of the measuring instrument is an important 0.300 has three significant figures. But zeros factor which leads to uncertainty in measurement. For at the end of a number without decimal point example, suppose the mass of an object has been are ambiguous. For example, 2500 may have determined to be 9.265 g. If the accuracy of the analytical two, three or four significant figures depending balance used is 0.001 g, this means that the actual upon whether the uncertainty in measure- mass of the object is 9.265 + 0.001 g, i.e., it may be ments is 100, 10 or 1 respectively. 9.264g or 9.266 g. Thus, in the reported mass, the first three digits (9, 2 and 6) are certain but the last digit(s) is uncertain and the uncertainty would be +1 in the last digit.All measured quantities are reported in same way by writing the certain digits and the last uncertain digit. This is done in terms of significant figures. The significant figures in a number are all the certain digits plus one uncertain digit. For example, in the above value (9.265 g), there are four significant figures. Exponential Numbers and Significant Figures Thus, the number of significant figures conveys the information that except the last digit, all other In exponential notations, the numerical portion digits are known with certainty. (all digits) represents the number of significant figures. Rules for Determining the Number of For example, a number, 0.000054 is expressed as 5.4 × 10–5 in terms of scientific notation. The number Significant Figures The following rules are followed to count the of significant figures in this number is 2. Similarly, number of significant figures in a particular number : the number of significant figures in the Avogadro’s number, 6.023 × 1023 is four. (i) All non-zero digits are significant. For example, 3.132 has four significant figures Now, if we want to write five thousand with three 1.76 has three significant figures significant figures, it can be written as 5.4 has two significant figures 5.00 × 103 (3 significant figures) 6.2316 has five significant figures. It cannot be written as 5000 because it has four The decimal place does not determine the and not three significant figures. number of significant figures. It must be remembered that the decimal point (ii) Zeros between two non zero digits are significant. does not count towards the number of significant figures. For example, For example, consider the number 54023. It has 3.01 has three significant figures five significant figures. We may write this number in

SOME BASIC CONCEPTS OF CHEMISTRY 1/15 different ways as: Solution: 54.023 cm (a) 6.626 × 10–34 J s = 4 significant figures 5.4023 decimetre (b) 6.023 × 1023 = 4 significant figures (c) 3.0 × 108 m s–1 = 2 significant figures or 0.54023 metre (d) 1.602 × 10–19 C = 4 significant figures. All these numbers have the same number of significant figures (i.e., 5) regardless of the position of the decimal point. The different numbers simply represent different ways (or units) of expressing the measurement. The above rules for calculating significant figures are illustrated below : ©  Example 4. Modern Publishers. All rights reserved. Calculate the number of significant figures in the following (i) 0.0025 (ii) 208 (iii) 5005 Significant (iv) 126,000 (v) 500.0 (vi) 2.0034 0.000 2040 4 significant figures Not significant Solution: 123.56 : 5 significant figures Number of significant figures : 5.407 : 4 significant figures (zero is (i) 0.0025 = 2 significant) (ii) 208 = 3 900.0 : 4 significant figures (all zeros are significant) (iii) 5005 = 4 0.0230 : 3 significant figures (only last zero (iv) 126000 = 3 (last three zeros are not significant) is significant) (v) 500.0 = 4  Example 1. State the number of significant figures in each of (vi) 2.0034 = 5 the following numbers : Calculations Involving Significant Figures (i) 207.35 (ii) 0.00368 (iii) 653 (iv) 3.653 × 104 (v) 0.378. The results of various measurements are to be added, subtracted, multiplied or divided, in most of Solution: the experiments to get desired results. But the (i) 207.35 has five significant figures. different numbers do not have the same precision in (ii) 0.00368 has three significant figures. The three measurement. In such cases, the final result cannot be more precise than the least precise number involved zeros in the beginning are not significant. in the calculations. The number of the significant (iii) 653. It has three significant figures. figures in such mathematical treatments are obtained (iv) 3.653 × 104. It has four significant figures. The by the following rules : number has been expressed in scientific notation. Rule 1. When addition or subtraction is to be carried out, the final result should be reported (v) 0.378 has three significant figures. upto the same number of decimal places as are  Example 2. present in the term having the least number of decimal places. Express the following in the scientific notation : (i) 0.0048 (ii) 234,000 (a) Addition of numbers. In the addition of (iii) 8008 (iv) 500.0 numbers 7.23, 2.1 and 0.312, the number having the least decimal places is 2.1. Therefore, the final result (v) 6.0012 after adding should also have digits only up to one decimal place. Solution: (i) 0.0048 = 4.8 × 10–3 7.23 (ii) 234,000 = 2.340 × 105 (iii) 8008 = 8.008 × 103 2.1 ←⎯ has only one decimal place (iv) 500.0 = 5.000 × 102 0.312 (v) 6.0012 = 6.0012 or 6.0012 × 10–0  Example 3. 9.642 ←⎯ answer should be reported upto one decimal place. Calculate the number of significant figures in the following values : Correct answer = 9.6 (a) Planck's constant = 6.626 × 10 −34 J s (b) Avogadro number = 6.023 × 10 23 Some more examples of addition of numbers are : (c) Velocity of light = 3.0 × 10 8 m s−1 (d) Electronic charge = 1.602 × 10 −19 C 89.232 Since 1.1 has one + 1.1 decimal place, therefore, answer should be reported 90.332 Correct answer = 90.3 to one decimal place.

1/16 MODERN’S abc + OF CHEMISTRY–XI 7.21 Since 7.21 has two noted that in calculations, the exact numbers 12.141 decimal places, therefore, will not have any effect on the number of answer should be reported significant figures. The number of significant 0.0028 upto two decimal places. figures in a calculation result depends only on the 19.3538 numbers of significant figures in quantities having Correct answer = 19.35 uncertainties. For example, we want to calculate the total mass of 8 pencils when each pencil has a mass of 3.0 g. The calculation is 3.0 × 8 = 24 Since 3.0 has two significant figures, the result should be 24 having two significant figures. The number 8 is exact and does not determine the number of significant figures. ©(b) Subtraction of numbers. The subtraction of Modern Publishers. All rights reserved.numbers is carried out in the same way as the addition. For example : 25.4630 − 24.21 ←⎯⎯ has two decimal places 1.2530 Correct answer = 1.25 (upto two decimal places) It may be noted that if a calculation involves a number of steps, the result should contain the Some other examples are : 5.2748 same number of significant figures as that of 18.4215 − 5.2721 the least precise number involved, other than the exact numbers. − 6.0 12.4215 0.0027 Since 6.0 has only Correct Retention of Significant Figures–Rounding off one digit after decimal answer = 0.0027 Figures Correct answer = 12.4 In solving problems, the final result often contains Rule 2. In multiplication or division, the final more digits than the number of significant figures. To result should be reported up to the same number retain the required number of significant figures, of significant figures as are present in the term rounding off procedure is applied. with the least number of significant figures. (c) Multiplication of numbers. If we multiply Rules for Rounding Off 5.1028 (having five significant figures) with 1.30 (having three significant figures), the value comes out (i) If the digit coming after the desired number of to be 6.63364. But according to the rule, it must be significant figures happens to be more than 5, reported with three significant figures. Therefore, the the preceding number is increased by 1. For correct answer is 6.63. example, if we have to remove 6 in 1.386, we have to round it to 1.39. 5.1028 × 1.30 (3 significant figures) (ii) If the digit coming after the desired number of significant figures is less than 5, it is neglected = 6.63364 as such i.e., the preceding number remains unchanged. For example, if we have to remove Correct answer = 6.63 (3 significant figures) 4 in 6.234, we have to round it to 6.23. (d) Division of numbers. If we divide 5.2765 (iii) If the digit coming after the desired number of (having five significant figures) by 1.25 (having three significant figures happens to be 5, then the significant figures), the result comes out to be 4.2212. preceding number is increased by 1 only in case it But according to the rule, it must be reported with three happens to be odd. In case of even number, the significant figures. Thus, the correct answer is 4.22. preceding number remains unchanged For example, if 6.75 is to be rounded off by removing 5.2765 ÷ 1.25 5, we have to increase 7 to 8 giving 6.8 as the result. However, if 6.45 is to be rounded off, it is (3 significant figures) rounded off to 6.4. = 4.2212 Thus, to express the results to three significant figures Correct answer = 4.22 (3 significant figures) KEY NOTE 8.312 is rounded off to 8.31 Calculations involving exact numbers 8.316 is rounded off to 8.32 Exact numbers arise when we count items or sometimes when we define a unit. For 8.375 is rounded off to 8.38 [rule (iii)] example, when we say 10 pencils, we exactly mean 10, not 9.9 or 10.1. Also when we say that there are 8.365 is rounded off to 8.36 [rule (iii)] 12 inches in a foot, we mean exactly 12. It may be It must be remembered that if the problem involves more than one step, the ‘rounding off ’ must be done only in the final answer. The intermediate steps of the calculations remain unchanged.

SOME BASIC CONCEPTS OF CHEMISTRY 1/17 Example 5. 0.0125 0.7864 Express the following numbers up to four 0.0215 it has 4 significant figures. significant figures : 0.8204 (i) 5.607892 (ii) 32.392800 Example 9. (iii) 0.007837 (iv) 1.78986 × 103 Express the results of the following calculations to the appropriate number of significant figures : ©(v) 60000 Modern Publishers. All rights reserved. 3.24 × 0.08666 (ii) (1.36 × 10−4 ) (0.5) Solution: (i) 5.006 2.6 (i) 5.607892 = 5.608 (ii) 32.392800 = 32.39 (iii) 0.582 + 324.65 (iv) 2.64 × 103 + 3.27 × 102 (iii) 0.007837 = 0.007837 (v) 943 × 0.00345 + 101. (iv) 1.78986 × 103 = 1.790 × 103 (v) 6.000 × 104 Solution: Example 6. (i) 3.24 × 0.08666 = 0.05608 = 0.0561 In the ca5lc.0u0la6tion, the number of significant figures in Express the following up to three significant places : the term 3.24 is 3, therefore, the result should have 3 significant figures. Therefore, the correct answer is 0.0561. (a) the height of a man, 5 feet 9 inches in The number after 0 is 8 and therefore it is rounded off to 1. centimetres (1 inch = 2.54 cm) (b) one millionth of one. (c) four thousand (d) decimal equivalent of 2/3 (ii) (1.36 × 10−4 ) (0.5) = 0.2615 × 10–4 Solution: 2.6 = 0.3 × 10–4. (a) 5 ft 9 inches = 69 inch = 69 × 2.54 = 175.26 cm. The answer should have one significant figure because With three significant figures answer is 175 cm. 0.5 has one significant figure. (b) One millionth of one is 1 . It can be written with (iii) 0.582 3 significant figures as110.06 0 × 10–6. + 324.65 (Second place of decimal) (c) four thousand = 4.00 × 103 325.232 = 325.23 (d) 2 = 0.667. Since 324.65 contains digits up to second place of decimal, 3 therefore, the answer must be reported up to second place. Example 7. (iv) 2.64 × 103 + 3.27 × 102 or 2.64 × 103 + 0.327 × 103 = 2.967 × 103 = 2.97 × 103 Calculate to proper significant figures : Since 2.64 has two digits after decimal place, the answer (a) 12.6 × 11.2 (b) 108/7.2 should be rounded off to two decimal places. Solution: (v) The first two numbers are to be multiplied initially (a) 12.6 × 11.2 = 141.12 as follows : Correct answer = 141 (upto 3 significant figures) 108 943 × 0.00345 = 3.25335 = 3.25. (b) = 15 According to the rules on multiplication of significant 7.2 figures, the answer has to be reported up to three significant figures. It should be 3.25. Upon further addition, the final Correct answer = 15 (upto 2 significant figures as answer may be calculated as : in 7.2). 3.25 Example 8. How many significant figures should be present + 101. (zero place of decimal) in the answer of the following calculations ? ————— 104.25 = 104 (i) 0.02856 × 298.15 × 0.112 ∴ The answer should have zero place of decimal. 0.5785 Example 10. (ii) 5 × 5.364 The mass of a piece of paper is 0.02 g and the (iii) 0.0125 + 0.7864 + 0.0215 mass of a solid substance and the piece of paper Solution: is 20.036 g. If the volume of the solid is 2.16 cm3, calculate the density of the substance up to proper (i) 0.02856 × 298.15 × 0.112 = 3 significant figures number of significant digits. Solution: 0.5785 The number of significant figures, in 0.112 (least Mass of piece of paper = 0.02 g number of significant figures) is 3, therefore, the result should have 3 significant figures. Mass of solid substance and (ii) 5 × 5.364 = 4 significant figures 5.364 has four significant figures (leaving the exact piece of paper = 20.036 g number 5) ∴ Mass of solid substance = 20.036 (iii) 0.0125 + 0.7864 + 0.0215 = 4 significant figures − 0.02 The result is to be reported upto four decimal ————— places as : 20.016 or = 20.02 (upto second decimal place).

1/18 MODERN’S abc + OF CHEMISTRY–XI Volume of solid = 2.16 cm3 7. Perform the following calculations upto proper number of significant figures : Density = Mass = 20.02 = 9.268 (i) (1.20 × 10–6) + (6.00 × 10–5) = ? Volume 2.16 (ii) (2.164 × 105)1/2 = ? (iii) (9.13 × 10–2) (7.006 × 10–3) = ? Since 2.16 has three significant figures, the answer (iv) 4.00 × 10–2 + 3.26 × 10–3 + 1 × 10–6 = ? should also contain three significant figures. 8. Calculate the number of significant figures up to ∴ Density up to proper number of significant figures which the following results will be expressed : = 9.27 g/cm3.© Modern Publishers. All rights reserved. (i) 2.36 × 0.07251 (ii) (28.2 − 21.2) (1.79 × 106) Example 11. 2.130 1.62 Perform the following calculations and express the result to proper number of significant figures : 9. Round up the following upto three significant (i) 144.3 m2 + (2.54 m × 8.4 m) figures : (ii) (4.05 × 102 mL) – (0.0225 × 102 mL) (i) 34.216 (ii) 10.4107 (iii) (3.50 × 102 cm) (4.00 × 106 cm) (iii) 0.04597 (iv) 2808 Solution: (i) (144.3 m2) + (2.54 m × 8.4 m) 10. If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by the light in 2.00 ns. 2.54 m × 8.4 m = 21.336 m2 or 21 m2 (upto 2 significant figures) 1. (a) Three (b) Two (c) Four (d) One (e) Three 2. (i) 6.02 × 1023 (ii) 6.00 × 103 (iii) 32.4 144.3 m2 21.0 m2 (iv) 5.60 (v) 0.667 (vi) 1.63 × 104 3. 4.0 × 102 g —––——— 4. (i) 9.00 × 105 (ii) 2.14 × 10–6 165.3 m2 or 165 m2 (iii) 4.07 × 105 (iv) 1.00 × 10–6 (ii) (4.05 × 102 mL) − (0.0225 × 102 mL) 5. (a) 15 (b) 2.6 × 104 (c) 1 × 10–7 (d) 0.0700 (e) 1.23 4.05 × 102 mL 6. (i) 1.4 m2 (ii) 17 cm3 (iii) 4.1 m 7. (i) 6.12 × 10–5 (ii) 4.652 × 102 − 0.0225 × 102 mL ————————— (iii) 6.40 × 10–4 (iv) 4.33 × 10–2 8. (i) Three (ii) Two 4.0275 × 102 mL or 4.03 × 102 mL 9. (i) 34.2 (ii) 10.4 (iii) 0.0460 (iv) 2.81 × 103 (upto second decimal 10. 0.60m place as in 4.05) Hints & Solutions on page 79 (iii) (3.50 × 102 cm) × (4.00 × 106 cm) = 14.0 × 108 cm2(upto 3 significant figures) DIMENSIONAL ANALYSIS Conversion of Units 1. Calculate the number of significant figures in the following : It is frequently necessary to convert one set of units to another in calculations. This can be done by a method (a) 1.00 × 106 (b) 0.0050 (c) 1.234 called factor label method or unit factor method or also called dimensional analysis. This is based on (d) 0.0006 (e) 0.368 the relationship between different units that expresses the same quantity. For example, suppose we want to 2. Express the following numbers to three significant convert 25.6 metres to centimetres. We know that figures : 1 m = 100 cm (i) 6.023 × 1023 (ii) 6000 (iii) 32.362400 or 1 = 100 cm (iv) 5.6034 (v) decimal equivalent of 2/3 1m 100 cm/1 m is called unit conversion factor or (vi) 1.6276 × 104 conversion factor because it converts a quantity expressed in one unit to a quantity expressed in another 3. The density of ice is 0.921 g cm–3. Calculate the mass unit. It is also called unit factor because the overall effect of a cubic block of ice which is 76 mm on each side. of multiplication by these factors is to multiply by 1. 4. Express the following numbers in exponential Now, multiply the given quantity by the notations to three significant figures : conversion factor, retaining the units of the physical quantity and that of the conversion factor in such a (i) 900035 (ii) 0.000002136 way that all units cancel out leaving behind only the required units i.e., numerator should have that part (iii) 406721 (iv) 0.000001 which is required in the desired result. 5. Perform the following calculations to proper Thus, 25.6 m = 25.6 m × 100 cm = 2560 cm. number of significant figures : 1m (a) 108/7.2 (b) (1.6 × 102)2 e j(c) 1.35 × 10−6 (0.4) (d) 3.25 × 0.08621 5.6 4.002 (e) (1.0042 – 0.0034) (1.23) 6. Calculate (i) area of a square whose side is 1.2 m (ii) volume of a sphere whose radius is 1.6 cm (iii) length of a rectangle having area 10.25 m2 and breadth 2.5 m.

SOME BASIC CONCEPTS OF CHEMISTRY 1/19 • It may be remembered that we choose the Let us convert litre atmosphere to joule (the SI conversion factor that has metres in the denominator. unit of energy). Let us convert 1 L atm to joules. This can be done by multiplying with two unit factors as : FHG JIK The advantage of this method is that if the 1 L = 10–3 m3 or 1 = 10−3 m 3 L equation is set up correctly, only then all the units will get cancelled except the required unit. If we do not get desired unit, then an error must have been made. © 1 atm = 101.325 × 103 Pa Modern Publishers. All rights reserved. Let us consider another example. Suppose we or 1 = 101.325 ×103 Pa want to convert 1.50 km3 of water into volume in litres. atm We know that Applying these unit factors : 1L = 1 dm3 FGH IJK1 L atm = 1 L atm × 10–3 m 3 So, we want to convert km3 to dm3 (litres). This 1L can be done in two steps as : × ⎛ 101.325 ×103 Pa ⎞ (i) Convert km3 to m3 ⎜⎝⎜ atm ⎟⎟⎠ (ii) Convert m3 to dm3 = 101.325 Pa m3 For converting km3 to m3, the unit factor is Now, Pa = N m2 1 km = 103 m HFG JKI∴ 101.325 Pa m3 = 101.325 so that 1 = ⎛ 103m ⎞ N m3 ∴ ⎜⎝⎜ 1 km ⎟⎟⎠ m2 Now 1.50 km3 = 1.50 km3 ×m3⎜⎛⎜⎝ 103m ⎞3 = 101.325 Nm = 1.50 × 109 1 km ⎟⎠⎟ = 101.325 J [∵ Nm = J] 1 m = 10 dm Thus, 1 L atm = 101.325 J Thus, we can easily apply several conversion or 1 = 10 dm factors in one step. For example, suppose we want to Applying unit factor, 1 m convert 2.50 miles into centimetres. The relationship between various units are : Volume of water =1.50 × 109 m3 ×dm⎜⎛⎝ 3101 dm ⎞3 1 mile = 1760 yards, 1 yard = 3 ft, 1 ft = 12 inches = 1.50 × 1012 m ⎟ ⎠ and 1 inch = 2.54 cm. The unit factors are : Both the steps can be performed in a single step as 1 = 1760 yards ; 1= 3 ft ; 1 = 12 inches ; ⎛ 103 m ⎞3 ⎞3 1 mile 1 yard 1 ft ⎜⎝⎜ 1 km ⎟⎠⎟ ⎟ Volume of water =1.50 km3 × × ⎛ 10 dm ⎠ and 1 = 2.54 cm ⎜ 1 m 1 inch ⎝ Applying these unit factors, Converts km3 Converts m3 into m3 into dm3 = 1.50 × 1012 dm3 2.50 miles ×1760 yards × 3 ft × 12 inches × 2.54 cm Since cubic decimetre is equal to litres, the volume 1 mile 1 yard 1 ft 1 inch of water is Converts mile Converts yard Converts ft Converts inch = 1.50 × 1012 litres into yard into ft into inch into cm It should also be noted that units can be handled = 402336 cm just like numerical part i.e., it can be cancelled, or = 4.02 × 105 cm multiplied, divided, squared, etc. Table 6. Relation between various units. Units of length Units of mass Units of volume 1 mile = 1760 yard 1 kg = 103 g 1 m3 = 103 L 1 yard = 3 ft 1g = 1000 mg 1 dm3 = 1 L 1 ft = 12 inch 1 lb = 453.59 g 1 cm3 = 10–3 L 1 inch = 2.54 cm 1 oz = 28.35 g 1 ft3 = 28.32 L 1 metre = 100 cm 1 metric ton = 1000 kg 1 quart = 0.9464 L 1 km = 1000 m 1 L = 1.056 quarts 1 mile = 1.609 km = 2205 lb 1 mile = 5280 ft 1 g = 15.4 grains 1 carat = 3.168 grains

1/20 MODERN’S abc + OF CHEMISTRY–XI Other units 1 dyne = 10–5 N 1 Å = 10–10 m 1 calorie = 4.184 J 1 atm = 760 mm or 760 torr 1 erg = 10–7 J = 101,325 Pa or Nm–2 1 eV = 1.6022 × 10–19 J = 1.013 × 106 dyne cm–2 1 bar = 105 Nm–2 = 105 Pa 1 mm or 1 torr = 133.322 Pa or Nm–2 © Modern Publishers. All rights reserved.Example 12. Solution: (i) The S I unit of distance is metre (m) 1 mile = 1.60 kilometre = 1.60 × 1000 m The density of vanadium is 5.96 g cm–3. Convert Unit factor = 1.60 × 1000 m = 1.6 × 103 m the density to SI units of kg m–3. 1 mile 1 mile Solution: Density = 5.96 g cm−3 ∴ 93 million miles = 93 × 106 miles × 1.6 × 103 m 1 mile We know 1 kg = 1000 g ∴ Unit factor = 1 kg = 93 × 1.6 × 109 m 1000 g = 148.8 × 109 m = 1.49 × 1011 m. 1 m = 100 cm (ii) 5 feet 2 inches = 62 inches ∴ Unit factor = 100 cm 1m 1 inch = 2.54 × 10−2 m Applying unit factor for g and cm3, we get Unit factor = 2.54 × 10−2 m 1 inch GFH JKI∴ 3 5.96 g 1 kg 100 cm 62 inches = 62 inches × 2.54 × 10−2 m 5.96 g cm–3 = cm 3 × 1000 g × 1m 1 inch = 5.96 × 103 kg m–3 = 62 × 2.54 × 10−2 m = 5960 kg/m3. = 157.48 × 10−2 m = 1.57 m. (iii) 1 mile = 1.60 km = 1.60 × 103 m Example 13. A jug contains 2 L of milk. Calculate the volume Unit factor = 1.60 × 103 m 1 mile of milk in m3. 1 hr = 60 × 60 s = 3.6 × 103 s Solution: We know 1 L = 1000 cm3 or Unit factor = 1000 cm3 Unit factor = 3.6 × 103s 1L 1 hr 1 m = 100 cm ∴ Speed = 100 miles hr Unit factor = 1 m 100 cm = 100 miles × 1.60 × 103 m × 1 hr hr 1 mile 3.6 ×103s Volume of milk = 2 L × ⎛ 1000 cm3 ⎞ ⎛ 1m ⎞3 ⎜ ⎟ × ⎜ 100 cm ⎟ = 44 m s–1. ⎝ 1L ⎠ ⎝ ⎠ = 2 × 10–3 m3 (iv) 1 Å = 10−10 m. Example 14. Unit factor = 10−10 m 1 A° Express each of the following in S I units : 0.74 A° × 10−10 m (i) 93 million miles (this is the distance between ∴ 0.74 Å = the earth and the sun). 1 A° (ii) 5 feet 2 inches (this is the average height of = 0.74 × 10−10 m or = 7.4 × 10−11 m an Indian female). (v) 0°C = 273.15 K (iii) 100 miles per hour (this is the typical speed of Rajdhani Express). 46°C = 273.15 K + 46 K = 319.15 K (iv) 0.74 Å (this is the bond length of hydrogen (vi) 1 pound = 454 × 10−3 kg molecule). Unit factor = 454 × 10−3 kg (v) 46°C (this is the peak summer temperature 1 pound in Delhi). ∴ 150 pound = 150 pound × 454 × 10−3 kg (vi) 150 pounds (this is the average weight of an 1 pound Indian male). = 150 × 454 × 10−3 kg = 68.1 kg.

SOME BASIC CONCEPTS OF CHEMISTRY 1/21 Example 15. or Unit factor, 1 = 10–3 kg 1g The mass of precious stones is expressed in terms of ‘carat’. What is the mass of a ring in grams HFG IKJ∴ which contains 0.600 carat diamond and 8.500 g 0.91 × 10–27 g = 0.91 × 10–27 g × 10–3 kg gold given that 1 carat = 3.168 grains and 1g 1 g = 15.4 grains ? © = 9.1 × 10–31 kg Modern Publishers. All rights reserved. Solution: (ii) Mass of human DNA molecule = 1 fg Weight of ring = 0.600 carat diamond + 8.500 g gold 1 fg = 10–15 g Now 1 carat = 3.168 grains or Unit factor, 1 = 10–15g 1 fg Unit factor = 3.168 grains 1 carat 1 g = 10–3 kg 1 g = 15.4 grains 10–3 kg 1g Unit factor = 1g or Unit factor, 1 = 15.4 grains Weight of diamond = 0.600 carat FGH IKJ∴ ⎛ 10–15g ⎞ 10–3 kg = 0.600 carat × 3.168 grains × 1 gram 1 fg = 1 fg × ⎝⎜⎜ 1 fg ⎟⎠⎟ × 1g 1 carat 15.4 grains = 0.123 g = 1 × 10–18 kg (b) (i) 1.4 Gm (diameter of Sun) Total mass of ring = 8.500 + 0.123 = 8.623 g We know 1 Gm = 109 m Example 16. A tennis ball was observed to travel at a speed of or Unit factor, 1 = 109m 96 miles per hour. Calculate the speed of the ball 1 Gm in metres per second. GFH KIJ∴ Solution: 1.4 Gm = 1.4 Gm × 109m = 1.4 × 109 m 1 Gm Speed of tennis ball = 96 miles per hour Now l mile = 1.60 km = 1.60 × 103 m (ii) 40 Em (thickness of Milky Way Galaxy) Unit factor = 1.60 × 103 m We know, 1 Em = 1018 m 1 mile or Unit factor, 1 = 1018m 1 hr = 60 × 60 s = 3.6 × 103 s 1Em Unit factor = 3.6 × 103 s 1 hr HFG JKI∴ 40 Em = 40 Em × 1018m = 40 × 1018 m 1 Em ∴ Speed = 96 mile or = 4.0 × 1019 m hr = 96 mile × 1.60 × 103m × 1 hr hr 1 mile 3.6 × 103s = 42.7 m s–1 11. The wavelength of a yellow line in spectrum of sodium atom is 5896 Å. Express it in nm. Example 17. (a) Convert the following in kilogram 12. How many cubic centimetres are there in 1 m3 ? 13. Convert (i) 0.91 × 10–27 g (mass of electron) (ii) 1 fg (mass of human DNA molecule) (i) 4.86 kg L–1 to grams per millilitre (b) Convert into metre (ii) 1.86 km to cm (i) 1.4 Gm (diameter of Sun) (iii) 6.92 × 10–7 m to micrometres and Angstroms (ii) 40 Em (thickness of Milky Way Galaxy) (iv) 9.2 × 10–3 cm3 to litres Solution: (a) (i) 0.91 × 10–27 g (mass of electron) 14. What is the capacity of a tank 0.8 m long 10 cm wide and 50 mm deep ? 1 g = 10–3 kg 15. How many cubic centimetres are there in 100 L ?

© 1/22 MODERN’S abc + OF CHEMISTRY–XI Modern Publishers. All rights reserved. 16. Convert the following in kilogram : 5. Law of combining volumes (or Gay Lussac’s law (i) 500 Mg (mass of jumbo jet loaded) of combining volumes). (ii) 3.34 × 10–24 g (mass of hydrogen molecule) The first four laws of chemical combination deal 17. Express the following in the designated units : with mass relationships while the fifth law deals with (i) 1.54 mm s–1 to pm μs–1 the volumes of the reacting gases. (ii) 25 gL–1 to mg dL–1 1. Law of Conservation of Mass (iii) 25 L to m3 This law deals with the relation between the mass (iv) 2.66 g cm–3 to μg μm–3 of reactants and the products during the chemical (v) 4.2 L h–2 to mL s–2 changes. It was postulated by a French chemist 18. Convert into metre : Antoine Lavoisier in 1789. (i) 7 nm (diameter of small virus) (ii) 41 pm (distance of nearest star) Law of conservation of mass states that 19. How many seconds are there in 2 days ? during any physical or chemical change, the 20. Convert the following into base units : total mass of the products is equal to the (i) 28.7 pm total mass of the reactants. (ii) 15.15 μs In other words, matter can neither be created nor (iii) 25365 mg destroyed during any physical or chemical change. 11. 589.6 nm Therefore, this law is also known as law of 12. 106 cm3 indestructibility of matter. 13. (i) 4.86 g/mL (ii) 1.86 × 105 cm Experimental verification of the law of (iii) 0.692 μm, 6920 Å (iv) 9.2 × 10–6 L conservation of mass. 14. 4L 15. 105 cm3 The law can be verified with the help of Landolt’s experiment. Landolt took the solutions of sodium 16. (i) 5.0 × 105 kg (ii) 3.34 × 10–27 kg chloride (NaCl) and silver nitrate (AgNO3) separately in two limbs of a ‘H’ shaped tube (known as Landolt’s 17. (i) 1.54 × 103 pm μs–1 (ii) 2.5 × 103 mg dL–1 tube). The tube was sealed and weighed. After (iii) 2.5 × 10–2 m3 (iv) 2.66 × 10–6 μg μm–3 weighing, the two solutions were mixed thoroughly (v) 3.2 × 10–4 mL s–2 by shaking the tube. As a result, the reaction occurred between silver nitrate and sodium chloride and a white 18. (i) 7 × 10–9 m (ii) 41 × 10–12 m ppt. of silver chloride is formed as: 19. 172800 s AgNO3(aq) + NaCl (aq) ⎯⎯→ AgCl(s) + NaNO3(aq) 20. (i) 2.87 × 10–11m (ii) 1.515 × 10–5s White ppt. (iii) 2.5365 × 10–2 kg After the reaction, the tube was again weighed. It Hints & Solutions on page 79 was observed that the weight remained practically unchanged. This verified the law of conservation of LAWS OF CHEMICAL COMBINATION mass. In the seventeenth century, the scientists had been NaCl Solution AgNO3 Solution trying to find out methods for converting one substance into another. During their quantitative studies of Fig. 8. Landolt’s experiment. chemical changes, they made certain generalisations. These generalisations are known as laws of chemical combination. These are : 1. Law of conservation of mass 2. Law of constant composition or definite propor- tions 3. Law of multiple proportions 4. Law of reciprocal proportions

SOME BASIC CONCEPTS OF CHEMISTRY 1/23 LEARNING PLUS© elements in the same proportion. This law is Modern Publishers. All rights reserved. sometimes referred to as law of definite composition Present Position of the Law in the Light or definite proportion. of Recent Developments Similarly, carbon dioxide can be obtained by a It may be noted that the law of conservation of number of methods, such as mass is not strictly valid for nuclear reactions. (i) By burning coal or candle According to Einstein theory of relativity, mass C + O2 ⎯⎯→ CO2 and energy are interconvertible. The mass (m) and energy (E) are related as, E = mc2 where c is the (ii) By heating limestone (CaCO3) velocity of light (3 × 108 m/sec). We know that CaCO3 ⎯⎯Hea⎯t → CaO + CO2 chemical reactions are generally accompanied by liberation of energy. Since energy and mass are (iii) By the action of dilute hydrochloric acid on related to each other, this means that the energy marble pieces. must be coming from the reactants. As a result, CaCO3 + 2HCl ⎯⎯→ CaCl2 + CO2 + H2O there should be decrease in mass of the reactants. However, the mass changing into energy for (iv) By heating sodium bicarbonate. ordinary chemical reactions (according to relation 2NaHCO3 ⎯⎯Hea⎯t → Na2CO3 + CO2 + H2O m = E/c2) is extremely small because the value of c is very large and, therefore, there is no measurable It has been observed that each sample of carbon change in mass during chemical processes. dioxide contains carbon and oxygen elements in the However, in case of nuclear reactions and ratio of 3 : 8 by weight. radioactive disintegrations, the change in mass is quite significant because tremendous amount of Similarly, pure water can be obtained from many energy is released during these reactions. sources. Irrespective of the source, water always Therefore, the law of conservation of mass does not contains hydrogen and oxygen elements combined hold good. In these reactions, some mass gets together in the ratio of 1 : 8 by weight. converted into energy. In such cases, mass and energy is totally conserved though mass and energy Experimental verification of the law. The law are not separately conserved. Thus, law of can be easily verified in the laboratory. For example, conservation of mass is modified and the modified copper oxide (CuO) can be prepared by the following law is known as law of conservation of mass- methods : energy which states that mass and energy are interconvertible, but the total mass and energy (i) by heating copper powder in oxygen : of the system remains constant. 2Cu + O2 ⎯⎯⎯→ 2CuO 2. Law of Constant Composition or Definite Copper powder Proportions (ii) by heating copper carbonate : The law of constant composition deals with the CuCO3 ⎯⎯⎯→ CuO + CO2 composition of various elements present in a compound. It was stated by a French chemist, Joseph Proust, in Copper carbonate 1799. Law of constant composition states that (iii) by heating copper nitrate : a pure chemical compound always contains 2Cu(NO3)2 ⎯⎯⎯→ 2CuO + 4NO2 + O2 same elements combined together in the same definite proportion by weight. Copper nitrate Proust worked with two samples of cupric carbonate; one of which was naturally occurring cupric The weighed quantities of three samples of copper carbonate and other was prepared in the laboratory. oxide as obtained above were reduced to copper He found that the composition of the elements present separately in a current of hydrogen : in two samples of cupric carbonate was same as shown below : CuO + H2 ⎯⎯⎯→ Cu + H2O The weight of copper left behind in each case is Cu Percentage noted. From the weight of copper left behind, the weight CO of oxygen in the samples of copper oxides was calculated. Naturally occurring 51.35 It was observed that in all the samples, the ratio cupric carbonate 9.74 38.91 of copper and oxygen by weight has been found to be the same i.e., 4 : 1. This illustrates the law of constant Cupric carbonate 51.35 9.74 38.91 composition. prepared in laboratory Limitations of Law of Constant Composition The above results show that irrespective of the source, a given compound always contains same (i) The law of constant composition does not hold good when a compound is obtained by using different isotopes of the combining elements. For example, wtOhhiesern1a2Cti:Oo322oif.sBCfou:rtmOwehidsen1fr4oCm:O32C2i.-s1Tf2ohrtumhsee, rddaiftfrifooemrbeenCttw-1ies4eointsooCpteoaspnoedf, the same element give different mass ratio between combining atoms. (ii) The elements may combine in the same ratio but the compounds formed may be different.

1/24 MODERN’S abc + OF CHEMISTRY–XI For example, in the compo unds uCl2aHr5OfoHrmaunlda In water, 2 parts by mass of hydrogen combine C2C42HH:366OO:C,1H6kno3row(1b2no:ta3hs: have same molec with 16 parts by mass of oxygen as : isomers) the ratio of C : H : O is Hydrogen + Oxygen ⎯⎯→ Water 8. Thus, the inverse of the law is not correct. 2 g 16 g 18g In hydrogen peroxide, 2 parts by mass of hydrogen combines with 32 parts by mass of oxygen as : 3. Law of Multiple Proportions Publishers. All rights reserved. This law was proposed by Dalton in 1803. Law of multiple proportions states that Hydrogen + Oxygen ⎯⎯→ Hydrogen peroxide 2 g 32 g 34g when two elements combine to form more than Therefore, the masses of oxgen (16 g, 32 g) which combine with fixed mass of hydrogen (2 parts) bear a one compound, then the masses of one of the simple ratio i.e. 16 : 32 or 1 : 2. elements which combine with a fixed mass of (ii) Nitrogen and oxygen combine to form five the other element are in a simple whole ononxxiittiyrrdgooeggeseennnnawtpmrhieoenixlctyihodnxecii(dtoNremo2u(ObNsi3n2)o,Oexni5id)wt.erioTt(gNhhee2ntOhdt)eie,ftfnrefoiirtxxerieindcdteo(wxwNieed2iOieggh4(h)NttasOnoo)dff, number ratio. The law may be illustrated by the following examples : (i) Hydrogen combines with oxygen to form two compounds namely water and hydrogen peroxide. nitrogen in all these oxides are calculated. Oxide Number of Number of Fixed weight Number of parts by weight parts by parts by of nitrogen of oxygen combining with N2O weight of weight of (14 parts) 14 parts by weight of NO nitrogen oxygen nitrogen N2O3 14 N2O4 28 16 14 8 N2O5 14 14 16 14 16 14 28 48 24 28 64 32 28 80 40 The ratio between the different weights of oxygen when two different elements combine separately in different compounds which combine with the same with a fixed mass of a third element, the ratio weight of nitrogen (14 parts) is : in which they do so will be the same or some simple multiple of the ratio in which they 8 : 16 : 24 : 32 : 40 combine with each other. or 1 : 2 : 3 : 4 : 5 This is a simple whole number and hence, supports In other words, the mass ratio of two elements A the law. and B which combine with the fixed mass of C Experimental Verification of Law of Multiple separately, is either the same or some simple whole Proportions number multiple of the mass ratio in which A and B cuprCicopopxeidrefo(rCmusOtw). o1o.0x0idgesmcuopf reoaucshoxoixdiede(Couf2cOo)papnedr Modern combine together. This law may be illustrated by the h(CyudOroagnedn.CuB2oOthretshpeectoixviedlye)sisrehaeacttewd iintha current of following examples : hydrogen Consider three elements sulphur, oxygen and hydrogen. Both sulphur and oxygen separately producing metallic copper. From the weight of copper c(tHoom2foObr)imnreessutpolepcfhotiurvmreldyhi.oyTxdhirdeoeyge(aSnlOsso2u)cloapmshibsdhienoew(Hwn2i:Sth) and water obtained, the respective weights of oxygen in the two each other compounds are obtained. Then, the different weights of oxygen which combine with the same weight of copper in the two compounds are calculated. These 2 H © weights are found to bear a simple whole number ratio. 16 Thus, the law has been verified. 2 H2S H2O 4. Law of Reciprocal Proportions or Law of Equivalent Proportions 32 This law was put forward by Richter in 1792. It S SO2 O states that 32 32

SOME BASIC CONCEPTS OF CHEMISTRY 1/25 According to the law, the ratio of weights of S and O which combine with the same weight of H will either be same or a simple multiple of the ratio in which S and O combine with each other. This may be verified as : In Twhyhadeterroragmteinoolesocufulpltehh(eiHdwe2O(eH)i,g2h2St)ps,a2ortfpssaburytlspwhbeuyirgwhaetniogdfhhot yxodyfgrhoeygndenrwochgoeimcnhbc,ioncmoembwibnitienhew1i6stehppa3ar2rtasptbaeylrytwswebiiygthhwtetohifgehoxtfiyxogfeedsnu.wlpehiguhr.t In ∴ (= 2 parts) of hydrogen is, © Modern Publishers. All rights reserved.32 : 16 or 2 : 1 ...(i) Now, let us calculate the ratio of sulphur and oxygen in SO2. ...(ii) ISnulspuhlpuhr uarnddiooxxyidgeen(ScOom2)bine in the ratio of 32 : 32 or 1 :1 The two ratios (i) and (ii) are related to each other as which are simple multiple of each other. 2 : 1 or 2 : 1 11 However, sulphur and oxygen also react to form sulphur trioxide (SO3) which is also in accordance with the law. ISnulspuhlpuhr uarndtriooxxyigdeen(ScOom3)bine in the weight ratio of 32 : 48 or 2:3 ...(iii) The two ratios (i) and (iii) are also related to each other as which are simple multiple of each other. 2 : 2 or 3 : 1 13 5. Gay Lussac’s Law of Combining Volumes Gay Lussac performed a number of experiments on reactions involving gases and found that some regularity exists between the volumes of the gaseous reactants and products. In 1808, he put forward a generalisation known as Gay Lussac’s law of combining volumes. It states that when gases react together or produced in a chemical reaction, they do so in a simple ratio by volume to one another and to the volumes of the products (if these are also gases) provided all gases are at the same temperature and pressure. This law may be illustrated by the following examples : 1 volume of hydrogen 1 volume of chlorine 2 volumes of hydrogen chloride It has been experimentally found that 1 volume of hydrogen reacts with 1 volume of chlorine to give 2 volumes of hydrogen chloride as : Thus, the volume ratio of hydrogen : chlorine : hydrogen chloride is 1 : 1 : 2. This is a simple whole number ratio and is also in agreement with their molar ratios when they are involved in the reaction. H2(g) + Cl2(g) ⎯⎯→ 2HCl(g) 1 volume 1 volume 2 volumes For example, 100 mL of hydrogen combine with 100 mL of chlorine to give 200 mL of hydrogen chloride. Similarly, we observe that 2 volumes of hydrogen combine with 1 volume of oxygen to give 2 volumes of water vapour as 2 volumes of hydrogen 1 volume of oxygen 2 volumes of water vapour

1/26 MODERN’S abc + OF CHEMISTRY–XI Thus, the volumes of hydrogen and oxygen which In the second compound, combine to form water is in the ratio : 2 : 1 : 2. Mass of carbon combining with 72.7 g of 2H2 (g) + O2 (g) ⎯⎯→ 2H2O (g) oxygen = 27.3 g 2 volumes 1 volume 2 volumes. Mass of carbon combining with 1.0 g of Similarly, it has been found that one volume of © oxygen = 27.3 = 0.375 g Modern Publishers. All rights reserved.nitrogen combines with three volumes of hydrogen to 72.7 give two volumes of ammonia gas. The ratio of mass of carbon combining with fixed mass of oxygen (i.e., 1 g) is N2(g) + 3H2(g) ⎯⎯→ 2NH3 1 volume 3 volumes 2 volumes 0.75 : 0.375 or 2 : 1. This is a simple ratio and therefore, illustrates the Example 18. law of multiple proportions. Isfo6lu.3tigoonf,NtahHe CreOs3idaureeaidsdfeodutnod15t.o0 gwoefiCgHh 31C8O.0OHg. Example 20. What is the mass of CO2 released in the reaction ? 2.0 g of a metal burnt in oxygen gave 3.2 g of its Solution: oxide. 1.42 g of the same metal heated in steam gave 2.27 g of its oxide. Which law is shown by NaHCO3 + CH3COOH ⎯ ⎯→ this data? Solution: In the first compound, 6.3 g 15.0 g 3.2 g of metal oxide contained 2.0 g of metal 100 g of metal oxide contained metal CH3COONa + H2O + CO2 = 2.0 × 100 = 62.5 g Residue xg 3.2 18.0 g ∴ % Metal in first compound = 62.5% Sum of the mass of reactants = MMasasssofofNCaHH3CCOO3O+H In the second compound, 2.27 g of metal oxide contained metal = 1.42 g = 6.3 + 15.0 = 21.3 g 100 g of metal oxide contained metal Sum of the mass of products = Mass of residue + = 1.42 × 100 = 62.55 g Mass of CO2 2.27 (18.0 + x) g ∴% Metal in second compound = 62.55% (where x is mass of CO2 released) Thus, the percentage of metal in metal oxide According to law of conservation of mass obtained from two experiments is nearly same. Hence, the above data illustrate the law of constant Mass of reactants = Mass of products composition. 21.3 = 18.0 + x Example 21. or x = 21.3 – 18.0 = 3.3 g Phosphorus and chlorine form two compounds. The first compound contains 22.54% by mass of ∴ Mass of CO2 released = 3.3 g phosphorus and 77.46% by mass of chlorine. In the second compound the percentages are 14.88 Example 19. for phosphorus and 85.12 for chlorine. Show that these data are consistent with the law of multiple Carbon and oxygen are known to form two proportions. compounds. The carbon content in one of these compounds is 42.9% while in the other, it is 27.3%. Solution: In the first compound, Show that the data are in the agreement with the Percentage of phosphorus = 22.54 law of multiple proportions. Percentage of chlorine = 77.46 Solution: In the first compound, Mass of carbon = 42.9 g Thus, 22.54 g of phosphorus combines with 77.46 g of Mass of oxygen = 100 – 42.9 = 57.1 g chlorine. In the second compound, In the second compound, Mass of carbon = 27.3 g Mass of oxygen = 100 – 27.3 = 72.7 g Percentage of phosphorus = 14.88 In the first compound. Percentage of chlorine = 85.12 Mass of carbon combining with 57.1 g of oxygen Thus, 14.88 g of phosphorus combine with 85.12 g of = 42.9 g chlorine. Mass of carbon combining with 1.0 g of oxygen = 42.9 × 1.0 = 0.75 g 57.1

SOME BASIC CONCEPTS OF CHEMISTRY 1/27 Let us fix the mass of phosphorus as 1 g and find the (ii) Brown oxide different masses of chlorine which combine with 1 g of phosphorus in two compounds. Mass of oxygen which combines with 1.035 g of lead = 0.160 g In the first compound, Mass of oxygen which combines with 1 g of lead Mass of chlorine which combines with 22.54 g of phosphorus = 77.46 g = 0.160 = 0.15 g 1.035 © The mass of chlorine which combines with 1 g of Modern Publishers. All rights reserved. (iii) Red oxide phosphorus = 77.46 = 3.44 g 22.54 Mass of oxygen which combines with 1.61 g of lead In the second compound, = 0.16 g Mass of chlorine which combines with 14.88 g of phosphorus = 85.12 g Mass of oxygen which combines with 1 g of lead Mass of chlorine which combines with 1 g of phosphorus = 0.16 = 0.10 g = 85.12 = 5.72 1.61 14.88 The ratio of different masses of oxygen which combine The ratio of the masses of chlorine which combines with with same mass of lead (1 g) in these oxides is : the fixed mass of phosphorus (1 g) in the two compounds is 0.075 : 0.15 : 0.10 3.44 : 5.72 1 : 1.66 or 3 : 5 (approximately) 3 :6:4 This is a simple whole number ratio. Therefore, the data is in agreement with the law of multiple proportions. This is a simple ratio. Example 22. Three oxides of lead on analysis were found to Hence, the data illustrate the law of multiple contain lead as under : proportions. (i) 3.45 g of yellow oxide contains 3.21 g of lead. (ii) 1.195 g of brown oxide contains 1.035 g of Example 23. lead. Two oxides of a metal contain 27.6% and 30% of (iii) 1.77 g of red oxide contains 1.61 g of lead. oxygen respectively. If the formula of the first Show that these data illustrate law of multiple compound is M3O4, find the formula of the second proportions. compound. Solution: The amounts of lead and oxygen in three oxides are : Solution: (i) Yellow oxide : Mass of lead = 3.21 g First oxide Second oxide Mass of oxygen = 3.45 – 3.21 = 0.24 g Oxygen = 27.6% Oxygen = 30% (ii) Brown oxide : Mass of lead = 1.035 g Metal = 72.4% Metal = 70% Mass of oxygen = 1.195 – 1.035 = 0.160 g Formula of first oxide = M3O4 Suppose the atomic weight of metal = x (iii) Red oxide : Mass of lead = l.61 g Mass of oxygen = 1.77 – 1.61 Percentage of metal in the compound M3O4 = 0.16 g = 3x × 100 Let us fix the mass of lead as 1 g and calculate the 3x + 64 different weights of oxygen which combine with 1 g of lead in these oxides. ∴ 3x × 100 = 72.4 3x + 64 (i) Yellow oxide Mass of oxygen which combines with 3.21 g of or 300 x = 217.2 x + 4633.6 lead = 0.24 g or 82.8 x = 4633.6 or x = 56 Mass of oxygen which combines with 1 g of lead Now in the second oxide, metal and oxygen are 70% and = 0.24 = 0.075 g 30%. Therefore, their atomic ratio will be 3.21 M :O 70 : 30 56 16 1.25 : 1.875 or 1 : 1.5 or 2 : 3 Therefore, formula of the compound = M2O3.

1/28 MODERN’S abc + OF CHEMISTRY–XI Example 24. Example 25. Hydrogen sulphide (H2S) contains 94.11% Water contains 88.90% oxygen and 11.10% of asunldphsuulrp, hwuartedrio(Hxi2dOe)(ScoOn2ta)icnosn1ta1i.n115%0%hyodxryoggeenn. hydrogen, ammonia contains 82.35% of nitrogen Show that the results are in agreement with law of and 17.65% of hydrogen and dinitrogen trioxide reciprocal proportions. contains 63.15% of oxygen and 36.85% of nitrogen. Show that these data illustrate law of reciprocal proportions. © Solution: Modern Publishers. All rights reserved. In water (H2O) Solution: Mass of hydrogen = 11.11 g In water (H2O) Mass of hydrogen Mass of oxygen = 100 – 11.11 g = 11.10 g = 88.89 g Mass of oxygen = 88.90 g In sulphur dioxide (SO2 ) = 50 g In ammonia (NH3) = 17.65 g Mass of oxygen Mass of hydrogen Mass of sulphur = 100 – 50 g = 50 g Mass of nitrogen = 82.35 g H NH3 H2O N O N2O3 Let us fix the mass of oxygen as 1 g. Let us fix the mass of hydrogen as 1 g. Now in H2O, 88.89 g of oxygen combine with hydrogen = 11.11 Now in H2O, 11.10 g of hydrogen combine with oxygen = 88.90 g 1 g of oxygen combines with hydrogen = 11.11 × 1 88.89 1 g of hydrogen combines with oxygen = 88.90 11.10 = 0.125 g In SO2, = 8.01 g 50 g of oxygen combine with sulphur = 50 g In NH3, 17.65 g of hydrogen combine with nitrogen = 82.35 g 1 g of oxygen combines with sulphur = 50 ×1=1g 50 1 g of hydrogen combines with nitrogen = 82.35 Thus, the ratio of the masses of hydrogen and sulphur 17.65 which combine with the fixed mass of oxygen (1 g) is = 4.67 g 0.125 : 1 or 1 : 8 ...(i) Thus, the ratio of nitrogen and oxygen which combine In hydrogen sulphide (H2S) with the fixed mass of hydrogen (1 g) is Mass of sulphur = 94.11 g 4.67 : 8.01 or 1 : 1.172 ...(i) Mass of hydrogen = 100 – 94.11 = 5.89 g In dinitrogen trioxide Therefore, the ratio by mass of hydrogen and sulphur in Mass of nitrogen = 36.85 g hydrogen sulphide is Mass of oxygen = 63.15 g 5.89 : 94.11 or 1 : 16 ...(ii) The ratios (i) and (ii) are Therefore, the ratio by mass of nitrogen and oxygen in dinitrogen trioxide is 1 : 1 or 2 : 1 8 16 36.85 : 63.15 or 1 : 1.172 ...(ii) which is simple whole number. Thus, the two ratios are the same. Hence, it illustrates Hence, the law of reciprocal proportions is verified. the law of reciprocal proportions.

SOME BASIC CONCEPTS OF CHEMISTRY 1/29 21. When 4.2 g of NaHCO3 is added to a solution of 28. Copper sulphide contains 66.5% Cu, copper oxide CH3COOH (acetic acid) weighing 10.0 g, it is contains 79.9% Cu and sulphur trioxide contains observed that 2.2 g of CO2 is released to the 40% S. Show that the data illustrates the law of atmosphere. The residue left is found to weigh 12.0 reciprocal proportions. g. Show that these observations are in agreement with the law of conservation of mass. 29. Two oxides of a certain metal were separately heated in a current of hydrogen until constant weights were Hint : Mass of reactants = 14.2 g : Mass of products obtained. The water produced in each case was = 14.2 g carefully collected and weighed. It was observed that 1 g of each oxide gave 0.1254 g and 0.2263 g of water 22. Hydrogen peroxide and water contain 5.93% and respectively. Show that the data illustrate the law 11.2% of hydrogen respectively. Show that the data of multiple proportion. illustrates law of multiple proportions. Hints & Solutions on page 79 Hint : The ratio of weights of oxygen combining with 1 g of hydrogen is 2 : 1. DALTON ATOMIC THEORY 23. Elements A and B combine to form two different To provide theoretical justification to the laws of compounds as : chemical combination, John Dalton postulated a © simple theory of matter. The basic postulates of the Modern Publishers. All rights reserved.0.3 g of A + 0.4 g of B ⎯⎯→ 0.7 g of compound Xtheory are given below : 18 g of A + 48 g of B ⎯⎯→ 66 g of compound Y. (i) Matter is made up of extremely small indivisible and indestructible ultimate particles called Show that the data illustrates the law of multiple atoms. proportions. (ii) Atoms of the same element are identical in all 24. Carbon combines with hydrogen to form compounds respects i.e. in shape, size, mass and chemical having the following compositions : properties. Compound Carbon (%) Hydrogen (%) (iii) Atoms of different elements are different in all respects and have different masses and chemical A 75 25 properties. B 85.7 14.3 (iv) Atom is the smallest unit that takes part in chemical combinations. C 92.3 7.7 (v) Atoms of two or more elements combine in a fixed Show that the data illustrate the law of multiple ratio to form compound atoms (now a days called proportions. molecules). Hint : Ratio between weights of H which combine (vi) Atoms can neither be created nor destroyed during with fixed weight of C is 4 : 2 : l. any physical or chemical change. 25. 1.375 g CuO was reduced by hydrogen and 1.098 g (vii) Chemical reactions involve only combination, Cu was obtained. In another experiment, 1.178 g of separation or rearrangement of atoms. Cu was dissolved in nitric acid and the resulting copper nitrate was converted into CuO by ignition. Limitations of Dalton’s Atomic Theory The weight of CuO formed was 1.476 g. Show that these results prove the law of constant composition. Dalton’ s atomic theory gave a powerful initiative to scientists in the beginning of the 19th century. The 26. Illustrate the law of definite proportions from the main achievement of the theory was that one could following data : derive the laws of chemical combination from it. It held the field for a century. But as a result of (i) 0.32 g of sulphur on burning in air produced discoveries in the beginning of 20th century by Sir J. 224 ml of SO2 at N.T.P. J. Thomson, Rutherford, Neils Bohr and others, the atomic theory was reviewed and modified. The main (ii) A metal sulphite reacts with a mineral acid to drawbacks of Dalton’s atomic theory are : give SO2 gas which contains 50% of sulphur and 50% of oxygen. (i) It could explain the laws of chemical combination by mass but failed to explain the law of gaseous 27. Copper sulphate crystals contain 25.45% Cu and volumes. 36.07% H2O. If the law of constant proportions is true then calculate the weight of Cu required to (ii) It could not explain why atoms of different obtain 40 g of crystalline copper sulphate. elements have different masses, sizes, valencies, etc. Ans. 10.18 g

1/30 MODERN’S abc + OF CHEMISTRY–XI (iii) It could not explain how and why atoms of AVOGADRO’S HYPOTHESIS different elements combine with each other to form compound atoms or molecules. While trying to correlate Gay Lussac’s law of gaseous volumes and Dalton’s atomic theory, Berzelius, (iv) It failed to explain the nature of forces that bind the Swedish scientist, put forward a generalisation, atoms together in a molecule. in 1811 known as Berzelius hypothesis. According to Berzelius hypothesis: © (v) It gave no satisfactory explanation between the Modern Publishers. All rights reserved. ultimate particle of an element and that of a equal volumes of all gases contain equal compound. number of atoms under similar conditions of temperature and pressure. Modern Atomic Theory However, when the Berzelius hypothesis was The main modifications made in the Dalton’s applied to some chemical reactions it was observed atomic theory as a result of new discoveries about atom that even a fraction of atoms was involved in some are : cases. The idea of fraction of atoms is against the concept of Dalton’s atomic theory : 1. Atom is no longer considered to be indivisible. It has been found that an atom has a For example, consider the formation of complex structure and is made up of subatomic hydrochloric acid gas (HCl). It has been observed particles such as electrons, protons and neutrons. experimentally that one volume of hydrogen combines with one volume of chlorine to produce two volumes 2. Atoms of same elements may not be similar of hydrochloric acid gas. This may be expressed as: in all respects. Atoms of same element may possess different relative masses. For example, there are two Hydrogen + Chlorine ⎯→ Hydrochloric acid gas different types of atoms of chlorine with atomic masses 35 a.m.u. and 37 a.m.u. respectively. Such atoms of 1 vol 1 vol 2 vol the same element which possess different atomic masses are called isotopes. According to Berzelius hypothesis, equal volumes will contain equal number of atoms. Let one volume 3. Atoms of different elements may have contains n atoms : similar one or more properties. There are certain atoms of different elements which possess same relative Hydrogen + Chlorine ⎯→ Hydrochloric acid gas masses. For example, atomic mass of calcium and argon is same (40 a.m.u.) but their chemical properties are or n atoms n atoms 2n compound atoms entirely different. Such atoms of different elements which possess same mass are called isobars. or 1 atom 1 atom 2 compound atoms 4. Atom is the smallest unit which takes part ½ atom ½ atom 1 compound atom in chemical reactions. Though an atom is composed of subatomic particles yet it is the smallest particle This means that one compound atom of which takes part in chemical reactions. hydrochloric acid gas is formed by the combination of ½ atom of hydrogen and ½ atom of chlorine. This means 5. The ratio in which the different atoms that fraction of atoms take part in chemical combination combine may be fixed and integral but may and therefore, atoms may undergo division during not always be simple. For example, in sugar chemical reactions. But this idea of fraction of atoms is molecule (C12H22O11), the ratio of C, H and O atoms is against the concept of Dalton’s atomic theory according 12 : 22 : 11, which is not simple. to which atom is the smallest unit of an element which takes part in chemical reactions. 6. Atom of one element may be changed into atoms of other element. For example, atoms of To solve this problem of conflict between the nitrogen can be changed into oxygen atoms by Dalton’s atomic theory and Berzelius hypothesis, iibnnetteecrraoacncttviioeonrntwewidtihthiαn-trnaoeyuspt.lrSuointmosin.laiTurhlmyi,su(rp2ar39no94icPueumss) (2it3s9h52rUcoa)ulclgaehnd Avogadro suggested that matter consists of two kinds transmutation. Thus, atom is no longer of ultimate particles. These are atoms and molecules. indestructible. According to him, 7. The mass of atom can be changed into the smallest particle of an element which may energy. According to Einstein’s equation, E = mc2 or may not have independent existence and takes (E = energy, m = mass and c = velocity of light), mass part in a chemical reactions is called an atom. and energy are interconvertible. Thus, atom is no longer indestructible. However, during chemical The smallest particle of a substance (element reactions, atom remains unchanged. or compound) capable of independent existence is called a molecule. Thus, the smallest particles of gases which can have independent existence are molecules and not the atoms and therefore, volume of gases should be related to number of molecules and not number of atoms. He could explain the results by considering the molecules to be polyatomic. He put forward a hypothesis known as Avogadro’s law. It may be stated as :

SOME BASIC CONCEPTS OF CHEMISTRY 1/31 under similar conditions of temperature and Now ½ molecule of hydrogen can exist because pressure, equal volumes of all gases contain one molecule of hydrogen contains 2 atoms of hydrogen equal number of molecules. and ½ molecule of hydrogen means one atom of hydrogen. Similarly, ½ molecule of chlorine contains Mathematically, we can say an atom of chlorine because chlorine is also a diatomic V∝N molecule. Thus, one molecule of hydrogen chloride is formed from one atom of hydrogen and one atom of chlorine. This is in agreement with Dalton’s atomic theory. ©where N is the number of molecules. Modern Publishers. All rights reserved. For example, if we enclose equal volumes of three gases hydrogen (oHf2)t,hoexysgaemne(Oca2)paancidtychulnordienre s(Ciml2i)lainr different flasks conditions of temperature and pressure, we find that all the flasks have the same number of molecules. Applications of Avogadro’s Law However, these molecules may differ in size and mass. 1. Calculation of Atomicity of Elementary Fig. 10. Illustration of Avogadro hypothesis. Gases The number of atoms present in its one molecule Gay-Lussac and Avogadro’s laws can be illustrated of the substance is called its atomicity. Avogadro’s as follows : law helps in determining the atomicity of elementary gases such as hydrogen, oxygen, chlorine, etc. For example, Atomicity of oxygen. The atomicity of oxygen can be calculated by considering the reaction between hydrogen and oxygen to form water vapour. It has been experimentally found that 2 volumes of hydrogen react with 1 volume of oxygen to form 2 volumes of water vapour. Hydrogen + Oxygen ⎯→ Water vapour 2H2(g) + O2(g) ⎯→ 2H2O (g) 2 volumes 1 volume 2 volumes Gay-Lussac’s law 2 vol 1 vol 2 vol Applying Avogardro’s hypothesis Avogadro’s law 2n molecules n molecules 2n molecules 2 n molecules n molecules 2n molecules 2 molecules 1 molecule 2 molecules N2(g) + 3H2(g) ⎯→ 2NH3 (g) 2 molecules 1 molecule 2 molecules or Gay-Lussac’s law 1 vol 3 vol 2 vol Avogadro’s law n molecules 3n molecules 2n molecules 1 molecule 2 1 molecule 3 molecules 2 molecules 1 molecule 1 molecule The Avogadro hypothesis can be successfully Thus, one molecule of water contains one molecule applied to different chemical reactions. of hydrogen (2 atoms) and ½ molecule of oxygen. The Formation of Hydrochloric acid gas molecular mass of water has been found to be 18 a.m.u. Consider the formation of hydrochloric acid gas. Since one molecule of hydrogen contains two atoms of It is observed that 1 volume of hydrogen combines with 1 volume of chlorine to give 2 volumes of hydrogen hydrogen, therefore, weight of oxygen in one molecule chloride (all volumes under same conditions). of water is 18 – 2 = 16 a.m.u. This corresponds to the Hydrogen + Chlorine ⎯→ Hydrogen chloride weight of one atom of oxygen. Thus, one atom is 1 vol 1 vol 2 vol present in half molecule of oxygen. Therefore, Applying Avogadro’s hypothesis assuming that 1 1 molecule of oxygen = 1 atom 2 volume contains n molecules, it follows that or 1 molecule of oxygen = 2 atoms Hydrogen + Chlorine ⎯→ Hydrogen chloride n molecules n molecules 2n molecules Thus, atomicity of oxygen is 2. Dividing throughout by 2n, we get 1 molecule 1 molecule 1 molecule 2. Determination of relationship between 2 2 vapour density and molar mass of a gas This means that 1 molecule of hydrogen chloride The vapour density of a gas is the ratio between the mass of a certain volume of the gas to the mass of contain ½ molecule of hydrogen and ½ molecule of the same volume of hydrogen gas under similar chlorine.

1/32© MODERN’S abc + OF CHEMISTRY–XI Modern Publishers. All rights reserved. conditions of temperature and pressure. Thus, ∴ Vapour density = Molar mass Vapour density (V.D.) of a gas 2 = Mass of certain volume of gas or Molar mass = 2 × Vapour density Mass of same volume of hydrogen Vapour density is also called relative density of the gas. (similar conditions) 3. Determination of relationship between According to Avogadro’s hypothesis, equal volumes mass and volume of gas. of all gases under similar conditions of temperature As discussed above, and pressure contain equal number of molecules. Let the given volume of the gas and hydrogen contain Molar mass = 2 × Vapour density n molecules at S.T.P. conditions. Vapour density = 2 × Mass of certain volume of gas at S.T.P. Mass of same volume of hydrogen at S.T.P. = Mass of n molecules of gas Mass of n molecules of hydrogen = 2 × Mass of 1L of gas at S.T.P. Mass of 1L of hydrogen at S.T.P. = Mass of 1 molecule of gas Mass of 1 molecule of hydrogen But mass of 1L of hydrogen gas is 0.089 g Since 1 molecule of hydrogen contains 2 atoms of ∴ Molar mass = 2 × Mass of 1L of gas at S.T.P. hydrogen, 0.089 Vapour density = Mass of 1 molecule of gas = 2 × Mass of 1L of gas at S.T.P. Mass of 2 atoms of hydrogen 0.089 But the ratio of the mass of one molecule of gas to = 22.4 × Mass of 1L of gas at S.T.P. the mass of an atom of hydrogen is called molar mass = Mass of 22.4L of gas at S.T.P. (discussed later). Thus, 22.4 L of any gas at S.T.P. weigh equal to molar mass of gas expressed in grams. This is also called Gram Molecular Volume (G.M.V.). 1 Q.1. How many significant figures are there in each of the following numbers : Ans. Q.2. (i) 1.00 × 106 (ii) 0.00010 (iii) π Ans. (i) Three (ii) Two (iii) an infinite number. Q.3. Convert 22.4 L in cubic metres. Ans. Q.4. 22.4 L × ⎛ 103cm3 ⎞ = 22.4 × 103 cm3 Ans. ⎝⎜ L ⎠⎟ = 22.4 × 103 cm3 × ⎛ 10−6 m3 ⎞ ⎝⎜ cm3 ⎟⎠ = 22.4 × 10–3 m3 What physical quantities are represented by the following units and what are their common names ? (i) kg m2 s–2 (ii) kg m s–2 (iii) dm3 (i) Energy, Joule (ii) Force, Newton (iii) Volume, Litre. The longest visible rays, at the end of the visible spectrum are 7.8 × 10–7 m in length. Express this length in (i) micrometers and (ii) nanometers. (i) 1 m = 106 μm Unit factor = 106 μm 1m ∴ 7.8 × 10–7 m = 7.8 × 10−7 m × 106 μm = 0.78 μm 1m

SOME BASIC CONCEPTS OF CHEMISTRY 1/33 (ii) 1m = 109 nm Q.5. Unit factor = 109 nm 1m Ans. Q.6. ∴ 7.8 × 10–7 m = 7.8 × 10–7 m × 109 nm = 780 nm. Ans. 1m Q.7. Which of the following mixture are homogeneous ? © Modern Publishers. All rights reserved.(a) tap water (b) air(c) soil (d) smoke (e) cloud Tap water, air, cloud. Is the molar volume of CO2 same or different from CO ? Same. At what temperature have the Celsius and Fahrenheit reading the same numerical value ? Ans. °C = 5 (F − 32) 9 x= 5 (x − 32) 9 4x = – 160 ∴ x = – 40 ∴ – 40°C, –40°F. Q.8. What is the difference between 0.006 and 6.00 × 10–3 g ? Ans. Q.9. 0.006 g contains one significant digit while 6.00 × 10–3 g contains 3 significant digits. Ans. Classify the following substances into elements, compounds and mixtures : Q.10. (i) Milk (ii) 22 carat gold (iii) Iodized table salt (iv) Diamond (v) Smoke (vi) Steel (vii) Brass Ans. (viii) Dry ice (ix) Mercury (x) Air (xi) Aerated drinks (xii) Glucose (xiii) Petrol (xiv) Glass (xv) Wood Q.11. Ans. Elements : (iv), (ix) Q.12. Compounds : (viii), (xii) Ans. Mixtures : (i), (ii), (iii), (v), (vi), (vii), (x), (xi), (xiii), (xiv), (xv) Given that density of water is 1 g mL–1. What is the density in SI units? ⎛ 1 kg ⎞ 1 ⎛ 100 cm ⎞ 3 ⎝⎜ 1000 g ⎠⎟ cm3 ⎝⎜ 1m ⎠⎟ 1 g mL–1 = 1 g cm–3 = 1g × × × = 1000 kg m–3. Is the law of constant composition true for all types of compounds? Explain why or why not? Law of constant composition is not true for all types of compounds. It is true only for the compounds obtained from one isotope. For example, carbon exists in two common isotopes, 12C and 14C. When CO2 is formed from 12C, the ratio of masses is 12 : 32 or 3 : 8, but when it is formed from 14C, the ratio will be 14 : 32 or 7 : 16, which is not same as in first case. Which postulate of the Dalton’s atomic theory was modified after the discovery of isotopes ? According to postulates of the Dalton’s atomic theory, atoms of same element are identical in all respects i.e., in shape, size, mass and chemical properties. However, after the discovery of isotopes, it was modified as : atoms of same element may not be similar in all respects. For example, there are two types of atoms of chlorine with atomic masses 35 amu and 37 amu respectively (called isotopes). ATOMS AND MOLECULES The atoms of certain elements, such as hydrogen, oxygen, nitrogen, etc., are not capable of independent Atom existence whereas atoms of helium, neon, argon, etc., Atom is the ultimate particle of an element. To are capable of independent existence. understand this, consider a piece of an element, say All elements are composed of atoms. There are gold. If we break this into smaller and smaller pieces, about 118 elements known and therefore, there are it will become so small that we will not be able to see 118 different types of atoms. it without a microscope. If we keep on breaking this Molecule particle, we will ultimately reach a stage when this particle cannot be further broken. This ultimate Just as the ultimate particle of an element is atom, particle of an element is called an atom. Thus, the ultimate particle of a chemical compound is called a molecule. When two or more atoms of different the smallest particle of an element which may elements combine, the molecule of a comound is or may not have independent existence is formed. For example, called an atom.

1/34 MODERN’S abc + OF CHEMISTRY–XI atom of atom of molecule of hydrogen hydrogen hydrogen (H2) © atom of atom of molecule of ATOMIC AND MOLECULAR MASSES Modern Publishers. All rights reserved.oxygenoxygen oxygen (O2) Atomic Mass The molecules are made up of atoms of the same or different elements and are capable of independent An atom of an element is so small particle that its existence. Thus, mass cannot be determined even with the help of most sensitive balance. For example, by an indirect method, the smallest particle of a substance (element the absolute mass of hydrogen atom has been found or compound) which is capable of to be 1.6736 × 10–24 g. Such extremely small numbers independent existence is called a molecule. are very inconvenient for calculations. The difficulty was overcome by expressing atomic masses, as Thus, a molecule contains two or more atoms. The relative masses. i.e., with respect to the mass of an properties of a substance are due to the properties of atom of a standard substance. its molecules. Molecules can be classified into two types : Earlier the chemists selected hydrogen atom as the standard substance because it is the lightest of all (i) Homoatomic molecules. known atoms. The atomic mass of hydrogen was taken as one. Thus, the atomic mass of an element is the These molecules are made up of the atoms of the number of times an atom of that element is heavier same element. Most of the elementary gases consist of than an atom of hydrogen. The relative atomic masses homoatomic molecules. For example, hydrogen gas were referred to as the atomic weights. For example, rSccehoismnlopsireiliscanttreislvyeg,olafnys.ittTwcrhoooengseasenitsomtamsonslodefcooutfxwlyehogsyemdanrtaoogymgabesseneosfu(fHarcrt2hhe)l.eoNrrS2ciinlmaaensisl(daCifrOilel2y2)d.,, it has been found that an atom of oxygen is 16 times as monoatomic, diatomic, triatomic, or polyatomic heavier than an atom of hydrogen. Thus, the atomic molecules according as they contain one, two, three or mass of oxygen relative to hydrogen is 16. Later on, more than three atoms respectively. The number of oxygen was fixed as the standard because it is more atoms in a molecule of an element is called atomicity. reactive than hydrogen and forms a large number of For example, He, Ne (monoatomic), aOto2,mNic2),, HS28 compounds. (diatomic), Oet3c.( t r iatomic), P4 (tetra (polyatomic), In 1961, the International Union of Chemists selected a new unit for expressing the atomic masses. OP They accepted the isotope of carbon (12C) with mass PP number 12 as the standard for comparing the atomic He HH OO P and molecular masses of elements and compounds. Carbon-12 is the stablest isotope of carbon and can be Helium Hydrogen Ozone Phosphorus P4 represented as 12C. (triatomic) (Tetraatomic) Thus, atomic mass of an element is defined as (monoatomic) (diatomic) the average relative mass of an atom of an (ii) Heteroatomic molecules. element as compared to the mass of an atom of carbon (12C) taken as 12. These molecules are made up of atoms of In other words, atomic mass is a number which different elements. They are also classified as di, tri, expresses as to how many times an atom of the element tetra...etc., depending upon the number of atoms is heavier than 1/12th of the mass of carbon atom (12C). present. For example mHo2Ole,cuHleCsl., CO2, NH3, CH4, Therefore, PCl5 are heteroatomic Atomic mass = Mass of an atom The molecules of some common compounds are shown below : 1 mass of a carbon atom (12C) 12 This scale of relative masses of atoms is called atomic mass unit scale and is abbreviated as a.m.u. However, the new symbol used is ‘u’ (known as unified mass) in place of a.m.u. In this system, 12C is assigned a mass of exactly 12 atomic mass unit and masses

SOME BASIC CONCEPTS OF CHEMISTRY 1/35 of all other atoms are given relative to this standard. Phosphorus P 31.0 Hence, one atomic mass unit is defined as Potassium K 39.0 Silicon Si 28.1 the quantity of mass equal to 1/12th of the Silver Ag 107.9 mass of an atom of carbon (12C). Sodium Na 23.0 Sulphur S 32.1 Mass of 1 amu has been calculated to be Tin Sn 118.7 1 amu = 1.66056 × 10–24 g Zinc Zn 65.4 Mass of an atom of hydrogen = 1.6736 × 10–24 g© Thus, in terms of amu, the mass of hydrogen atomModern Publishers. All rights reserved.Concept of Average Atomic Mass The word average has been used in the above = 1.6736 × 10−24 g = 1.008 amu definition and is very significant. It has been found 1.66056 × 10−24 g that the majority of the elements occur in nature as mixtures of several isotopes. Isotopes are the different Thus, the atomic mass of hydrogen is atoms of the same element possessing different atomic 1.008 u. Similarly, the mass of oxygen – 16 (16O) atom masses but same atomic number. The atomic mass of is 15.995 u or 16 u; of nitrogen is 14 u and of sulphur such an element is the average of the atomic masses is 32 u. This means that an atom of hydrogen is 1.008 of the naturally occurring isotopes of the element times heavier than 1/12th of the mass of carbon taking into account their relative abundance. This is atom (12C); an atom of oxygen is 16 times heavier than called average atomic mass. 1/12th of the mass of carbon atom (12C) and so on. For example, carbon occurs as three isotopes with relative abundances and masses as : The atomic masses of some common elements are given in Table 7. It may be noted that only the Isotope Relative Atomic approximate values of atomic masses which are abundance (%) mass (amu) normally used in the chemical calculations, are given. For example, atomic mass of hydrogen is 1.008 g but 12C 98.892 12 it is taken as 1. 13C 1.108 13.00335 14C 2 × 10–10 14.00317 Nowadays, atomic masses of the elements have been determined accurately using an instrument Therefore, average atomic mass of carbon called mass spectrometer. Table 7. Atomic mass of a few common elements referred to 12C = 12.0 Element Symbol Atomic mass = 98.892 × 12 + 1.108 × 13.00335 + 2 × 10−10 × 14.00317 100 Aluminium Al 27.0 Antimony Sb 121.8 = 12.001 u Arsenic As 74.9 Similarly, chlorine occurs in nature in the form of Argon Ar two isotopes with atomic mass 35 and 37 in the ratio Beryllium Be 18 of 3 : 1 respectively. Therefore, Bismuth Bi 9.01 Boron B 209 Average atomic mass of chlorine = 35 × 3 + 37 × 1 = 35.5 u Bromine Br 10.8 3+1 Calcium Ca 79.9 Carbon C 40.1 Thus, we say that on an average, an atom of Chlorine Cl 12.0 chlorine is 35.5 times heavier than 1/12th of the mass Copper Cu 35.5 of a carbon atom (C12). Fluorine F 63.5 Helium He 19.0 Similarly, neon occurs as three isotopes and their Hydrogen H 4.0 relative abundance are as follows : Iodine I 1.0 Iron Fe 126.9 Isotope Fractional abundance Lead Pb 55.8 Lithium Li 207.2 20Ne 0.9051 Manganese Mn 6.94 Magnesium Mg 54.9 21Ne 0.0027 Mercury Hg 24.3 Neon Ne 200.6 22Ne 0.0922 Nickel Ni 20.2 Nitrogen N 58.7 Average atomic mass of neon Oxygen O 14.0 16.0 = 20 × 0.9051 + 21 × 0.0027 + 22 × 0.0922 = 20.179 u Since the atomic mass is the ratio of the masses, it has no units. However, we express it as a.m.u., which only signifies that it is taken on atomic mass unit scale in which 1/12th of carbon atom is fixed as 1 u.