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  • Std. XI Sci.

    Perfect Chemistry - I

    No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical including photocopying; recording or by any information

    storage and retrieval system without permission in writing from the Publisher.

    Edition: June 2014

    Prof. Santosh B. Yadav (M. Sc., SET, NET)

    Prof. Anil Thomas (M.Sc., Chemistry)

    Published by

    Target PUBLICATIONS PVT. LTD. Shiv Mandir Sabhagriha, Mhatre Nagar, Near LIC Colony, Mithagar Road, Mulund (E), Mumbai - 400 081 Off.Tel: 022 6551 6551 Email: [email protected] Website: www.targetpublications.org

    Price : ` 230/-

    Printed at: India Printing Works 42, G.D. Ambekar Marg, Wadala, Mumbai 400 031 Target Publications PVT. LTD. Allrightsreserved

  • Std. XI Sci.

    Perfect Chemistry - I

    Written according to the New Syllabus (2012-2013) published by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.

    TEID : 734

    Salient Features :

    Exhaustive coverage of syllabus in Question Answer Format. Covers answers to all Textual, Intext and NCERT Questions. Simple and Lucid language. Neat, Labelled and authentic diagrams. Quick review for instant revision and summary of the chapter. Solved & Practice Numericals duly classified. Multiple Choice Questions for effective preparation.

  • PREFACE In the case of good books, the point is not how many of them you can get through, but rather how many can get through to you.

    Std. XI Sci. : PERFECT CHEMISTRY - I is a complete and thorough guide critically analysed and extensively drafted to boost the students confidence. The book is prepared as per the Maharashtra State board syllabus and provides answers to all textual and intext questions. Sub-topic wise classified question and answer format of this book helps the student to understand each and every concept thoroughly. Neatly labelled diagrams have been provided wherever required.

    National Council Of Educational Research And Training (NCERT) questions and problems based on Maharashtra board syllabus have been provided along with solutions for a better grasp of the concept and preparing the students on a competitive level.

    Additional information about a concept is provided in the form of Note. Definitions, statements and laws are specified with italic representation. Formulae are provided in chapters which involve numericals to help the students to tackle difficult problems. Solved problems are provided to understand the application of different concepts and formulae. Quick Review has been provided which gives an overview of the chapters. Additional theory questions have been provided to help the student gain insight on the various levels of theory-based questions.

    Practice problems and multiple choice questions help the students to test their range of preparation and the amount of knowledge of each topic.

    The journey to create a complete book is strewn with triumphs, failures and near misses. If you think weve nearly missed something or want to applaud us for our triumphs, wed love to hear from you. Please write to us on : [email protected]

    A book affects eternity; one can never tell where its influence stops.

    Best of luck to all the aspirants! Yours faithfully

    Publisher

    Contents

    No. Topic Name Page No.

    1 Some Basic Concepts of Chemistry 1

    2 States of Matter 38

    3 Structure of Atom 88

    4 Periodic Table 144

    5 Redox Reactions 173

    6 Chemical Equilibrium 223

    7 Surface Chemistry 288

    8 Nature of Chemical Bond 320

    `Chapters 9 to 17 are a part of Std. XI Chemistry -II'

    Note: All the Textual questions are represented by * mark All the Intext questions are represented by # mark

  • Target Publications Pvt. Ltd. Chapter 01: Some Basic Concepts of Chemistry

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    01 Some Basic concepts of chemistry 1.0 Prominent scientists

    Scientists Contributions Joseph Louis Gay-Lussac (1778 1850) (French chemist and physicist)

    i. ii. iii.

    Formulated the gas law. Collected samples of air at different heights and recorded temperatures and moisture contents. Discovered that the composition of atmosphere does not change with increasing altitude.

    Amedeo Avogadro (1776 1856) (Italian scholar)

    i. ii.

    Published article in French journal on determining the relative masses of elementary particles of bodies and proportions by which they enter combinations. Published a research paper titled New considerations on the theory of proportions and on determination of the masses of atoms.

    Note: In order to give a tribute to Avogadros contributions related to molecular theory, the number of elementary

    entities (atoms, molecules, ions or other particles) in 1 mole of a substance, 6.022 1023 is known as Avogadro number.

    1.1 Introduction Q.1. Chemistry has played an important role in the fulfillment of basic needs of man. Explain. Ans: Increasing population has led to an increase in the demands of basic needs of man (food, clothing and

    shelter). Developments in the field of chemistry have helped to cope up with these necessities as follows: i. Food: a. The population of a country requires nutritious and hygienic food in sufficient quantity. To

    achieve the same, there is a need to manufacture good quality fertilizers and insecticides. b. The advancement of chemistry has helped many countries to become not only self sufficient but

    also an exporter of food commodities. ii. Clothing: a. Good quality clothes are required for every individual to adjust with changing environmental

    conditions. b. Because of the production of synthetic fibres like nylon, rayon, etc. in factories, this need has

    been fulfilled. iii. Shelter: a. The human population needs comfortable and well-built houses. Iron, cement and steel are

    required in large quantities for construction of such houses. b. Chemistry has played an important role in the extraction of these metals from their respective

    ores. Q.2. Define chemistry. Ans: Chemistry is defined as the study of the composition, structure and properties of matter and the reactions

    by which one form of matter may be converted into another form. Q.3. Give reason : Chemistry is called as a central science. Ans: i. Chemistry is an active evolving science and is of vital importance to the entire world. Although the

    subject has very ancient roots, it is a modern science. ii. The basic knowledge of chemistry is essential for development of subjects like physics, biology,

    geology, engineering, environmental science and many others. Therefore, it is called as a central science.

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    Q.4. What are the various branches of chemistry? Ans: The several branches of chemistry are as follows: i. Physical chemistry: It is the branch of chemistry that deals with the structure of matter, the energy changes and the

    theories, laws and principles that explain the transformation of matter from one form to another. ii. Inorganic chemistry: It is the branch of chemistry that deals with chemistry of elements other than carbon and their

    compounds. iii. Organic chemistry: This branch of chemistry deals with reactions of the compounds of carbon. iv. Analytical chemistry: This is the branch of chemistry which deals with the separation, identification and quantitative

    determination of the compositions of different substances. v. Biochemistry: This is the branch of chemistry that deals with substances which are constituents of living organisms. Note: Pharmaceutical, environmental and nuclear chemistry are also branches of chemistry. 1.2 Importance and scope of chemistry Q.5. Explain the importance and scope of chemistry. Ans: Chemistry has a wide scope and importance in various fields. i. Processes based on chemical technology help to extract, purify, synthesize and analyse materials like

    iron, steel, aluminium, zinc, alloys like brass, amalgams as well as precious metals like silver, gold, platinum.

    ii. All drugs are synthesized in chemical / pharmaceutical laboratories. eg.

    Drugs Treatment of diseases i L-dopa For treatment of Parkinsons disease. ii Human insulin For treatment of diabetes. iii Cisplatin and taxol Life saving drugs to give relief to cancer patients. iv Azidothymidine (AZT) For treatment of AIDS. v Tamiflue For treatment of swine flu.

    iii. Photosynthesis is the process through which trees and plants prepare their food using chlorophyll

    (green pigment) in presence of sunlight. The process is a simple chemical reaction and takes place naturally.

    CO2 + H2O sunlightchlorophyll food grains/fruits/flowers/cotton/medicine etc.

    iv. Fossil fuels like coal, petroleum, natural gas, etc. are combustible chemicals which are used to produce energy which is used to drive trains, trucks, buses and all automobiles. The energy is also used to generate electricity. Several electrochemical cells like Daniel cell, lead storage cell, dry cell, nickel cadmium cell, lithium ion cell, fuel cell, etc., are used as a source of energy. These cells are less polluting and more efficient. There are attempts being made to convert solar energy into electrical energy using photovoltaic cells, the solar cells. Attempt is also made to obtain hydrogen from water, which is used in fuel cells to generate electricity.

    v. With the help of chemistry it is easy to design and generate large number of materials like polymers, plastic, liquid crystals, adhesives and surface coating materials like latex paints. Knowledge of chemistry can also be used to synthesize new materials that can act as super conductors at or near room temperature due to which loss of electricity will get reduced by almost 20%. Microprocessors used in computers are silicon chips formulated and developed by chemists.

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    #Q.6.Give five applications of subject chemistry which are not mentioned in the book. Ans: i. Warfare: With the knowledge of chemistry, various destructive gases and bombs have been invented which are

    used during wars. ii. Cosmetics: Chemistry has helped to produce good quality cosmetics. iii. Health: Chemistry plays an important role in maintaining ones good health by providing knowledge about

    proper intake of proteins, carbohydrates, fats, minerals, vitamins, etc. iv. Education: Chemistry provides inter-relationship to study the para-chemistry subjects such as Bio-chemistry,

    Pharmacy, Herbal Science, Toxicology, Archaeology, Environmental Science, etc. v. In recent years, chemistry has given us new materials such as superconducting ceramics, conducting

    polymers, optical fibres, micro alloys, carbon fibres, etc. which are used for various purposes. 1.3 Historical approach to particulate nature of matter Q.7. Define matter. Ans: Matter is anything which has mass and occupies space. Q.8. Explain the classification of matter on the basis of its chemical composition. Ans: Matter on the basis of chemical composition can be classified as follows: i. Pure substances: Substances which always have a fixed composition are called pure substances. They are of two types: a. Elements: Elements are the pure substances which are made up of only one component. eg. Gold, silver, copper, mercury, bromine, oxygen etc. b. Compounds: Compounds are the pure substances which are made up of two or more components. eg. Water, ammonia, methane etc. ii. Mixtures: A mixture is a simple combination of two or more substances in which the constituent substances

    retain their separate identities. The composition of mixture can be varied to any extent. Therefore, mixtures do not have fixed

    composition. eg. Mixture of ethyl alcohol and water, salt in water, mixture of gases, etc. Mixtures are of two types: a. Homogeneous mixture: A mixture in which the concentration of the constituents remains uniform throughout the

    mixture and all the constituents are present in one phase, is called a homogeneous mixture. eg. Mixture of salt and water. b. Heterogeneous mixture: If two or more phases are present in a mixture, it is called a heterogeneous mixture. eg. Phenol - water system, silver chloride-water system, iron fillings-sand system, etc. Q.9. What is Phlogiston theory? Ans: Phlogiston theory: i. According to this theory, a combustible substance contains phlogiston (a mysterious matter) and some

    clax. During combustion, phlogiston evolves and is lost in the atmosphere. Clax remains in the form of an ash.

    ii. Combustion reactions were explained by phlogiston theory. eg. During the combustion of a candle in a closed container, the air inside the container is saturated

    with phlogiston. Since air cannot accommodate more phlogiston, the candle gets extinguished.

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    Q.10. What is dephlogisticated air? Who named it? Ans: i. Joseph Priestley (a British scientist) focussed sunrays on a substance (mercuric oxide) to heat it. ii. A gas evolved, in which substances could burn more vigorously than in air. iii. A burning candle became brighter in this gas. iv. Priestley was of the view that this gas is the normal air without phlogiston. Hence he named it

    dephlogisticated air. Q.11. What was the contribution of Sir Henry Cavendish in the phlogiston theory? Ans: i. Sir Henry Cavendish carried out the reaction of a dilute acid with metals such as zinc, iron, etc. He

    named the gas evolved as flammable air. It was found that this gas burnt in air and in dephlogisticated air and produced water.

    ii. Cavendish suggested that flammable air is water associated with phlogiston. This is in continuation with the idea of phlogiston.

    Q.12. Who ruled out the theory of phlogiston? Why? Ans: i. The theory of phlogiston was ruled out by Antoine Lavoisier (a French Scientist). ii. He proved that a part of air is used in chemical reaction during combustion. This part of air was called

    oxygen. It means acid forming. iii. He also proved that oxygen was the gas formed in Joseph Priestleys experiment. iv. He also proved that the flammable air produced by Cavendish was a new gas, which he named as

    hydrogen, meaning water forming. Q.13. Define and explain the following with the help of examples. i. Elements ii. Compounds Ans: i. Elements: a. An element is defined as a substance which cannot be separated into simpler substances by any

    chemical process. eg. Gold, silver, copper, carbon, etc. b. Out of 118 elements that have been identified till recently, most are naturally occuring and a

    few are prepared in the laboratory (man-made). c. Chemists use one or two letter symbols to represent elements. d. All the chemical symbols have first letter capital and second letter small (if present). e. The symbols of the elements are derived either from English names or from Latin names of the elements. eg. Aluminium Al , Einsteinium Es, Gold Au (Aurum) f. Elements contain only one type of atom. Elements are further classified as metals, non-metals, metalloids. ii. Compounds: a. Compounds are defined as substances of definite compositions which can be decomposed into

    two or more substances by a simple chemical process. eg. Water, sodium chloride, sugar, alcohol, etc. b. The properties of compounds differ from the properties of the substances and elements obtained

    from decomposition of the compounds. eg. Hydrogen and oxygen are obtained from decomposition of water. Water can be used to

    extinguish fire whereas oxygen supports combustion and hydrogen is combustible. c. Compounds are classified into two subclasses organic compounds and inorganic compounds. d. Compounds contain two or more components.

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    Note: Names and symbols of some elements

    Element Sym-bol

    Element Sym-bol

    Element Sym-bol

    Element Sym-bol

    Element Sym-bol

    Aluminium Al Cadmium Cd Mercury Hg Neobium Nb Rhodium Rh Argon Ar Caesium Cs Holmium Ho Neodymium Nd Rhenium Re Silver Ag Cerium Ce Iodine I Neptunium Np Sulphur S Gold Au Curium Cm Irridium Ir Oxygen O Scandium Sc Actinium Ac Calefornium Cf Krypton Kr Osmium Os Selenium Se Americium Am Erbium Er Lithium Li Potassium K Strontium Sr Beryllium Be Einsteinium Es Lanthanum La Phosphorous P Sodium Na Boron B Fluorine F Lutetium Lu Lead Pb Technicium Tc Barium Ba Francium Fr Lawrencium Lr Palladium Pd Uranium U Bismuth Bi Iron Fe Magnesium Mg Platinum Pt Tungston W Carbon C Galium Ga Manganese Mn Promethium Pm Vanadium V Chlorine Cl Germanium Ge Molybdenum Mo Protoactinium Pa Xenon Xe Calcium Ca Gadolinium Gd Mendelivium Md Plutonium Pu Ytterbium Yb Chromium Cr Hydrogen H Nitrogen N Radium Ra Zirconium Zr Cobalt Co Helium He Neon Ne Rubidium Rb Copper Cu Hafnium Hf Nickel Ni Ruthenium Rn

    Q.14. Classify the following substances into elements, compounds, homogeneous mixtures and heterogeneous mixtures.

    Sand in water, Sodium chloride, Nitrogen, Sodium chloride in water, Pumice stone, Air, Phenol-water system, Carbon dioxide, Gold

    Ans: i. Sand in water : Heterogeneous mixture ii. Sodium chloride : Compound iii. Nitrogen : Element iv. Sodium chloride in water : Homogeneous mixture v. Pumice stone : Heterogeneous mixture vi. Air : Homogeneous mixture vii. Phenol-water system : Heterogeneous mixture viii. Carbon dioxide : Compound ix. Gold : Element Q.15. Distinguish between i. Mixtures and compounds. ii. Compounds and elements. Ans: i. Mixtures and compounds:

    Mixtures Compounds i. The constituents of a mixture may be

    present in any ratio. The constituents of a compound are always present in a fixed ratio.

    ii. Mixtures may or may not be homogeneous in nature.

    Compounds are always homogeneous in nature.

    iii. The properties of a mixture are in between those of its constituents.

    The properties of a compound are entirely different from those of its constituent elements.

    iv. The constituents of a mixture can be easily separated by simple physical means.

    The constituents of a compound cannot be easily separated by simple physical means but can be separated by chemical processes.

    v. The melting and boiling points of mixtures are usually not sharp.

    Chemical compounds possess sharp melting and boiling points.

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    ii. Compounds and elements: Compounds Elements

    i. A compound is a substance composed of two or more different chemical elements.

    An element is a pure chemical substance made of same type of atoms.

    ii. A compound can be separated into simpler substances by chemical reactions.

    Elements cannot be broken down into simpler substances.

    iii. Compounds contain different elements in a fixed ratio arranged in a defined manner through chemical bonds.

    Elements are distinguished by their atomic number (number of protons in their nucleus).

    iv. A compound is represented using a formula. An element is represented using symbols. v. The list of compounds is endless but can

    broadly be classified as ionic and covalent. There are about 118 elements that have been identified and can be classified as metals, non-metals or metalloids.

    eg. Sodium chloride (NaCl), Sodium bicarbonate (NaHCO3), etc.

    eg. Iron, copper, silver, gold, etc.

    Q.16. Define a unit. Ans: The arbitrarily decided and universally accepted standards used in the measurement of physical quantities

    are called units. eg. meter (m), kilogram (kg). Q.17. State the need of units. Ans: i. In chemistry, several experiments are carried out which involve observation and collection of both

    qualitative and quantitative data. ii. Measurement of physical properties such as mass, length, volume, temperature, pressure, time, etc.,

    comprise of the quantitative data. iii. For this purpose, the magnitude or size of physical quantity is compared with a suitable standard.

    These units are arbitrarily chosen on the basis of universally accepted standards. iv. To express any measured property, a number and an appropriate unit has to be used. Only number

    does not give any idea about the property. Q.18. What are the various systems in which units are expressed? Ans: Units are expressed in various systems like CGS (centimetre for length, gram for mass and second for

    time), FPS (foot, pound, second) and MKS ( metre, kilogram, second) systems, etc. Note: i. During calculations, confinement to one single system of unit is advisable. ii. NASAs Mars climate orbiter (first weather satellite for mars) was destroyed due to heat. The mission

    failed as there was a confusion while estimating the distance between earth and mars in miles and kilometers.

    Q.19. What are SI units? Name the fundamental SI units. Ans: SI Units: In 1960, the general conference of weights and measures proposed revised metric system, called International

    system of Units i.e. SI system abbreviated from its French name Systeme Internationale d Units. The seven fundamental SI units are as given below:

    No. Fundamental quantity SI unit Symbol i. Length Metre m ii. Mass Kilogram kg iii. Time Second s iv. Temperature Kelvin K v. Amount of substance Mole mol vi. Electric current Ampere A vii. Luminous intensity Candela cd

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    Q.20. What are derived units? Ans: The units of all physical quantities can be derived from the seven fundamental SI units. These units are

    known as derived units. eg. i. Area = Length squared (m2) ii. Concentration = mole per cubic metre (mol m3) Note: The table given below shows some common derived units.

    No. Physical quantity Relationship with fundamental unit Unit i. Area Length squared m2 ii. Volume Length cubed m3 iii. Density Mass per unit volume kg m3 iv. Velocity Distance travelled in unit time ms1 v. Acceleration Velocity change per unit time ms 2 vi. Force Mass Acceleration kg m s2 (newton, N) vii. Pressure Force per unit area kg m1 s2 viii. Electric charge Current Time As (coulomb, C) ix. Electric potential or Potential

    difference Energy per unit charge kg m2 s2 A1

    (J A1 s1 or Volt,V or JC1)

    x. Energy (work or heat)

    Force distance travelled kg m2 s2 (Joule s1)

    xi. Concentration Mole per cubic metre mol m3 xii. Heat capacity Cp = dH / dT

    Cv = dE / dT JK1 mol1

    xiii. Electrochemical equivalent Z = E/F kg C1 (kg/Coulomb) Some common SI prefixes used for expressing big and small numbers:

    Prefix Symbol Magnitude Meaning (multiply by) Tera T 1012 1 000 000 000 000 Giga G 109 1 000 000 000 Mega M 106 1 000 000 myria my 104 1 000 0 (this is now obsolete) kilo k 103 1 000 hecto h 102 100 deka da 10 10 deci d 101 0.1 centi c 102 0.01 milli m 103 0.001 micro 106 0.000 001 nano n 109 0.000 000 001 pico p 1012 0.000 000 000 001 femto f 1015 0.000 000 000 000 001

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    1.4 Laws of chemical combination Q.21. What is a chemical combination? Ans: The process in which the elements combine with each other chemically, to form compounds, is called as

    chemical combination. *Q.22.State and explain the laws of chemical combination. Ans: Laws of chemical combination: One of the most important aspects of the subject of chemistry is the study of chemical reactions. These

    chemical reactions take place according to certain laws called as Laws of chemical combination. They are:

    i. Law of conservation of mass: a. The law of conservation of mass states that, mass is neither created nor destroyed during

    chemical combination of matter. b. It was first stated by Russian scientist Lomonosove (1765) and later independently stated by

    French chemist Antoine Lavoisier (1783) who performed careful experimental studies for various combustion reactions.

    c. Lavoisier observed that the total masses of the reactants (before the reaction) were in agreement with the total masses of the products (after the reaction).

    eg. Consider the reaction for the burning of carbon to form carbon dioxide. Carbon + Oxygen Carbon dioxide The sum of the masses of carbon and oxygen (reactants) is always equal to the mass of the

    carbon dioxide (product). This is in accordance with the law of conservation of mass. ii. Law of definite composition / proportions: a. The law of definite composition/ proportions was stated by French chemist Joseph Proust. b. The law states that any pure compound always contains the same elements in a definite

    proportion by weight irrespective of its source or method of preparation. c. Proust worked with two samples of cupric carbonate; one of which was naturally occuring

    cupric carbonate and other was prepared in the laboratory. He found that the composition of elements present in both the samples of cupric carbonate was same as shown below:

    Percentage Cu C O

    Naturally occurring cupric carbonate 51.35 38.91 9.74 Cupric carbonate prepared in the laboratory 51.35 38.91 9.74

    d. The law was further supported by different samples of pure water which contained same

    amount of oxygen (88.81% by weight) and hydrogen (11.19% by weight) and different samples of pure sugar which contained same amount of carbon (42.1% by weight), hydrogen (6.5% by weight) and oxygen (51.4% by weight). This was irrespective of the source.

    iii. Law of multiple proportions: a. John Dalton (British scientist) proposed the law of multiple proportions in 1803. b. The law states that, if two elements chemically combine with each other forming two or more

    compounds with different compositions by mass, then the ratios of masses of the two interacting elements in the two compounds are small whole numbers.

    eg. Chemical reaction of carbon with oxygen gives two compounds carbon monoxide and carbon

    dioxide. Carbon monoxide is a poisonous, combustible gas. However carbon dioxide is a nonpoisonous, noncombustible gas.

    1g of carbon reacts with 1.33 g of oxygen to form carbon monoxide. 1g of carbon reacts with 2.66 g of oxygen to form carbon dioxide. The ratio of weights of oxygen to that of carbon for carbon dioxide is

    2.66gof oxygen1.00gof carbon

    = 2.66

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    And the ratio of weights of oxygen to that of carbon for carbon monoxide is

    1.33gof oxygen1.00gcarbon

    = 1.33

    The two ratios are in the proportion 2.661.33

    = 2 i.e., 2:1

    Therefore, the ratio of the masses of oxygen that combine with the same mass of carbon is 2:1 i.e., a simple ratio.

    iv. Gay Lussacs law of combining volumes of gases: a. Joseph Louis Gay Lussac (a French chemist) observed that there exists a definite relationship

    between volumes of gaseous reactants and the volumes of gaseous products. He generalized his observations in the form of a law of combining volumes of gases.

    b. Gay Lussacs law states that, when gases react together to produce gaseous products, the volumes of reactants and products bear a simple whole number ratio with each other, provided volumes are measured at same temperature and pressure.

    c. eg. Under identical conditions of temperature and pressure, 1L of hydrogen gas reacts with 1 L of chlorine gas to produce 2 L of hydrogen chloride gas i.e.

    Hydrogen + Chlorine Hydrogen chloride [ 1 L ] [ 1 L ] [ 2 L ] [ 1 vol ] [ 1 vol ] [ 2 vol ] Thus, the ratio of volumes is 1:1:2 d. Volumes may be measured in any convenient unit such as L, mL, cm3, dm3, etc. eg. Consider the reaction for the conversion of sulphur dioxide to sulphur trioxide. Sulphur dioxide + Oxygen Sulphur trioxide [ 200 mL ] [ 100 mL] [ 200 mL ] [ 1 vol ] [ 12 vol ] [ 1 vol ] The ratio of volumes is 2 : 1: 2. Note: The first three laws deal with the mass relationships whereas the fourth law deals with the volume of the

    reacting gases. Q.23. Who opposed the law of definite proportion? How were the objections ruled out? Ans: i. Berthollet (French scientist) opposed Prousts law of definite proportion. ii. He gave examples of the substances that contained different proportions of elements. iii. However, the experimental work of analysis explained by Berthollet was based on impure samples or

    incomplete reactions. Hence, Berthollets objections were ruled out. Q.24. Is the law of constant composition true for all types of compounds? Justify your answer. Ans: No, law of constant composition is not true for all types of compounds. It is true for only those compounds

    which are obtained from one isotope. eg. Carbon exists in two common isotopes: 12C and 14C. When it forms 12CO2, the ratio of masses is

    12 : 32 or 3 : 8. However, when it is formed from 14C i.e., 14CO2, the ratio will be 14 : 32 i.e., 7 : 16 , which is not same as in the first case.

    Q.25. Verify the law of multiple proportions for the chemical reaction between hydrogen and oxygen. Ans: The chemical reaction of hydrogen with oxygen gives two compounds, water and hydrogen peroxide. Water

    contains 88.89% by weight of oxygen and 11.11% by weight of hydrogen. The ratio of the percentages by weight of oxygen to that of hydrogen is equal to

    88.89%by weight of oxygen11.11%by weight of hydrogen

    = 8 = 8:1

    And hydrogen peroxide contains 94.12% by weight of oxygen and 5.88% by weight of hydrogen. The ratio of the percentages by weights is equal to

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    94.12% by weight of oxygen5.88% by weight of hydrogen

    = 16 = 16:1

    The ratios of the two compounds are in the proportion 16 :18:1

    = 2:1

    Therefore, the ratio of weights of oxygen that combine with the same weight of hydrogen in the two compounds is equal to 2:1.

    This is in accordance with the law of multiple proportions. Q.26. Give two examples which support the Gay Lussacs law of combining volumes of gases. Ans: i. Under identical conditions of temperature and pressure, 2 L of hydrogen gas reacts with 1L of oxygen

    gas to produce 2 L of steam (water vapour). i.e., Hydrogen + Oxygen Steam (water vapour) 2 L 1 L 2 L (2 vol) (1 vol) (2 vol) The ratio of volumes, is 2 : 1 : 2. ii. Under identical conditions of temperature and pressure, 1L of nitrogen gas reacts with 3 L of

    hydrogen gas to produce 2 L of ammonia gas, i.e. Nitrogen + Hydrogen Ammonia 1L 3 L 2 L (1 vol) (3 vol) (2 vol) Here the ratio of volumes is 1 : 3 : 2 From these two examples, it can be concluded that there exists a simple ratio of whole numbers of volumes

    of the gaseous reactants with gaseous products. Hence, these examples support the Gay Lussacs law. Note: i. Gay Lussacs law of combining volumes is applicable only to reactions involving gases and not to solids

    and liquids. ii. The volumes of gases in the chemical reaction are not additive. For example, in case of reaction between

    hydrogen and chlorine gases it appears to be additive. However in case of reaction between hydrogen and oxygen, 2 volumes of hydrogen and 1 volume of oxygen, equal to 3 volumes of reactants get converted into 2 volumes of the product, steam.

    iii. Similarly, in case of formation of ammonia, 1 volume of nitrogen and three volumes of hydrogen, equal to 4 volumes of reactants, react to get converted into 2 volumes of the product, ammonia.

    1.5 Daltons atomic theory *Q.27. State and explain Daltons atomic theory. Ans: To provide theoretical justification to the laws of chemical combination, John Dalton postulated a simple

    atomic theory of matter (1808). According to him, Atom is the smallest indivisible particle of a substance. The basic assumptions of the

    theory are given below: i. All matters are made up of tiny, indestructible, indivisible unit particles called atoms. Atoms are the

    smallest particles of the element and molecules are the smallest particles of a compound. ii. All atoms of the same element have same size, shape and mass and all other properties. Atoms of

    different elements have different properties. iii. Compounds are formed when atoms of different elements combine. The atoms in a compound unite

    in small whole number ratios like 1: 1, 1 : 2, 1: 3, 2 : 1, 2 : 3, etc. iv. A chemical reaction involves only the separation, combination or rearrangement of integer number of

    atoms. During a chemical reaction, atoms are neither created nor destroyed.

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    Q.28. How was the law of conservation of mass explained by Dalton on the basis of his atomic theory? Ans: i. Based on the assumptions of atomic theory, Dalton explained the law of conservation of mass. ii. He stated that, during a chemical reaction, atoms can rearrange and change their partners. Molecules

    could be decomposed into atoms. iii. However, total number of atoms in the reactants and products should be same. Thus, mass is

    conserved during a reaction. Q.29. How was the law of multiple proportion explained by Dalton on the basis of his atomic theory? Ans: i. The law of multiple proportion states that, if two elements chemically combine with each other

    forming two or more compounds with different compositions by mass, then the ratios of masses of two interacting elements in the two compounds are small whole numbers.

    ii. According to Daltons atomic theory, compounds are formed when atoms of different elements combine.

    iii. Dalton explained that under certain conditions, atoms of two types combine in the ratio 1:1 to form a molecule. Under some other conditions, they may combine in the ratio 1:2 or 1:3 or 2:3, etc.

    Note: Dalton proposed some symbols for some common atoms and molecules. They are shown in the following

    figure: 1.6 Concepts of elements atoms and molecules Q.30. Explain the following terms: i. Atoms ii. Molecules Ans: i. Atoms: a. The smallest indivisible particle of an element is called an atom. Thus, it has properties similar

    to that of the element. b. Atoms may or may not exist freely. Atoms of almost all the elements can react with one another

    to form compounds. c. Every atom of an element has definite mass of the order of 1027 kg and has a spherical shape of

    radius of the order of 1015 m. d. The smallest atom of the element is that of hydrogen with mass of 1.667 1027 kg. eg. Water consists of 3-atoms; 2-hydrogen atoms and 1-oxygen atom. Note: The properties of the constituent atoms are different from those of their respective compounds. eg. Carbon dioxide gas is used as a fire extinguisher although carbon is combustible and oxygen supports

    combustion.

    Oxygen Sulphur Nitrogen Hydrogen

    Carbon Potassium Water

    Methane Carbon monoxide

    Hydrogen peroxide Carbon dioxide

    Daltons symbols for some elements and compounds

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    ii. Molecules: a. A molecule is an aggregate of two or more atoms of definite composition which are held

    together by chemical bonds. b. A molecule may contain atoms of same elements or different elements. c. It is the smallest independent unit of a compound which exists freely. It has all the properties of

    the original compound. d. It cannot be divided into constituent atoms with the help of simple methods but decomposition

    into constituent atoms can be achieved under drastic conditions. eg. Water molecule (H2O). 1.7 Atomic and molecular masses Q.31. Why do different elements have different atomic masses? OR Justify: It is not possible to weigh a single atom. Ans: i. Each and every atom of an element consists of fixed number of protons, neutrons and electrons,

    which are the subatomic particles. ii. The number of protons differs from atom to atom. iii. As the number of subatomic particle changes, the mass of the atom changes. iv. The size of one atom is of the order of 1015 m. The mass is of the order of 1027 kg. Thus it is not possible to weigh a single atom. Q.32. How is atomic mass measured? OR Why is atomic mass called relative mass? Ans: i. The size of a single atom is of the order of 1015 m and the mass is of the order of 1027 kg. Thus it is

    not possible to weigh a single atom. ii. The masses of the atoms of the elements can be determined experimentally by using mass

    spectrograph. iii. For this purpose, a standard element is chosen and assigned appropriate value of mass of an atom. iv. By international agreement in 1961, for determining atomic masses, carbon-12, a distinct atom of carbon,

    was chosen as the standard with its atomic mass as 12.000 atomic mass unit (abbreviated as amu). v. Atomic mass unit is defined as 1/12 of the mass of an atom of carbon-12 equal to 1.6605 1024 g. vi. The masses of all other atoms are determined relative to the mass of an atom of carbon-12. Thus,

    atomic mass is called as relative mass. Note: Recently, the unit of atomic mass, amu is replaced by u which means unified mass. Q.33. Define isotopes. Ans: Isotopes are the atoms of the same element having same atomic number but different mass number. Therefore

    isotopes of an element contain same number of protons and electrons but different number of neutrons. *Q.34. Explain the need of the term average atomic mass? Ans: i. Several naturally occurring elements exist in the form of two or more isotopes. They have different

    isotopic masses. ii. In any sample of an element, the isotopes are present in different quantities. iii. The observed atomic mass of the atom of the element is the average atomic mass of the element. The

    natural abundances of the isotopes are taken into consideration for this purpose. The natural abundance is the percentage occurrence.

    iv. Thus, it is the average weight of an atom of the element which is used in calculating the atomic weight of the element.

    v. Chemistry is a macroscopic science and involves a large number of atoms (or molecules). Hence, an average mass of an atom or average atomic mass is an inevitable term.

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    eg. Chlorine has two isotopes, Cl35 and Cl37, present in 75% and 25% proportion respectively. Hence,

    the atomic weight of chlorine is the weighed average of these two isotopic weights i.e., (35.0 0.75) + (37.0 0.25) = 35.5.

    Note: i. The variation of the exact atomic mass of isotope of elements with its relative abundance is obtained

    by using the mass spectrometer with higher precision. ii. By using Astons mass spectrometer, it was shown that Neon exists in nature in the form of a mixture

    of the following three isotopes: a. Neon-20 with atomic mass 19.9924 u with natural abundance 90.92% b. Neon - 22 with atomic mass 21.9914 u with natural abundance 8.82%. c. Neon - 21 with atomic mass 20.9940 u with natural abundance 0.26%. The observed atomic mass of Neon is its average atomic mass which is calculated as shown below: Average atomic mass of Ne

    =

    (Atomic massof Ne-20 % of Ne-20)Atomic massof Ne-22 % of Ne-22

    100Atomic massof Ne-21 % of Ne-21

    + +

    = [19.9924 u 90.92 + 21.9914 u 8.82 + 20.994 u 0.26] /100 = 20.1713u *Q.35. Define the term Molecular Mass.

    Ans: Molecular mass of a substance is defined as the ratio of mass of one molecule of a substance to th1

    12of the

    mass of one atom of Carbon-12. It is also the algebraic sum of atomic masses of constituent atoms which constitute the molecule. Q.36. What are the characteristics of molecular mass (molar mass)? Ans: i. Molecular mass is also expressed in amu.

    ii. It indicates comparative mass of a molecule of a compound with respect to th1

    12of the mass of one

    atom of Carbon-12. iii. The molar mass expressed in gram is known as gram molar mass. 1 gram molar mass is also known

    as 1 gram molecule or 1 gram mole or 1 mole. iv. 1 mole of the element is the amount of the element equal to its atomic mass in gram. It is also called 1

    gram atom of the element.

    19 20 21 22 23

    2210 Ne (8.82%)2110 Ne (0.26%)

    2010 Ne (90.92%)

    Atomic mass (amu)

    Inte

    nsity

    (per

    cent

    age)

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    1.8 Avogadros law *Q.37.State and explain Avogadros law. Ans: i. In the year 1811, Avogadro combined Gay Lussacs law and Daltons theory and thereby proposed

    Avogadros law. ii. Avogadros law states that, equal volumes of all gases, under identical conditions of temperature

    and pressure, contain equal number of molecules. iii. Mathematically, Avogadros law is stated as, at constant pressure and temperature, volume of a gas

    is directly proportional to the number of molecules. iv. V number of molecules (P, T constant) Since number of molecules is proportional to the number of moles of gas n, V n (P, T constant)

    or Vn

    = constant, where n = massof gasmolar massof gas

    v. At standard temperature and pressure (STP condition), i.e., at 273.15 K and 1 atmosphere, the volume of 1 mole of a gas i.e., molar volume of a gas can be calculated by gas equation,

    PV = nRT

    Volume per mol (molar volume) = Vn

    Vn

    = RTP

    , (R = 0.08206 L atm mol1 K1)

    = 0.08206 273.151atm = 22.414 L mol1.

    vi. Thus 1 mole of any pure gas occupies a volume of 22.414 L (or 0.022414 m3) at standard temperature and pressure.

    The value 22.414 L mol1 is called Avogadros molar volume or molar gas volume at STP. Q.38. Describe the significance of Avogadros number. Ans: The significance of Avogadros number is as follows: i. Avogadros number is equal to the number of molecules present in one gram mole or one gram

    molecular weight of any compound. Gram molecular weight of any substance is the weight in grams of Avogadros number, i.e., 6.0221023 molecules.

    ii. It is equal to the number of molecules in one mole or number of atoms in one gram atomic weight of an element. Gram atomic weight of an element is the weight of Avogadro number of atoms.

    iii. It is equal to the number of molecules in 22.414 dm3 of any gas at STP. iv. The actual weight of a molecule of a compound or an atom of an element can be calculated using this

    number. Q.39. What is Atomicity? Ans: Atomicity of a molecule is the total number of atoms of constituent elements combined to form a molecule. eg. Atomicity of Oxygen (O2) is two, while that of Ozone (O3) is three. Note: He and Ne are monoatomic O2, N2 and H2 are diatomic O3 is triatomic P4 is tetraatomic S8 is Polyatomic.

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    *Q.40. Explain how an atom of an element is distinguished from its molecule. Ans: i. Consider the formation of hydrogen chloride from hydrogen and chlorine. ii. Hydrogen and chlorine do not exist in free atomic state but exist in molecular state and molecules

    take part in the reaction to form hydrogen chloride. iii. The reaction may be written as Hydrogen + Chlorine Hydrogen chloride According to Gay Lussacs law: [1 volume] [1 volume] [2 volumes] According to Avogadros law: (V n) [n molecules] [n molecules] [2n molecules]

    Dividing by 2n 1 molecule2

    1 molecule2

    [1 molecule]

    iv. This shows that 12

    molecule of hydrogen and 12

    molecule of chlorine give 1 molecule of HCl. If

    molecules of hydrogen and chlorine are considered to be monoatomic, it indicates that atoms are divisible and this is contrary to Daltons atomic theory which states that atom is indivisible.

    So, the molecule of hydrogen and chlorine are considered to be diatomic. *Q.41. Explain mole concept. Ans: i. The mass of one atom of an element or one molecule of a compound is negligibly small and is

    difficult to weigh. ii. The amount of a substance equal to its atomic mass or molar mass in grams is 1 mole of a substance. iii. Thus, one mole of a substance is defined as the amount of the substance that contains the number of

    particles (atoms, molecules, ions or electrons, etc.) as present in 0.012 kg of carbon12. iv. This number of particles is determined to be equal to 6.022 1023 particles. Q.42. What is Avogadro Number (NA)? Ans: The number of atoms, molecules, ions, or electrons, etc. present in 1 mole of a substance is found to be

    equal to 6.022 1023, which is called Avogadro Number (NA). Thus, NA = 6.022 1023 mol1 Note: i. Number of electrons equal to NA make an electrical charge of one Faraday. ii. One mole of various gases (of different molar masses) occupy 22.414 L at STP and contain

    avogadros number of molecules.

    + =

    2X molecules of

    hydrogen chloride(2 Volume)

    X molecules of

    chlorine (1 Volume)

    X molecules of

    hydrogen (1 Volume)

    Formation of HCl molecule

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    Q.43. Give the relationship between mass of a gas, number of moles and volume of the gas at STP, with the

    help of mole triangle.

    Ans: The mole triangle representing the relationship between the mass of gas, number of moles, volume of gas at

    STP and the number of molecules.

    Note:

    i. Number of moles (n) = massof a substancemolar massof thesubstance

    = WM

    where, W is mass and M is molar mass of the gas.

    ii. One mole of a gas contains 6.022 1023 molecules.

    Number of molecules = n Avogadro number = n NA = n 6.022 1023.

    iii. One mole of a gas at STP (273.15 K and 1 atmosphere) occupies 22.414 L

    Volume of a gas at STP = n 22.414 L.

    Mole triangle

    Number of fundamental

    particles

    Multiplied by Avogadros number

    Divided by Avogadros number

    Divided by molecular weight

    (molar mass) Multiplied by 22.4 dm3 Mass of substance Multiplied by

    molecular weight(molar mass)

    Divided by 22.4 dm3

    Volume occupied by gas at STP

    in dm3

    Number of moles

    6.022 1023 molecules of

    O2

    6.022 1023molecules of

    CO2

    6.022 1023molecules of

    CH4

    6.022 1023molecules of

    Ar

    22.4 L 32.0 g

    22.4 L39.9 g

    22.4 L44.0 g

    22.4 L16.0 g

    1 mole of different gases having differentmasses containing 6.022 1023 molecules andoccupying volume of 22.414 L at STP

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    1.9 Percentage composition and molecular formula *Q.44. Explain the terms i. Percentage composition ii. Empirical formula iii. Molecular formula Ans: i. Percentage composition: The percentage composition of a compound is defined as the percentage by weight of each element

    present in the compound.

    Percentage (by weight) = Massof theelement in1moleof thecompoundGrammolecular weight of thecompound

    100 %

    eg. One mole of ethanol C2H5OH (molecular mass 46) contains, 2 moles of carbon atom = 12 2 = 24 g 1 mole of oxygen atom = 16 1 = 16 g 6 moles of hydrogen atom = 1 6 = 6 g Thus 46 g of ethanol contains 24 g, 16 g and 6 g of carbon, oxygen and hydrogen respectively. Hence,

    the percentages of constituent elements are:

    Percentage of carbon = 2446

    100 = 52.17%

    Percentage of oxygen = 1646

    100 = 34.78%

    Percentage of hydrogen = 646

    100 = 13.04%

    Each 100 g sample of ethanol contains 52.17 g carbon, 34.78 g oxygen and 13.04 g hydrogen. ii. Empirical formula: The empirical formula of a compound is defined as a chemical formula indicating the relative number

    of constituent atoms in a molecule in the simplest ratio. eg. The empirical formula of ethanol (C2H5OH) is C2H6O and that of benzene (C6H6) is CH. iii. Molecular formula: The formula which gives the actual number of each kind of constituent atoms in one molecule of the

    compound is called the molecular formula of the compound. It is an integral multiple of empirical formula. eg. The molecular formula of ethanol is C2H5OH and that of benzene is C6H6. Q.45. Write the steps for determination of empirical and molecular formula of a compound. Ans: Steps for determination of empirical and molecular formula of a compound: i. First the percentage of all the elements present in the compound are experimetally determined. If the

    sum of the percentages of the constituent elements is less than 100%, then oxygen is present. The difference between 100 and sum of the percentages of the constituent elements is the percentage of oxygen.

    ii. The number of moles of each constituent element present in 100g of the substance is obtained by dividing percentage of the element by its atomic mass.

    iii. The ratio of number of moles of constituent elements is determined. It is then converted into smallest simple whole number ratio.

    eg. For ethanol, C2H5OH, the smallest simple whole number ratio in the order of C : H : O is 2 : 6 : 1.

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    iv. For fractional ratios, the whole number ratio is obtained by multiplying it with suitable integer. For this, minor fractions may be neglected.

    v. The empirical formula of the compound is obtained by writing the whole numbers of the ratio of number of moles of elements as the subscripts to the right side of the corresponding atoms of the elements.

    vi. Molecular mass of the substance is experimentally determined. The ratio (r) of molar mass of the substance to the empirical formula mass of the substance is determined, then

    Molecular formula = r empirical formula of compound. #Q.46. Write empirical and molecular formulae of following compounds: i. water ii. carbon iii. methane iv. hydrogen peroxide v. acetylene vi. glucose vii. diborane viii. tetraphosphorous decoxide. Ans:

    Compound Empirical formula Molecular formula i. Water H2O H2O ii. Carbon C C iii. Methane CH4 CH4 iv. Hydrogen peroxide HO H2O2v. Acetylene CH C2H2vi. Glucose CH2O C6H12O6 vii. Diborane BH3 B2H6 viii. Tetraphosphorous decoxide P2O5 P4O10

    1.10 Chemical reactions and stoichiometry Q.47. What is a chemical reaction? Ans: Chemical reaction is a process in which a single substance or many substances interact with each other to

    produce one or more substances. They are represented in terms of chemical equation. eg. The chemical reaction for the formation of water (product) from gaseous hydrogen and oxygen (reactants)

    may be written as, H2(g) + O2(g) H2O(g) This is as shown below:

    Q.48. Describe in brief representation of chemical symbols in a chemical equation. Ans: Chemical equation use chemical symbols called formulae of reactants and products. For example, symbol

    of atomic hydrogen is H. It exists in gaseous state as a diatomic molecule. Therefore, chemical formula of hydrogen molecule is H2, its gaseous state is indicated by the suffix (g) and written completely as H2(g). Similarly other gases are represented as, oxygen O2(g), nitrogen N2(g), fluorine F2(g), chlorine Cl2(g), bromine Br2(g) and Iodine I2(g).

    At room temperature, bromine exists in liquid state, hence, bromine in liquid state is represented as Br2(l) while iodine exists in solid state and hence represented as I2(S).

    +

    Two hydrogen molecules + One oxygen molecule Two water molecules2H2 + O2 2H2O

    Formation of water molecules

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    Q.49. What are reactants and products in a chemical reaction? Ans: The starting material which takes part in chemical reaction is called as reactant, the substance formed after

    the chemical reaction is called product. eg. Gaseous hydrogen and oxygen react to form water. H2(g) + O2(g) H2O(g) Here, H2(g) and O2(g) are reactants and H2O(g) is product obtained in the reaction. Q.50. Explain the balancing of mass in the following reaction: H2(g) + O2(g) H2O(g) Ans: i. In the given reaction, H2(g) + O2(g) H2O(g) conservation of mass in not observed. ii. Hence, balancing of mass is essential. iii. In order to conserve the mass of oxygen the reaction is balanced as 2H2(g) + O2(g) 2H2O(g) iv. This balanced chemical equation suggests that two molecules of hydrogen react with one molecule of

    oxygen gas to produce two water molecules. *Q.51. Write a note on chemical stoichiometry. Ans: i. Chemical stoichiometry is a process of making calculations based on formulae and balanced

    chemical equations. ii. Consider the balanced chemical reaction, 1N2(g) + 3H2(g) 2NH3(g) Numbers 1, 3 and 2 are coefficients representing the number of moles of N2(g), H2(g) and NH3(g)

    respectively. iii. From this chemical equation, it can be seen that 1 mole of N2(g) (28 grams) react with 3 moles H2(g) (6 grams) giving 2 moles of NH3(g) (34 grams). Hence from the amount of the reactants, the amount of products that would be formed can be calculated. iv. From stoichiometry, the amounts of reactants required to obtain definite amount of products can be estimated. Q.52. What are the steps involved in writing a balanced chemical equation by stoichiometry? OR What are the steps involved in stoichiometric calculation? Ans: Following are the steps involved in the stoichiometric calculations: i. The correct formula of reactants and products must be written. The valencies of the atoms of

    compounds should be satisfied. ii. In order to write balanced chemical equation, following three steps must be followed: Step I : The names of the reactants are written on the left hand side. A + sign is used to separate the reactants. An

    arrow is drawn from left to right and on the right side of the arrow, the names of the products are written. A + sign is used to separate the products.

    eg. The reaction for the combustion of methane to form carbon dioxide and water is written as, methane + oxygen carbon dioxide + water Step II : The chemical equation is rewritten in terms of chemical formula of each substance. CH4 (g) + O2 (g) CO2 (g) + H2O(g) Step III : To balance the mass of the chemical reaction, proper whole number coefficients are selected for each

    reactant and product. CH4(g) + 2O2 (g) CO2 (g) + 2H2O(g) This is the balanced chemical equation.

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    Note: Some Common Compounds:

    No. Compound Formula No. Compound Formula i. Phosphoric acid H3PO4 xi. Ferrous chloride FeCl2 ii. Sodium phosphate Na3PO4 xii. Ferric chloride FeCl3 iii. Ferric phosphate FePO4 xiii. Stannous chloride SnCl2 iv. Aluminium phosphate AlPO4 xiv. Stannic chloride SnCl4 v. Copper phosphate Cu3(PO4)2 xv. Sulphuric acid H2SO4 vi. Ferrous phosphate Fe3(PO4)2 xvi. Sodium sulphate Na2SO4 vii. Hydrogen chloride HCl xvii. Copper sulphate CuSO4 viii. Sodium chloride NaCl xviii. Ferrous sulphate FeSO4 ix. Cuprous chloride Cu2Cl2 xix. Ferric sulphate Fe2(SO4)3 x. Cupric chloride CuCl2

    Q.53. Show that Law of conservation of mass is fully justified in a balanced chemical equation. Ans: i. A balanced chemical reaction can be used to establish the weight relationships of reactants and

    products. ii. This is in accordance with the law of conservation of mass, which states that, total mass of reactants

    is always equal to total mass of the products. eg. The mass relationship is given in following balanced reaction. (Atomic masses: Fe = 55.85 u, Cl = 35.45 u) 2Fe(s) + 3Cl2(g) 2FeCl3(s) (2 atoms) (3 molecules) (2 molecules) [2 55.85 g] [3 35.45 2 g] [2 (55.85 + 3 35.45) g] [111.7 g ] [ 212.7 g] [324.4 g] 324.4 g 324.4 g Hence, the mass is conserved during the reaction. *Q.54. What are limiting and excess reactants? Ans: i. The stoichiometric coefficients of reactants and products in the balanced chemical equation

    determines the amounts of reactants required and products formed in a chemical reaction. ii. Usually, to save cost cheaper reactant is taken in excess while the costlier reactant is used in lesser

    amount. iii. As and when the reactant, which is taken in lesser amount, gets consumed the reaction stops. For the

    reactant, which is taken in excess, only a part of it is consumed while the rest is left behind with the products.

    iv. Thus, the reactant taken in inadequate quantity is called as limiting reactant. It is the reactant that reacts completely, but limits further progress of the reaction.

    v. The excess reactant is the reactant which is taken in excess compared to the limiting reactant. Hence, some amount of it remains unreacted.

    Q.55. In the combustion of methane in air, what is the limiting reactant and why? Ans: In the combustion of methane in air, methane is the limiting reactant because the other reactant is oxygen of

    the air which is always present in excess. Thus, the amounts of carbon dioxide and water formed will depend upon the amount of CH4 burnt.

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    Quick Review The classification of matter (On basis of chemical composition): Different laws and the contributing scientists:

    No. Laws Contributing Scientists i. Law of conservation of mass i. Lomonosove (Russian scientist)

    ii. Antoine Laviosier (French chemist) ii. Law of definite composition Joseph Proust (French chemist) iii. Law of multiple proportions John Dalton (British scientist) iv. Law of combining volumes of gases Joseph Louis GayLussac (French chemist)

    Important Formulae

    1. One mole of atoms = Massof elementAtomicmass

    2. Mass of one atom = 23Atomicmass6.022 10

    3. Mass of one molecule = 23Molecular mass

    6.022 10 4. Number of moles (n) = Massof substance

    Molar massof substance 5. Number of molecules = n Avogadro number 6. Volume of gas at STP = n 22.414 L.

    7. Percentage (by weight) = Massof theelement in1moleof thecompoundGram molecular weight of thecompound

    100

    8. Molecular formula = r empirical formula (where r is ratio of molecular mass to empirical mass).

    Mixtures A simple combination of two

    or more substances in whichthe constituent substancesretain their separate identities.

    Pure substances

    Substances which always have a fixed composition.

    Elements Pure substances which are made up of only one component. eg. Gold, silver, etc.

    Homogeneous mixture A mixture in which the concentration of the constituents remains uniform throughout the mixture and all the constituents are present in one phase. eg. Mixture of salt and water

    Matter Anything which has massand occupies space.

    Compounds Pure substances which are made up of two or more components. eg. Water, ammonia, etc.

    Heterogeneous mixtureA mixture in which two or more phases are present.eg. Phenol-water system

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    Solved Examples Type I : Problems based on average atomic mass Example 1 Calculate the average atomic mass of naturally occuring argon from the following data: (NCERT)

    Solution: To find: Average atomic mass of naturally occuring argon Formula: Average atomic mass of argon

    =

    36 36

    38 38

    40 40

    atomicmassof Ar percentageof Aratomicmassof Ar percentageof Ar 100atomicmassof Ar percentageof Ar

    + +

    Calculation: Average atomic mass of argon

    = 35.96755 0.337 37.96272 0.063 39.9624 99.60100

    + +

    = 39.947 g mol1. Ans: Average atomic mass of argon = 39.947 g mol1. Example 2 Calculate the atomic mass (average) of chlorine using the following data: (NCERT)

    % Natural abundance Atomic mass

    35C1 75.77 34.9689 37C1 24.23 36.9659

    Solution: To find: Average atomic mass of chlorine Formula: Average atomic mass of chlorine

    =

    35 35

    37 37

    atomic massof Cl percentageof Clatomic massof Cl percentageof Cl

    100

    +

    Calculation: Average atomic mass of chlorine

    = 34.9689 75.77 36.9659 24.23100

    +

    = 35.4528 g mol1 Ans: Average atomic mass of chlorine = 35.4528 g mol1.

    Isotope Isotopic mass (g mol1) abundance 36Ar 35.96755 0.337% 38Ar 37.96272 0.063% 40Ar 39.9624 99.600%

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    Example 3 Boron occurs in nature in the form of two isotopes having atomic mass 10 and 11. What are the percentage abundances of two isotopes in the sample of boron having average atomic mass 10.8? Solution: Given: Atomic masses of two isotopes of boron = 10 and 11, Average atomic mass of boron = 10.8 To find: The percentage abundances of two isotopes in the sample of boron having average atomic mass 10.8

    Formula: Average atomic mass =10

    11

    atomicmassof B percentage100

    atomicmassof B percentage

    +

    Calculation: Let the % abundance of 10B isotope = x. % abundance of 11B isotope = 100 x. Average atomic mass = 10.8

    From formula, Average atomic mass = 10 (100 ) 11100

    + x x = 10.8

    10x + 1100 11x = 10.8 100 x = 1100 + 1080 x = 20 Percentage abundance: 10B = 20%, 11B = (100 20) = 80% Ans: Percentage abundance: 10B = 20%, 11B = 80%. Type II : Problems based on Avogadro number and mole concept *Example 4 Calculate the number of moles and molecules of ammonia present in 5.6 dm3 of its volume. Solution: Let number of moles present in 5.6 dm3 of ammonia = x Number of moles present in 22.414 dm3 of ammonia = 1

    x = 5.622.414

    = 0.25 mole

    1 mole of NH3 (ammonia) = 6.022 1023 molecules Number of molecules in 0.25 mole of NH3 (ammonia) = 0.25 6.022 1023 = 1.5055 1023 molecules Ans: The number of moles and molecules in 5.6 dm3 of NH3 are 0.25 mole and 1.5055 1023 molecules,

    respectively. *Example 5 3.49 g of ammonia at STP occupies volume of 4.48 dm3. Calculate molar mass of ammonia. Solution: Let x grams be the molar mass of NH3. Volume occupied by 3.49 g of NH3 at S.T.P = 4.48 dm3 Volume occupied by x g of NH3 at S.T.P = 22.414 dm3 .[1 mole of any gas occupies 22.414 dm3 at S.T.P]

    x = 22.414 3.494.48 = 17.46 g mol1.

    Ans: Molar mass of ammonia is 17.46 g mol1.

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    *Example 6 Calculate the mass of potassium chlorate required to liberate 6.72 dm3 of oxygen at STP, molar mass of potassium chlorate is 112.5 g mol1. Solution: The molecular formula of potassium chlorate is KClO3. Required chemical equation: 2KClO3 2KCl + 3O2 [2 112.5 = 225 g] [3 22.414 = 67.242 dm3] Thus, 225 g of potassium chlorate will liberate 67.242 dm3 of oxygen gas. Let x gram of KClO3 liberate 6.72 dm3 at S.T.P.

    x = 225 6.7267.242 = 22.49 g

    Ans: Mass of potassium chlorate required is 22.49 g. *Example 7 Calculate the volume of oxygen required for complete combustion of 0.25 mole of methane at STP. Solution: Required chemical equation: CH4 + 2O2 CO2 + 2H2O [1 mole] [2 moles] Thus, 1 mole of CH4 requires 2 moles or 2 22.414 dm3 of O2 for complete combustion. 0.25 mole of CH4 requires x dm3 of O2 for complete combustion. x = 0.25 2 22.414 = 0.5 22.4 = 11.207 dm3 Ans: The volume of O2 required is 11.207 dm3. *Example 8 Calculate the volume of hydrogen required for complete hydrogenation of 0.25 dm3 of ethyne at STP. Solution: Required chemical equation: C2H2 + 2H2 C2H6 [1 mole] [2 moles] 1 mole of C2H2 occupies 22.414 dm3 2 moles of H2 occupies 44.828 dm3 Thus, 22.414 dm3 of C2H2 requires 44.828 dm3 of H2 for complete hydrogenation. Let 0.25 dm3 of C2H2 require x dm3 of H2 for complete hydrogenation.

    x = 0.25 44.82822.414 = 0.5 dm3

    Ans: The volume of hydrogen required is 0.5 dm3. *Example 9

    Calculate the number of atoms of hydrogen present in 5.6 g of urea (molar mass of urea = 60 g mol1). Also calculate the number of atoms of N, C and O. Solution: Given: Mass of urea = 5.6 g Molar mass of urea = 60 g mol1 To find: The number of atoms of hydrogen, nitrogen, carbon and oxygen Calculation: Molecular formula of urea: CO(NH2)2 1 molecule of urea has total 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of C and 1 of O. 1 mole or 60 g of urea contains 6.022 1023 molecules

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    60 g of urea has 4 6.022 1023 atoms of hydrogen.

    Number of H atoms in 5.6 g of urea = 235.6 4 6.022 10

    60 = 2.248 1023 atoms of hydrogen.

    Similarly, 60 g of urea has 2 6.022 1023 atoms of nitrogen

    Number of N atoms in 5.6 g of urea = 235.6 2 6.022 10

    60 = 1.124 1023 atoms of nitrogen.

    Similarly, 60 g of urea has 1 6.022 1023 atoms of carbon

    Number of C atoms in 5.6 g of urea = 235.6 1 6.022 10

    60 = 0.562 1023 atoms of carbon.

    Similarly, 60 g of urea has 1 6.022 1023 atoms of oxygen

    Number of O atoms in 5.6 g of urea = 235.6 1 6.022 10

    60 = 0.562 1023 atoms of oxygen.

    Ans: 5.6 g of urea contains 2.248 1023 atoms of H, 1.124 1023 atoms of N, 0.562 1023 atoms of C and 0.562 1023 atoms of O. *Example 10 Calculate the number of atoms of C, H and O in 72.5 g of isopropanol, C3H7OH (molar mass 60) Solution: Given: Mass of isopropanol (C3H7OH) = 72.5 g Molar mass of isopropanol = 60 g Total number of atoms = 12, out of which, there are 3 carbon atoms, 8 hydrogen atoms and 1

    oxygen atom. To find: The number of atoms of C, H, and O Calculation: In 60 g of C3H7OH, there are 3 6.022 1023 atoms of carbon. In 72.5 g of C3H7OH, let there be x atoms of carbon

    x = 2372.5 3 6.022 10

    60

    = 21.829 1023 = 2.183 1024 atoms of carbon. Similarly, In 60 g of C3H7OH, there are 8 6.022 1023 atoms of hydrogen. In 72.5 g of C3H7OH, let there be y atoms of hydrogen.

    y = 2372.5 8 6.022 10

    60 = 58.21 1023 atoms of hydrogen

    = 5.821 1024 atoms of hydrogen. Similarly, In 60 g of C3H7OH, there are 1 6.022 1023 atoms of oxygen In 72.5 g of C3H7OH, let there be z atoms of oxygen.

    z = 2372.5 1 6.022 10

    60 = 7.277 1023 atoms of oxygen.

    Ans: 72.5 g of isopropanol contains 2.183 1024 atoms of C, 5.821 1024 atoms of H and 7.277 1023 atoms of O.

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    Example 11 Calculate the number of moles and the volume in litres of the following gases at STP: i. 1.6 g of oxygen ii. 3. 5 103 kg of nitrogen iii. 85 103 kg of hydrogen sulphide Solution: i. For Oxygen: One mole of O2 = 2 16 = 32 g (molar mass) = 22.414 dm3 at STP

    Number of moles of O2 = weight(w)

    molar mass (M) = 1.6

    32

    = 0.05 Volume of oxygen (at STP) = moles 22.414 L = 0.05 22.414 L = 1.12 L or dm3 ii. For Nitrogen: One mole of N2 = 2 14 = 28 g = 28103 kg (molar mass) = 22.414 dm3 at STP

    Number of moles of N2 = wM

    = 3

    3

    3.5 1028 10

    = 0.125 Volume of N2 (at STP) = moles 22.414 L = 0.125 22.414 = 2.8 L or dm3 iii. For Hydrogen sulphide: One mole of H2S = (2 1) + (1 32) = 34 g i.e., 34 103 kg (molar mass) = 22.414 dm3 at STP

    Number of moles of H2S = wM

    = 3

    3

    85 1034 10

    = 2.5 Volume of H2S (at STP) = moles 22.414 L = 2.5 22.414 = 56 L or dm3 Ans: i. Number of moles and volume of oxygen in 1.6 g of oxygen are 0.05 and 1.12 L respectively. ii. Number of moles and volume of nitrogen in 3.5 103 kg of nitrogen are 0.125 and 2.8 L

    respectively. iii. Number of moles and volume of hydrogen sulphide in 85 103 kg of hydrogen sulphide are 2.5 and

    56 L respectively. Example 12 Calculate the number of moles and number of molecules of 12 g of carbon, 64 g of oxygen and 72 g of water. (Atomic masses: C = 12, O = 16, H = 1) Solution: Given: Atomic masses of C = 12, O = 16 and H = 1. To find: The number of moles and number of molecules of 12 g of carbon, 64 of oxygen and 72 g of water. Formula:

    1. Number of moles (n) = Mass of the substance(w)Atomicmass or molecular massof substance(M)

    2. Number of molecules = n Avogadro number

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    Calculation: a. For Carbon:

    i. Number of moles = wM

    = 1212

    ( Atomic mass of C = 12)

    = 1 ii. Number of molecules = 1 6.022 1023 = 6.022 1023 b. For Oxygen:

    i. Number of moles = 6432

    ( Molecular mass of O2 = 2 16 = 32)

    = 2 ii. Number of molecules = 2 6.022 1023

    = 12.044 1023 = 1.2044 1024 c. For Water:

    i. Number of moles = 7218

    ( Molecular mass of H2O = (2 1) + (1 16) = 18)

    = 4 ii. Number of molecules = 4 6.022 1023

    = 24.088 1023= 2.4088 1024 Ans: i. Number of moles and number of molecules in 12 g of carbon are 1 and 6.022 1023 respectively. ii. Number of moles and number of molecules in 64 g of oxygen are 2 and 1.2044 1024 respectively. iii. Number of moles and number of molecules in 72 g of water are 4 and 2.4088 1024 respectively. Example 13 Calculate the mass of the following i. 0.25 mole of iron ii. 2.5 moles of ammonia iii. 250 molecules of sodium chloride iv. 1.2 moles of methane Solution: i. 0.25 mole of iron (Fe): Atomic mass of iron = 56 1 mole of Fe 56 g of Fe 56 103 kg of Fe 0.25 mole of Fe 56 103 0.25 14 103 kg of Fe Mass of 0.25 mole of Fe = 1.4 102 kg. ii. 2.5 moles of ammonia (NH3): Molecular mass of ammonia = (1 14) + (3 1) = 17 (N) (3H) 1 mole of ammonia 17 g of ammonia 17.0 103 kg of ammonia 2.5 moles of ammonia 17.0 103 2.5 42.5 103 kg of ammonia Mass of 2.5 moles of ammonia = 4.25 102 kg. iii. 250 molecules of sodium chloride (NaCl): Molecular mass of NaCl = 23 + 35.5 = 58.5 (Na) (Cl) 1 mole of NaCl 58.5 g of NaCl 58.5 103 kg of NaCl 6.022 1023 molecules of NaCl 58.5 103 kg of NaCl

    250 molecules of NaCl 3

    23

    58.5 10 2506.022 10

    2.429 1023 kg

    Mass of 250 molecules of NaCl = 2.429 1023 kg.

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    iv. 1.2 moles of methane (CH4): Molecular mass of CH4 = 12 + (41) = 16 (C) (4H) 1 mole of methane = 16 g of methane = 16.0 103 kg of methane 1.2 moles of methane = 16.0 103 1.2 = 19.2 103kg. Mass of 1.2 moles of methane = 1.92 102 kg. Ans: i. Mass of 0.25 moles of iron is 1.4 102 kg. ii. Mass of 2.5 moles of ammonia is 4.25 102 kg. iii. Mass of 250 molecules of sodium chloride is 2.429 1023 kg. iv. Mass of 1.2 moles of methane is 1.92 102 kg. Example 14 What will be the mass of one 12C atom in g? Solution: 1 mole of 12C atom 6.022 1023 atom 12 g Mass of 6.022 1023 atoms of 12C = 12 g

    Mass of 1 atom of 12C = 2312

    6.022 10 g = 1.9927 1023 g

    Ans: Mass of one 12C atom is 1.9927 1023 g. Type III : Problems based on percentage composition, molecular formula and empirical formula *Example 15 Phosphoric acid is widely used in carbonated beverages, detergents, toothpastes and fertilizers. Calculate the mass percentages of H, P and O in phosphoric acid if atomic masses are H = 1, P = 31 and O = 16. Solution: Given: Atomic mass of H = 1, P = 31 and O = 16 To find: The mass percentage of hydrogen, phosphorous, oxygen in H3PO4

    Formula: %(by weight) = Massof element in onemoleof compound 100Gram molecular weight of compound

    Calculation: Molecular formula of phosphoric acid: H3PO4 Molar mass of H3PO4 = 3 (1) + 1 (31) + 4 (16) = 3 + 31 + 64 = 98 g mol1

    Percentage of Hydrogen = 398

    100

    = 3.06 %

    Percentage of phosphorus = 3198

    100

    = 31.63 %

    Percentage of oxygen = 6498

    100

    = 65.31 % Ans: Mass percentage of hydrogen, phosphorous and oxygen in phosphoric acid are 3.06%, 31.63% and

    65.31% respectively.

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    *Example 16 Calculate the mass percentage composition of the elements in nitric acid (H = 1, N = 14, O = 16). Solution: Given: Atomic mass of H = 1, N = 14 and O = 16 To find: The mass percentage of H, N and O in HNO3

    Formula: %(by weight) = Massof element in onemoleof compound 100Gram molecular weight of compound

    Calculation: Molecular formula of nitric acid : HNO3 Molar mass = 1 (1) + 1 (14) + 3 (16) = 1 + 14 + 48 = 63 g mol1

    Percentage of hydrogen = 163

    100

    = 1.59 %

    Percentage of nitrogen = 1463

    100

    = 22.22%

    Percentage of oxygen = 4863

    100

    = 76.19% Ans: Mass percentage of hydrogen, nitrogen and oxygen in nitric acid are 1.59%, 22.22% and 76.19%

    respectively. Example 17 Calculate the mass percentage of different elements present in sodium sulphate (Na2SO4). (NCERT) Solution: Given: Molecular formula of sodium sulphate = Na2SO4 To find: The mass percentage of Na, S and O in sodium sulphate

    Formula: %(by weight) = Massof element in onemoleof compound 100Gram molecular weight of compound

    Calculation:Atomic mass of Na = 23, S = 32, O = 16 Molar mass of Na2SO4 = 2 (23) + 1 (32) + 4 (16) = 46 + 32 + 64 = 142 g mol1

    Percentage of sodium = 46 100142

    = 32.39 %

    Percentage of sulphur = 32 100142

    = 22. 54 %

    Percentage of oxygen = 64 100142

    = 45.07 %

    Ans: Mass percentage of sodium, sulphur and oxygen in sodium sulphate are 32.39, 22.54 and 45.07 respectively.

    *Example 18 Analysis of vitamin C shows that it contains 40.92% carbon by mass, 4.58% hydrogen and 54.50% oxygen. Determine the empirical formula of vitamin C. Solution: Given: Analysis of vitamin C shows, Percentage mass of carbon = 40.92% Percentage mass of hydrogen = 4.58% Percentage mass of oxygen = 54.50%

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    To find: The empirical formula of vitamin C Calculation:

    Moles of carbon = % of carbonAtomic massof carbon

    = 40.9212

    = 3.41

    Moles of hydrogen = % of hydrogenAtomic massof hydrogen

    = 4.581

    = 4.58

    Moles of oxygen = % of oxygenAtomic massof oxygen

    = 54.5016

    = 3.406 3.41

    Ratio of number of moles of C:H:O = 3.413.41

    = 1 : 4.583.41

    = 1.34 : 3.413.41

    = 1

    Ratio = 1 : 1.34 : 1 Multiply by 3 to get whole number Ratio = 3 : 4.02 : 3 3 : 4 : 3 The empirical formula of compound Vitamin C is C3H4O3. Ans: Empirical formula of Vitamin C is C3H4O3. Example 19 Determine the empirical formula of an oxide of iron which contains 69.9% iron and 30.1% oxygen by mass.

    (NCERT) Solution: Given: Percentage mass of iron = 69.9 % Percentage mass of oxygen = 30.1 % To find: The empirical formula of an oxide of iron Calculation:

    Moles of iron = %of iron 69.9 1.25Atomicmassof iron 55.85

    = =

    Moles of oxygen = %of oxygen 30.1 1.88Atomicmassof oxygen 16

    = =

    Ratio of number of moles of Fe : O = 1.25 1.881: 1.501.25 1.25

    = =

    Ratio = 1 : 1.50 Multiply by 2 to get whole number Ratio = 2 : 3 The empirical formula is Fe2O3 Ans: The empirical formula of an oxide of iron is Fe2O3. Example 20 Calculate the percentage of water of crystallization in the sample of blue vitriol (CuSO4.5H2O) Solution: To find: The percentage of water of crystallization in the sample of blue vitriol Formula:

    % of H2O = Weight of water in onemoleof vitriol

    Molar massof vitriol

    Calculation: Molecular mass of CuSO4.5H2O = 63.5 + 32 + (4 16) + (5 18) = 249.5 For formula,

    % of H2O = 90

    249.5 100 = 36.07%

    Ans: The percentage of water of crystallization in blue vitriol is 36.07%.

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    #Example 21 An inorganic compound contained 24.75% (w/w) potassium and 34.75% (w/w) manganese and some other common elements. Give the empirical formula of the compound (K = 39 u, Mn = 59 u, O = 16 u) Solution: Given: Atomic mass of K = 39 u, Mn = 59 u, and O = 16 u. Percentage of potassium and manganese = 24.75 % and 34.75 % respectively. To find: The empirical formula of the given inorganic compound Calculation: Percentage of potassium = 24.75 % Percentage of manganese = 34.75 % Total percentage = 59.50 % Remaining must be that of oxygen Percentage of oxygen = 100 59.50 = 40.50 %

    Moles of potassium = % of potassium Atomic massof potassium

    = 24.7539

    = 0.635

    Moles of manganese = % of manganese Atomicmassof manganese

    = 34.7559

    = 0.589

    Moles of oxygen = % of oxygen Atomic massof oxygen

    = 40.5016

    = 2.53

    Ratio of K : Mn : O = 0.6350.589

    : 0.5890.589

    : 2.530.589

    = 1.08 : 1 : 4.29 1 : 1 : 4

    Ans: The empirical formula of given inorganic compound is KMnO4. #Example 22 Phosphoric acid used in carbonated beverages contain 3.086% (w/w) hydrogen and 31.61% (w/w) phosphorous and remaining oxygen. If the atomic masses of hydrogen, phosphorous and oxygen are 1.01 u, 31.0 u and 16 u respectively and if the molar mass of phosphoric acid is 98.03 g mol1, what is the molecular formula of phosphoric acid? Solution: Given: Atomic mass of H = 1.01 u, P = 31.0 u and O = 16 u The molar mass of phosphoric acid = 98.03 g mol1 Percentage of hydrogen and phosphorus = 3.086 % and 31.61 % respectively. To find: The molecular formula of phosphoric acid Calculation: Percentage of hydrogen = 3.086 % Percentage of phosphorous = 31.61 % Total percentage = 34.696 % Remaining amount is oxygen Percentage of oxygen = 100 34.696 = 65.304 %

    Moles of hydrogen = % mass of hydrogen Atomic massof hydrogen

    = 3.0861.01

    = 3.05

    Moles of phosphorous = % massof phosphorous Atomic massof phosphorous

    = 31.6131

    = 1.019

    Moles of Oxygen = % massof oxygen Atomic massof oxygen

    = 65.30416

    = 4.08

    Ratio of H: P: O = 3.051.019

    : 1.0191.019

    : 4.081.019

    = 3 : 1 : 4

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    Hence, the empirical formula of phosphoric acid = H3PO4 Empirical formula mass = 3 (1.01) + 1 (31) + 4(16) = 3.03 + 31 + 64 = 98.03 g mol1

    The ratio r = Molar mass Empiricalformula mass

    = 98.0398.03

    = 1

    Molecular formula = r Empirical formula = 1 H3PO4 = H3PO4 Ans: The molecular formula of phosphoric acid is H3PO4. Type IV : Problems based on stoichiometry Example 23 Calculate the mass of iron which will be converted into oxide (Fe3O4) by the action of 18 g of steam on it. Solution: The chemical equation representing the reaction is: 3Fe + 4H2O Fe3O4 + 4H2

    [3 56] [4 18] [168 g] [72 g] Thus, 72 g of steam reacts with 168 g of iron.

    18 g of steam will react with 16872

    18 = 42 g of iron

    Ans: Mass of iron which will be converted into oxide by action of 18 g of steam is 42 g. Example 24 How much copper can be obtained from 100 g of copper sulphate (CuSO4)? (NCERT) Solution: 1 mole of copper can be obtained from 1 mole of copper sulphate Atomic mass of copper = 63.5 u Molar mass of copper sulphate = 1 (63.5) + 1 (32) + 4 (16) = 63.5 + 32 + 64 = 159.5 g mol1 63.5 g of copper can be obtained from 159.5 g of copper sulphate

    Amount of copper that can be obtained from 100 g of copper sulphate is 63.5 100159.5

    = 39.81 g

    Ans: 39.81 g of copper can be obtained from 100 g of copper sulphate. Example 25 1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO4 was obtained as dry precipitate. Calculate the percentage purity of the sample. Solution: 1.5 g of impure Na2SO4

    Treatedwith BaCl2 1.74 g of BaSO4 The chemical equation representing the reaction is: Na2SO4 + BaCl2 BaSO4 + 2NaCl [(2 23) + 32 + (4 16)] [137 + 32 + (4 16)] [46 + 32 + 64] [137 + 32 + 64] [142 g] [233 g]

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    To calculate the mass of Na2SO4 from which 1.74 g of BaSO4 is obtained: 233 g of BaSO4 is produced from 142 g of Na2SO4.

    Mass of Na2SO4 from which 1.74 g of BaSO4 would be obtained = 142233

    1.74 = 1.06 g

    The mass of pure Na2SO4 present in 1.5 g of impure sample = 1.06 g To calculate the percentage purity of the impure sample: 1.5 g of impure sample contains 1.06 g of pure Na2SO4

    100 g of the impure sample will contain 1.061.5

    100 = 70.67 g of pure Na2SO4

    Ans: Percentage purity of the sample is 70.67 %. Example 26 Calculate the amount of lime Ca(OH)2 , required to remove hardness of 50,000 L of well water which has been found to contain 1.62 g of calcium bicarbonate per 10 L . Solution: Calculation of total Ca(HCO3)2 present: 10 L of water contains 1.62 g of Ca(HCO3)2

    50,000 L of water will contain 1.6210

    50,000 = 8100 g of Ca(HCO3)2

    Calculation of lime required: The balanced equation for the reaction: Ca(HCO3)2 + Ca(OH)2 2CaCO3 + 2 H2O [1 mole] [1 mole] [40 + (1 + 12 + 48 ) 2] [40 + (16 + 1) 2] [40 + 122] [40 + 34] [162] [74] 162 g of Ca(HCO3)2 requires 74 g of lime.

    Mass of lime required by 8100 g of Ca(HCO3)2 = 74

    162 8100 g = 3700 g = 3.7 kg

    Ans: The amount of lime required to remove hardness of 50,000 L of well water, with 1.62 g of calcium bicarbonate per 10 L is 3.7 kg.

    Type V : Problems based on limiting and excess reactants Example 27 Gold reacts with chlorine at 150 C as per balanced chemical reaction, 2Au + 3Cl2 2AuCl3. 10 g each of gold and chlorine are sealed in a container and heated at 150 C till the reaction is complete. Name the limiting and excess reactants. Also calculate the amount of AuCl3 formed and the mass of the excess reactants left behind. (Atomic masses: Au = 196.97 u, Cl = 35.45 u) Solution: The reaction of gold with chlorine to produce AuCl3 is 2Au + 3Cl2 2AuCl3. 2 moles of gold reacts with 3 moles of Cl2 to given 2 moles of AuCl3 will be obtained. 2Au + 3Cl2 2AuCl3 [2 196.97 g] + [3 35.45 2] [2(196.97 + 3 35.45)] [393.94 g] + [212.7 g] [606.64] [606.64 g] [606.64 g]

    Theoretical ratio of moles of gold to chlorine = 23

    = 0.667

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    Actual amount is 10 g of Au and 10 g Cl2

    Number of moles of Au in 10 g = 10196.97

    = 0.0508

    Number of moles of Cl2 in its 10 g =10

    70.90= 0.141 mol of Cl2

    Actual ratio of moles of gold to chlorine = 0.05080.141

    = 0.36

    Theoretical ratio of moles of Au to Cl2 is 0.667 and actual ratio of moles of Au to Cl2 is 0.36 Thus actual ratio is smaller than theoretical ratio Au is the limiting reactant, and Cl2 is the excess reactant. 2 moles of Au react with 3 moles of Cl2

    0.0508 moles of Au will react with 0.0508 32 = 0.0762 moles of Cl2

    Moles of Cl2 remaining = 0.141 0.0762 = 0.0648 moles of Cl2 Mass of Cl2 reacted = 0.0762 70.90 = 5.40 g (70.90 g is molar mass of Cl2) Mass of Cl2 left behind = 10 5.40 = 4.6 g 2 moles of Au give 2 moles of AuCl3 0.0508 mole of Au will give = 0.0508 mole of AuCl3 Mass of AuCl3 formed = 0.0508 303.32 (Molar mass of AuCl3 = 303.32 g) = 15.409 g AuCl3 Ans: Au is the limiting reactant, while Cl2 is the excess reactant. The amount of AuCl3 formed is 15.409 g and

    the mass of the excess reactant (Chlorine) left behind is 4.6 g. Additional Theory Questions

    Q.1. Define: i. physical chemistry. Refer Q.4.i. ii. inorganic chemistry. Refer Q.4.ii. iii. organic chemistry. Refer Q.4.iii. iv. analytical chemistry. Refer Q.4.iv. v. biochemistry. Refer Q.4.v. Q.2. Explain the following: i. Pure substance. Refer Q.8.i. ii. Mixtures. Refer Q.8.ii. Q.3. Define the following with suitable examples: i. Homogeneous mixtures. Refer Q.8.ii.a. ii. Heterogeneous mixtures. Refer Q.8.ii.b. *Q.4. State and explain the law of conservation of mass. Refer Q.22.i. Q.5. State and explain the law of definite composition / proportions. Refer Q.22.ii. Q.6. State and explain the law of multiple proportions. Ref