Std. 11th Chemistry, Maharashtra Board, English Medium - edoqs

Std. XI Sci.

Perfect Chemistry - I

Edition: June 2014

Prof. Santosh B. Yadav

Prof. Anil Thomas

(M. Sc., SET, NET)

(M.Sc., Chemistry)

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

Std. XI Sci.

Perfect Chemistry - I

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.

TEID : 734

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 we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you. Please write to us on : [email protected] A book affects eternity; one can never tell where its influence stops.

Yours faithfully

Best of luck to all the aspirants!

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

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01 1.0

Chapter 01: Some Basic Concepts of Chemistry

Some Basic concepts of chemistry

Prominent scientists Scientists Joseph Louis Gay-Lussac (1778 – 1850) (French chemist and physicist)

Amedeo Avogadro (1776 – 1856) (Italian scholar)

Contributions Formulated the gas law. Collected samples of air at different heights and recorded temperatures and moisture contents. iii. Discovered that the composition of atmosphere does not change with increasing altitude. i. Published article in French journal on determining the relative masses of elementary particles of bodies and proportions by which they enter combinations. ii. Published a research paper titled “New considerations on the theory of proportions and on determination of the masses of atoms.” i. ii.

Note: In order to give a tribute to Avogadro’s 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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Std. XI Sci.: Perfect Chemistry

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 i ii iii iv v 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. sunlight CO2 + H2O ⎯⎯⎯⎯→ food grains/fruits/flowers/cotton/medicine etc. chlorophyll

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. 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.

v.

2

L-dopa Human insulin Cisplatin and taxol Azidothymidine (AZT) Tamiflue

Treatment of diseases For treatment of Parkinson’s disease. For treatment of diabetes. Life saving drugs to give relief to cancer patients. For treatment of AIDS. For treatment of swine flu.

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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 one’s 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 super–conducting 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 Priestley’s 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.

ii.

4

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. 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 Aluminium Argon Silver Gold Actinium Americium Beryllium Boron Barium Bismuth Carbon Chlorine Calcium Chromium Cobalt Copper

Symbol Al Ar Ag Au Ac Am Be B Ba Bi C Cl Ca Cr Co Cu

Element Cadmium Caesium Cerium Curium Calefornium Erbium Einsteinium Fluorine Francium Iron Galium Germanium Gadolinium Hydrogen Helium Hafnium

Symbol Cd Cs Ce Cm Cf Er Es F Fr Fe Ga Ge Gd H He Hf

Element Mercury Holmium Iodine Irridium Krypton Lithium Lanthanum Lutetium Lawrencium Magnesium Manganese Molybdenum Mendelivium Nitrogen Neon Nickel

Symbol Hg Ho I Ir Kr Li La Lu Lr Mg Mn Mo Md N Ne Ni

Element Neobium Neodymium Neptunium Oxygen Osmium Potassium Phosphorous Lead Palladium Platinum Promethium Protoactinium Plutonium Radium Rubidium Ruthenium

Symbol Nb Nd Np O Os K P Pb Pd Pt Pm Pa Pu Ra Rb Rn

Element Rhodium Rhenium Sulphur Scandium Selenium Strontium Sodium Technicium Uranium Tungston Vanadium Xenon Ytterbium Zirconium

Symbol Rh Re S Sc Se Sr Na Tc U W V Xe Yb Zr

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, Phenolwater 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 i. The constituents of a mixture may be present in any ratio. ii. Mixtures may or may not be homogeneous in nature. iii. The properties of a mixture are in between those of its constituents. iv. The constituents of a mixture can be easily separated by simple physical means. v. The melting and boiling points of mixtures are usually not sharp.

Compounds The constituents of a compound are always present in a fixed ratio. Compounds are always homogeneous in nature. The properties of a compound are entirely different from those of its constituent elements. The constituents of a compound cannot be easily separated by simple physical means but can be separated by chemical processes. Chemical compounds possess sharp melting and boiling points.

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Compounds and elements: Compounds A compound is a substance composed of two or more different chemical elements. ii. A compound can be separated into simpler substances by chemical reactions. iii. Compounds contain different elements in a fixed ratio arranged in a defined manner through chemical bonds. iv. A compound is represented using a formula. v. The list of compounds is endless but can broadly be classified as ionic and covalent. i.

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

Elements An element is a pure chemical substance made of same type of atoms. Elements cannot be broken down into simpler substances. Elements are distinguished by their atomic number (number of protons in their nucleus).

An element is represented using symbols. There are about 118 elements that have been identified and can be classified as metals, nonmetals or metalloids. Sodium 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. ii.

During calculations, confinement to one single system of unit is advisable. NASA’s 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. i. ii. iii. iv. v. vi. vii. 6

Fundamental quantity Length Mass Time Temperature Amount of substance Electric current Luminous intensity

SI unit Metre Kilogram Second Kelvin Mole Ampere Candela

Symbol m kg s K mol A cd

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Chapter 01: Some basic concepts of chemistry

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 m−3) Note: The table given below shows some common derived units. No. i. ii. iii. iv. v. vi. vii. viii. ix.

Physical quantity Area Volume Density Velocity Acceleration Force Pressure Electric charge Electric potential or Potential difference

Relationship with fundamental unit Length squared Length cubed Mass per unit volume Distance travelled in unit time Velocity change per unit time Mass × Acceleration Force per unit area Current × Time Energy per unit charge

x.

Energy (work or heat) Concentration Heat capacity

Force × distance travelled

xi. xii.

xiii. Electrochemical equivalent

Mole per cubic metre Cp = dH / dT Cv = dE / dT Z = E/F

Unit m2 m3 kg m−3 ms−1 ms −2 kg m s−2 (newton, N) kg m−1 s−2 As (coulomb, C) kg m2 s−2 A−1 (J A−1 s−1 or Volt,V or JC−1) kg m2 s−2 (Joule s–1) mol m−3 JK−1 mol−1 kg C−1 (kg/Coulomb)

Some common SI prefixes used for expressing big and small numbers: Prefix Tera−

Symbol T

Magnitude 1012

Meaning (multiply by) 1 000 000 000 000

Giga−

G

109

1 000 000 000

M

10

6

1 000 000

10

4

1 000 0 (this is now obsolete)

10

3

1 000

2

100

Mega− myria− kilo−

my k

hecto−

h

10

deka−

da

10

−

−

deci−

− d

centi−

c

10−2

milli− micro− nano− pico− femto−

m µ n p f

10

10 −1

− 0.1 0.01

10

−3

0.001

10

−6

0.000 001

10

−9

0.000 000 001

10

−12

0.000 000 000 001

10

−15

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:

Naturally occurring cupric carbonate

51.35

Percentage C 38.91

Cupric carbonate prepared in the laboratory

51.35

38.91

Cu

d.

iii.

8

O 9.74 9.74

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.

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 non−poisonous, non−combustible 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 2.66g of oxygen = 2.66 1.00g of carbon

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And the ratio of weights of oxygen to that of carbon for carbon monoxide is 1.33g of oxygen = 1.33 1.00g carbon The two ratios are in the proportion

iv.

2.66 1.33

= 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. Gay Lussac’s 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 Lussac’s 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 [1L] [1L] [2L] [ 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 ] [ 1 2 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 Proust’s 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, Berthollet’s 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 oxygen = 8 = 8:1 11.11% by weight of hydrogen 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 oxygen = 16 = 16:1 5.88% by weight of hydrogen 16 :1 = 2:1 8 :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.

The ratios of the two compounds are in the proportion

Q.26. Give two examples which support the Gay Lussac’s 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) 2L 1L 2L (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 3L 2L (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 Lussac’s law. Note: i. Gay Lussac’s 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

Dalton’s atomic theory

*Q.27. State and explain Dalton’s 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. 10

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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 Dalton’s 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:

Oxygen

Sulphur

Carbon

Potassium

Methane

Hydrogen peroxide

Nitrogen

Hydrogen

Water

Carbon monoxide

Carbon dioxide

Dalton’s symbols for some elements and compounds 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 10–27 kg and has a spherical shape of radius of the order of 10–15 m. d. The smallest atom of the element is that of hydrogen with mass of 1.667 × 10–27 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.

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1.7


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).

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 10−15 m. The mass is of the order of 10−27 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 10−15 m and the mass is of the order of 10−27 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 × 10–24 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. 12

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eg. Chlorine has two isotopes, Cl−35 and Cl–37, 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.

Intensity (percentage)

20 10

Ne (90.92%)

21 10

19

Ne (0.26%)

22 10

Ne (8.82%)

21 20 22 Atomic mass (amu)

23

ii.

By using Aston’s 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 mass of Ne-20 × % of Ne-20) ⎤ ⎢ ⎥ ⎢+ Atomic mass of Ne-22 ×% of Ne-22⎥ ⎥ = ⎢ ⎢+ Atomic mass of Ne-21× % of Ne-21⎥ 100 ⎣ ⎦ = [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”. th Ans: Molecular mass of a substance is defined as the ratio of mass of one molecule of a substance to 1 of the

12

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. th

ii. iii. iv.

It indicates comparative mass of a molecule of a compound with respect to

1 of the mass of one 12

atom of Carbon-12. 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. 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.

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1.8


Avogadro’s law

*Q.37.State and explain Avogadro’s law. In the year 1811, Avogadro combined Gay Lussac’s law and Dalton’s theory and thereby proposed Ans: i. Avogadro’s law. ii. Avogadro’s law states that, “equal volumes of all gases, under identical conditions of temperature and pressure, contain equal number of molecules”. iii. Mathematically, Avogadro’s law is stated as, “at constant pressure and temperature, volume of a gas is directly proportional to the number of molecules”. iv. V α number of molecules (P, T constant) Since number of molecules is proportional to the number of moles of gas n, V α n (P, T constant) V mass of gas or = constant, where n = n molar mass of gas

v.

∴

vi.

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 V Volume per mol (molar volume) = n V RT = , (R = 0.08206 L atm mol–1 K–1) n P 0.08206 × 273.15 = = 22.414 L mol−1. 1atm 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 mol−1 is called Avogadro’s molar volume or molar gas volume at STP.

Q.38. Describe the significance of Avogadro’s number. Ans: The significance of Avogadro’s number is as follows: i. Avogadro’s 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 Avogadro’s number, i.e., 6.022×1023 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. 14

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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. iii.

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. The reaction may be written as Hydrogen

+

Chlorine

⎯→

Hydrogen chloride

According to Gay Lussac’s law:

[1 volume]

[1 volume]

[2 volumes]

According to Avogadro’s law: (V∝ n)

[n molecules]

[n molecules]

[2n molecules]

⎡1 ⎤ ⎢ molecule⎥ ⎢⎣ 2 ⎥⎦

⎡1 ⎤ ⎢ molecule⎥ ⎢⎣ 2 ⎥⎦

[1 molecule]

Dividing by 2n iv.

1 1 molecule of hydrogen and molecule of chlorine give 1 molecule of HCl. If 2 2 molecules of hydrogen and chlorine are considered to be monoatomic, it indicates that atoms are divisible and this is contrary to Dalton’s atomic theory which states that atom is indivisible. So, the molecule of hydrogen and chlorine are considered to be diatomic. This shows that

+

X molecules of hydrogen (1 Volume)

=

X 2X molecules of molecules of chlorine hydrogen chloride (1 Volume) (2 Volume)

Formation of HCl molecule *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. iii.

The amount of a substance equal to its atomic mass or molar mass in grams is 1 mole of a substance. 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 carbon–12.

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 mol−1 Note:

i. ii.

Number of electrons equal to NA make an electrical charge of one Faraday. One mole of various gases (of different molar masses) occupy 22.414 L at STP and contain avogadro’s number of molecules.

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6.022 × 1023 molecules of O2

6.022 × 1023 molecules of CO2

6.022 × 1023 molecules of CH4

6.022 × 1023 molecules of Ar

22.4 L 32.0 g

22.4 L 44.0 g

22.4 L 16.0 g

22.4 L 39.9 g

1 mole of different gases having different masses containing 6.022 × 1023 molecules and occupying volume of 22.414 L at STP 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.

Number of fundamental particles Multiplied by Avogadro’s number

Mass of substance

Divided by molecular weight (molar mass) Multiplied by molecular weight (molar mass)

Divided by Avogadro’s number

Number of moles

Multiplied by Volume 22.4 dm3 occupied by gas at STP in dm3 Divided by 3 22.4 dm

Mole triangle Note:

i.

Number of moles (n) =

W mass of a substance = M molar mass of thesubstance

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.

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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 thecompound × 100 % Gram molecular weight of thecompound

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:

24 × 100 = 52.17% 46 16 Percentage of oxygen = × 100 = 34.78% 46 6 × 100 = 13.04% Percentage of hydrogen = 46 Each 100 g sample of ethanol contains 52.17 g carbon, 34.78 g oxygen and 13.04 g hydrogen. 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. 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. Percentage of carbon =

∴ ii.

iii.

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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For fractional ratios, the whole number ratio is obtained by multiplying it with suitable integer. For this, minor fractions may be neglected. 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. 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 i. Water H2O ii. Carbon C iii. Methane CH4 iv. Hydrogen peroxide HO v. Acetylene CH vi. Glucose CH2O vii. Diborane BH3 viii. Tetraphosphorous decoxide P2O5

Molecular formula H2O C CH4 H2O2 C2H2 C6H12O6 B2H6 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:

⎯→

+

Two hydrogen molecules + One oxygen molecule ⎯→ Two water molecules 2H2

+

O2

⎯→

2H2O

Formation of water molecules 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). 18

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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) In the given reaction, H2(g) + O2(g) → H2O(g) conservation of mass in not observed. Ans: i. 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 i. Phosphoric acid ii. Sodium phosphate iii. Ferric phosphate iv. Aluminium phosphate v. Copper phosphate vi. Ferrous phosphate vii. Hydrogen chloride viii. Sodium chloride ix. Cuprous chloride x. Cupric chloride

Formula H3PO4 Na3PO4 FePO4 AlPO4 Cu3(PO4)2 Fe3(PO4)2 HCl NaCl Cu2Cl2 CuCl2

No. xi. xii. xiii. xiv. xv. xvi. xvii. xviii. xix.

Compound Ferrous chloride Ferric chloride Stannous chloride Stannic chloride Sulphuric acid Sodium sulphate Copper sulphate Ferrous sulphate Ferric sulphate

Formula FeCl2 FeCl3 SnCl2 SnCl4 H2SO4 Na2SO4 CuSO4 FeSO4 Fe2(SO4)3

Q.53. Show that “ Law of conservation of mass is fully justified in a balanced chemical equation”. A balanced chemical reaction can be used to establish the weight relationships of reactants and Ans: i. 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 + 3 × 35.45) g] [2 × 55.85 g] [3 × 35.45 × 2 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? The stoichiometric coefficients of reactants and products in the balanced chemical equation Ans: i. 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. The excess reactant is the reactant which is taken in excess compared to the limiting reactant. Hence, v. 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. 20

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Chapter 01: Some basic concepts of chemistry Quick Review

The classification of matter (On basis of chemical composition): Matter Anything which has mass and occupies space.

Pure substances Substances which always have a fixed composition.

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

Mixtures A simple combination of two or more substances in which the constituent substances retain their separate identities.

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

Different laws and the contributing scientists: No. Laws i. Law of conservation of mass

ii. iii. iv.

Law of definite composition Law of multiple proportions Law of combining volumes of gases

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

Contributing Scientists i. Lomonosove (Russian scientist) ii. Antoine Laviosier (French chemist) Joseph Proust (French chemist) John Dalton (British scientist) Joseph Louis Gay−Lussac (French chemist)

Important Formulae

1. 2. 3. 4. 5. 6. 7. 8.

Mass of element Atomic mass Atomic mass Mass of one atom = 6.022 ×1023 Molecular mass Mass of one molecule = 6.022 ×1023 Mass of substance Number of moles (n) = Molar mass of substance Number of molecules = n × Avogadro number Volume of gas at STP = n × 22.414 L. Mass of the element in1mole of the compound Percentage (by weight) = × 100 Gram molecular weight of the compound Molecular formula = r × empirical formula (where r is ratio of molecular mass to empirical mass).

One mole of atoms =

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Std. XI Sci.: Perfect Chemistry 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: Isotope 36

Ar Ar 40 Ar 38

Solution: To find: Formula:

Isotopic mass (g mol−1)

abundance

35.96755 37.96272 39.9624

0.337% 0.063% 99.600%

(NCERT)

Average atomic mass of naturally occuring argon Average atomic mass of argon ⎡ atomic mass of 36 Ar × percentage of 36 Ar ⎤ ⎢ ⎥ = ⎢ + atomic mass of 38 Ar × percentage of 38 Ar ⎥ 100 ⎢ + atomic mass of 40 Ar × percentage of 40 Ar ⎥ ⎣ ⎦

Calculation: Average atomic mass of argon

35.96755 × 0.337 + 37.96272 × 0.063 + 39.9624 × 99.60 100 −1 = 39.947 g mol . =

Ans: Average atomic mass of argon = 39.947 g mol−1. Example 2 Calculate the atomic mass (average) of chlorine using the following data: Atomic % Natural abundance mass 35 C1 75.77 34.9689 37 C1 24.23 36.9659 Solution: To find: Average atomic mass of chlorine Formula: Average atomic mass of chlorine ⎡ atomic mass of 35 Cl × percentage of 35 Cl ⎤ ⎢ ⎥ + atomic mass of 37 Cl × percentage of 37 Cl ⎦ ⎣ = 100 Calculation: Average atomic mass of chlorine 34.9689 × 75.77 + 36.9659 × 24.23 = 100 = 35.4528 g mol−1 Ans: Average atomic mass of chlorine = 35.4528 g mol−1.

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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 ⎡ atomic mass of 10 B × percentage ⎤ Formula: Average atomic mass = ⎢ ⎥ 100 11 ⎣⎢ + atomic mass of 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 =

x × 10 + (100 − x) × 11 = 10.8 100

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 5.6 x= = 0.25 mole ∴ 22.414 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]

22.414× 3.49 = 17.46 g mol−1. 4.48 Ans: Molar mass of ammonia is 17.46 g mol−1.

∴

x=

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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 mol−1. 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. 225× 6.72 x= = 22.49 g ∴ 67.242 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: + 2O2 ⎯→ CO2 + 2H2O CH4 [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: 2H2 ⎯→ C2H6 C2H2 + [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. 0.25 × 44.828 ∴ x= = 0.5 dm3 22.414 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 mol−1). 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 mol−1 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 24

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60 g of urea has 4 × 6.022 × 1023 atoms of hydrogen. 5.6 × 4 × 6.022 × 1023 = 2.248 × 1023 atoms of hydrogen. ∴ Number of ‘H’ atoms in 5.6 g of urea = 60 Similarly, 60 g of urea has 2 × 6.022 × 1023 atoms of nitrogen 5.6 × 2 × 6.022 × 1023 = 1.124 × 1023 atoms of nitrogen. ∴ Number of N atoms in 5.6 g of urea = 60 Similarly, 60 g of urea has 1 × 6.022 × 1023 atoms of carbon 5.6 × 1 × 6.022 × 1023 = 0.562 × 1023 atoms of carbon. ∴ Number of C atoms in 5.6 g of urea = 60 Similarly, 60 g of urea has 1 × 6.022 × 1023 atoms of oxygen 5.6 × 1 × 6.022 × 1023 = 0.562 × 1023 atoms of oxygen. ∴ Number of O atoms in 5.6 g of urea = 60 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: Mass of isopropanol (C3H7OH) = 72.5 g Given: 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 72.5 × 3 × 6.022 × 1023 x = ∴ 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. 72.5 × 8 × 6.022 × 1023 y= = 58.21 × 1023 atoms of hydrogen ∴ 60 = 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. 72.5 × 1 × 6.022 × 10 23 z= = 7.277 × 1023 atoms of oxygen. ∴ 60 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 × 10−3 kg of nitrogen iii. 85 × 10−3 kg of hydrogen sulphide Solution: i. For Oxygen: One mole of O2 = 2 × 16 = 32 g (molar mass) = 22.414 dm3 at STP weight(w) 1.6 ∴ Number of moles of O2 = = 32 molar mass (M) = 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 = 28×10−3 kg (molar mass) = 22.414 dm3 at STP w 3.5 × 10−3 = ∴ Number of moles of N2 = M 28 × 10−3 = 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 × 10−3 kg (molar mass) = 22.414 dm3 at STP 85×10−3 w = Number of moles of H2S = M 34 ×10−3 = 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 × 10–3 kg of nitrogen are 0.125 and 2.8 L respectively. iii. Number of moles and volume of hydrogen sulphide in 85 × 10–3 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: Mass of the substance(w) 1. Number of moles (n) = Atomic mass or molecular mass of substance(M) 2. Number of molecules = n × Avogadro number 26

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Calculation: a. For Carbon:

w 12 = (∵ Atomic mass of C = 12) M 12 =1 ii. Number of molecules = 1 × 6.022 × 1023 = 6.022 × 1023 For Oxygen: 64 i. Number of moles = (∵ Molecular mass of O2 = 2 × 16 = 32) 32 =2 ii. Number of molecules = 2 × 6.022 × 1023 = 12.044 × 1023 = 1.2044 × 1024 i.

b.

c.

For Water:

72 (∵ Molecular mass of H2O = (2 × 1) + (1 × 16) = 18) 18 =4 ii. Number of molecules = 4 × 6.022 × 1023 = 24.088 × 1023= 2.4088 × 1024 Number of moles and number of molecules in 12 g of carbon are 1 and 6.022 × 1023 respectively. Number of moles and number of molecules in 64 g of oxygen are 2 and 1.2044 × 1024 respectively. Number of moles and number of molecules in 72 g of water are 4 and 2.4088 × 1024 respectively. i.

Ans: i. ii. iii.

Number of moles =

Number of moles =

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 × 10−3 kg of Fe ∴ 0.25 mole of Fe ≡ 56 × 10−3 × 0.25 ≡ 14 × 10−3 kg of Fe Mass of 0.25 mole of Fe = 1.4 × 10−2 kg. ∴ ii.

∴ ∴ ∴ iii.

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 × 10−3 kg of ammonia 2.5 moles of ammonia ≡ 17.0 × 10−3 × 2.5 ≡ 42.5 × 10−3 kg of ammonia Mass of 2.5 moles of ammonia = 4.25 × 10−2 kg.

∴

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 × 10−3 kg of NaCl 6.022 × 1023 molecules of NaCl ≡ 58.5 × 10−3 kg of NaCl

∴

250 molecules of NaCl ≡

∴

Mass of 250 molecules of NaCl = 2.429 × 10−23 kg.

∴

58.5 × 10−3 × 250 ≡ 2.429 × 10−23 kg 6.022 × 1023

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iv.

1.2 moles of methane (CH4): Molecular mass of CH4 = 12 + (4×1) = 16 (C) (4H)

∴

1 mole of methane = 16 g of methane = 16.0 × 10−3 kg of methane

∴

1.2 moles of methane = 16.0 × 10−3 × 1.2 = 19.2 × 10−3kg.

∴

Mass of 1.2 moles of methane = 1.92 × 10–2 kg.

Ans: i. ii. iii. iv.

Mass of 0.25 moles of iron is 1.4 × 10–2 kg. Mass of 2.5 moles of ammonia is 4.25 × 10–2 kg. Mass of 250 molecules of sodium chloride is 2.429 × 10–23 kg. Mass of 1.2 moles of methane is 1.92 × 10–2 kg.

Example 14 What will be the mass of one 12C atom in g? Solution:

1 mole of

12

C atom ≡ 6.022 × 1023 atom ≡ 12 g

∴

Mass of 6.022 × 1023 atoms of

∴

Mass of 1 atom of

12

C =

12

C = 12 g

12 g = 1.9927 × 10−23 g 23 6.022 × 10

Ans: Mass of one 12C atom is 1.9927 × 10−23 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: Atomic mass of H = 1, P = 31 and O = 16 Given: To find: The mass percentage of hydrogen, phosphorous, oxygen in H3PO4 Mass of element in one mole of compound × 100 Formula: %(by weight) = Gram 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 mol−1 3 Percentage of Hydrogen = × 100 98 = 3.06 % 31 Percentage of phosphorus = × 100 98 = 31.63 % 64 Percentage of oxygen = × 100 98 = 65.31 % Ans: Mass percentage of hydrogen, phosphorous and oxygen in phosphoric acid are 3.06%, 31.63% and 65.31% respectively. 28

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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 Mass of element in one mole of compound × 100 Formula: %(by weight) = Gram molecular weight of compound Calculation: Molecular formula of nitric acid : HNO3 ∴ Molar mass = 1 × (1) + 1 × (14) + 3 × (16) = 1 + 14 + 48 = 63 g mol−1 1 ∴ Percentage of hydrogen = × 100 63 = 1.59 % 14 Percentage of nitrogen = × 100 63 = 22.22% 48 Percentage of oxygen = × 100 63 = 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 Mass of element in one mole of compound × 100 Formula: %(by weight) = Gram 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 mol−1 46 × 100 = 32.39 % ∴ Percentage of sodium = 142 32 × 100 = 22. 54 % ∴ Percentage of sulphur = 142 64 × 100 = 45.07 % ∴ Percentage of oxygen = 142 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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The empirical formula of vitamin C To find: Calculation: 40.92 % of carbon Moles of carbon = = = 3.41 12 Atomic mass of carbon 4.58 % of hydrogen Moles of hydrogen = = = 4.58 1 Atomic mass of hydrogen Moles of oxygen = ∴ ∴

54.50 % of oxygen = = 3.406 ≈ 3.41 16 Atomic mass of oxygen

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

3.41 4.58 3.41 =1: = 1.34 : =1 3.41 3.41 3.41

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: % of iron 69.9 Moles of iron = = = 1.25 Atomic mass of iron 55.85 % of oxygen 30.1 Moles of oxygen = = = 1.88 Atomic mass of oxygen 16 1.25 1.88 = 1: =1.50 ∴ Ratio of number of moles of Fe : O = 1.25 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: Weight of water in one mole of vitriol % of H2O = Molar mass of vitriol Calculation: Molecular mass of CuSO4.5H2O = 63.5 + 32 + (4 × 16) + (5 × 18) = 249.5 For formula, 90 % of H2O = × 100 = 36.07% 249.5 Ans: The percentage of water of crystallization in blue vitriol is 36.07%. 30

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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 % % of potassium Moles of potassium = Atomic mass of potassium

24.75 = 0.635 39 34.75 % of manganese = = 0.589 Moles of manganese = 59 Atomic mass of manganese 40.50 % of oxygen = = 2.53 Moles of oxygen = 16 Atomic mass of oxygen 0.635 0.589 2.53 : : = 1.08 : 1 : 4.29 ≈ 1 : 1 : 4 Ratio of K : Mn : O = 0.589 0.589 0.589 =

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 mol−1, 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 mol−1 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 % 3.086 % mass of hydrogen Moles of hydrogen = = = 3.05 1.01 Atomic mass of hydrogen 31.61 % mass of phosphorous = = 1.019 Moles of phosphorous = 31 Atomic mass of phosphorous 65.304 % mass of oxygen = = 4.08 Moles of Oxygen = 16 Atomic mass of oxygen

∴

Ratio of H: P: O =

3.05 1.019 4.08 : : =3:1:4 1.019 1.019 1.019

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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 mol−1 98.03 Molar mass The ratio r = = =1 98.03 Empiricalformula mass

= r × Empirical formula = 1 × H3PO4 = H3PO4 Ans: The molecular formula of phosphoric acid is H3PO4. Molecular formula

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. 168 × 18 = 42 g of iron ∴ 18 g of steam will react with 72 Ans: Mass of iron which will be converted into oxide by action of 18 g of steam is 42 g. Example 24 (NCERT) How much copper can be obtained from 100 g of copper sulphate (CuSO4)? 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 mol−1 ∴ 63.5 g of copper can be obtained from 159.5 g of copper sulphate 63.5 × 100 = 39.81 g ∴ Amount of copper that can be obtained from 100 g of copper sulphate is 159.5 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: Treated with BaCl2 → 1.74 g of BaSO4 1.5 g of impure Na2SO4 ⎯⎯⎯⎯⎯⎯

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] 32

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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. 142 ∴ Mass of Na2SO4 from which 1.74 g of BaSO4 would be obtained = × 1.74 = 1.06 g 233 ∴ 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 1.06 ∴ 100 g of the impure sample will contain × 100 = 70.67 g of pure Na2SO4 1.5 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 1.62 ∴ 50,000 L of water will contain × 50,000 = 8100 g of Ca(HCO3)2 10 Calculation of lime required: The balanced equation for the reaction: + Ca(OH)2 ⎯→ 2CaCO3 + 2 H2O Ca(HCO3)2 [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. 74 Mass of lime required by 8100 g of Ca(HCO3)2 = × 8100 g = 3700 g = 3.7 kg 162 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] ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 2 Theoretical ratio of moles of gold to chlorine = = 0.667 3

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Actual amount is 10 g of Au and 10 g Cl2 10 Number of moles of Au in 10 g = = 0.0508 196.97 10 = 0.141 mol of Cl2 Number of moles of Cl2 in its 10 g = 70.90 0.0508 = 0.36 Actual ratio of moles of gold to chlorine = 0.141 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 × 3 ∴ 0.0508 moles of Au will react with = 0.0762 moles of Cl2 2 ∴ 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. Refer Q.22.iii. *Q.7. State and explain Gay Lussac’s law of combining volumes of gases. Refer Q.22.iv. *Q.8. Define elements, compounds and molecules. Refer Q.13.i.a, Q.13.ii.a. and Q.30.ii.a. Q.9. Show that molar gas volume at STP is 22.414 L or 0.022414 m3. OR Prove that 1 mole of any gas at STP always has a volume of 22.414 L. Refer Q.37.(iii to vi) 34

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get Publications Pvt. Ltd. Practice Problems


13.

Type I

1.

The relative abundance of various isotopes of silicon is: Si (28) = 92.25%, Si (29) = 4.65% and Si (30) = 3.10%. Calculate the average atomic mass of silicon.

Type II

2.

How many molecules and atoms of sulphur are present in 0.1 mole of S8 molecules?

3.

Calculate the number of moles of iodine in a sample containing 1.0 × 1022 molecules.

4.

Calculate moles of Ammonia in 0.1284 kg of Ammonia.

5.

O2 is present in one litre flask at a pressure of 7.6 × 10–10 mm of Hg. Calculate no. of O2 molecules at 0 °C.

6.

Calculate the number of moles in the following: i. 9.0 × 10−2 kg water ii. 89.6 dm3nitrogen at STP iii. 0.5 g hydrogen iv. 7.1 × 10−2 kg chloride ions.

7.

8.

9.

Calculate the number of moles in the following: i. 0.108 kg of water ii. 0.277 kg of carbon iii. 1.058 × 10−5 kg of sodium ions iv. 1.7 ×10−2 kg of hydrogen sulphide Calculate the weights of the following: i. 0.3 mole of carbon ii. 0.1 mole of hydrogen chloride iii. 2.4 moles of bromide ions Calculate the number of molecules in the following i. 4.5 × 10−5 kg of chlorine ii. 0.28 dm3 of oxygen at STP

Type III

10.

Calculate the percentage composition of the following compound: i. Urea CO(NH2)2 ii. CuSO4.5H2O

11.

Calculate the mass percentage composition of copper pyrites CuFeS2

12.

Calculate the percentage composition of glucose (C6H12O6).

A compound containing sodium, sulphur and oxygen has the following percentage composition Na = 29.11%, S = 40.51% and O = 30.38% and molecular weight 158. Find the empirical formula.

Type IV

14.

1.0 g of a mixture of carbonates of calcium and magnesium gave 240 cm3 of CO2 at STP calculate the percentage composition of the mixture.

Type V

15.

50.0 kg of N2(g) and 10.0 kg of H2(g) are mixed to produce NH3(g) i. Calculate the NH3(g) formed. ii. Identify the limiting reagent in this reaction if any. Multiple Choice Questions

1.0

Prominent scientists

1.

The law of combining volumes of gases was formulated by _______. (A) Avogadro (B) Gay-Lussac (C) Aristotle (D) Joseph Priestley

1.1

Introduction

2.

The branch of chemistry which deals with carbon compounds is called _____ chemistry. (A) organic (B) inorganic (C) carbon (D) bio

1.2

Importance and scope of chemistry

3.

_______ is used for the treatment of parkinson’s disease. (A) L-dopa (B) Taxol (C) Cisplatin (D) Tamiflue

1.3

Historical approach to particulate nature of matter

4.

The phlogiston theory was suggested for _______. (A) neutralisation reaction (B) oxidation reaction (C) reduction reaction (D) combustion reaction

5.

Electrochemical equivalent has unit ____. (A) kg m s−1 (B) kg m2 s−1 −1 (C) kg C (D) kg m−1 s−2

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6.

7.

8.

get Publications Pvt. Ltd. A ______ is a simple combination of two or more substances in which the constituent substances retain their separate identities. (A) compound (B) mixture (C) alloy (D) amalgam

1.5

Dalton’s atomic theory

13.

_____ proposed some symbols for some common atoms and molecules. (A) Democritus (B) Newton (C) Thomson (D) Dalton

1.6

Concepts of elements, atoms and molecules

14.

A/An _____ is an aggregate of two or more atoms of definite composition which are held together by chemical bonds. (A) ion (B) molecule (C) compound (D) mixture

1.7

Atomic and molecular masses

15.

One mole of oxygen weighs______. (A) 8 g (B) 32 g (C) 16 g (D) 6.023 × 1023 g

16.

The number of atoms present in a molecule of a substance is called ________. (A) atomicity (B) volume (C) density (D) mass

1.8

Avogadro’s law

A sample of calcium carbonate (CaCO3) has the following percentage composition : Ca = 40 %; C = 12 %; O = 48 % If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate from another source will be _______. (A) 0.016 g (B) 0.16 g (C) 1.6 g (D) 16 g

17.

The number of molecules in 22.4 cm3 of nitrogen gas at STP is _______. (B) 6.023 × 1023 (A) 6.023 × 1020 20 (D) 22.4 × 1023 (C) 22.4 × 10

18.

The number of molecules present in 8 g of oxygen gas are_______. (A) 6.022 × 1023 (B) 3.011 × 1023 23 (C) 12.044 × 10 (D) 1.505 × 1023

Two elements, A and B, combine to form two compounds in which ‘a’ g of A combines with ‘b1’ and ‘b2’g of B respectively. According to law of multiple proportion _____. (A) b1 = b2 (B) b1 and b2 bear a simple whole number ratio (C) a and b1 bear a whole number ratio (D) no relation exists between b1 and b2

19.

11.2 cm3 of hydrogen gas at contains_______ moles (A) 0.0005 (B) 0.01 (C) 0.029 (D) 0.5

20.

4.4 g of an unknown gas occupies 2.24 L of volume under STP conditions. The gas may be_______. (A) CO2 (B) CO (D) SO2 (C) O2

21.

Under similar conditions, oxygen and nitrogen are taken in the same mass. The ratio of their volumes will be_______. (A) 7 : 8 (B) 3 : 5 (C) 6 : 5 (D) 9 : 2

22.

Atomicity of ozone is_____. (A) 1 (B) 2 (C) 3 (D) 4

Which one of the following is NOT a mixture? (A) Iodized table salt (B) Gasoline (C) Liquefied Petroleum Gas (LPG) (D) Distilled water Magnitude of ‘pico−’ is _______. (A) 10−12 (B) 10−15 12 (C) 10 (D) 1015

1.4

Laws of chemical combination

9.

The sum of the masses of reactants and products is equal in any physical or chemical reaction. This is in accordance with _______. (A) law of multiple proportion (B) law of definite composition (C) law of conservation of mass (D) law of reciprocal proportion

10.

11.

12.


At constant temperature and pressure, one litre of nitrogen gas reacts with three litres of hydrogen gas to produce two litres of ammonia gas. This is in accordance with _______. (A) law of multiple proportion (B) law of definite composition (C) law of conservation of mass (D) law of combining volumes of gases 36

STP,

Tar



23.

19.7 kg of gold was recovered from a smuggler. How many atoms of gold were recovered? (Au = 197) (A) 197 (B) 6.02 × 1023 (D) 6.02 × 1025 (C) 6.02 × 1024

24.

The number of atoms in 4.25 g of NH3 is approximately_______. (B) 1.5. × 1023 (A) 1 × 1023 23 (D) 6.022 × 1023 (C) 2 × 10

25.

26.

What is the mass of 0.5 mole of ozone molecule? (A) 8 g (B) 16 g (C) 24 g (D) 48 g Mole triangle is the relationship between the mass of a gas, the number of moles, the volume of gas at S.T.P. and the _______. (A) number of electrons (B) number of molecules (C) pressure at S.T.P (D) temperature at S.T.P

1.9

Percentage formula

27.

The percentage of oxygen in NaOH is _______. (A) 40 (B) 60 (C) 8 (D) 10

28.

29.

composition

and

molecular

Which of the following has same molecular formula and empirical formula? (A) CO2 (B) C6H12O6 (D) all of these (C) C2H4 The empirical formula of C2H2 is _____. (B) CH (A) C2H4 (D) C4H10 (C) CH4

1.10 Chemical reactions and stoichiometry

30.

31.

_____ reactant is the reactant that reacts completely but limits further progress of the reaction. (A) Oxidizing (B) Reducing (C) Limiting (D) Excess If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is________. (A) 0.7 (B) 0.5 (C) 0.3 (D) 0.1

32.

Two elements X (atomic mass = 75) and Y (atomic mass = 16) combine to give a compound having 75.8% of X . The formula of the compound is _______. (A) XY (B) X2Y (C) XY2 (D) X2Y3 Answers to Practice Problems

1.

28.11 u.

2. 3.

6.022 × 1022 molecules, 4.82 × 1023 atoms of sulphur 0.0166 mol

4.

7.55 moles

5.

2.68 × 1010

6.

i. iii.

5 moles 0.25 moles

ii. iv.

4.0 moles 2 moles

7.

i. iii.

6 moles 4.6 × 10−4 mole

ii. iv.

23.08 moles 0.5 mole

8.

i. iii.

3.6 × 10−3 kg 0.192 kg

ii.

3.65 × 10−3 kg

9.

i.

3.82 × 1020

ii.

7.52 × 1021

10.

i.

% of carbon = 20% % of oxygen = 26.67% % of nitrogen = 46.67% % of hydrogen = 6.67%

ii.

% of Cu = 25.45% % of S = 12.83% % of O = 57.72% % of H = 4.00

11.

Cu = 34.64%, Fe = 30.44%, S = 34.9%

12.

C = 40.0%, H = 6.67%, O = 53.33%

13.

Na2S2O3

14.

CaCO3: 57.5%, MgCO3: 42.5%

15.

i. ii.

56.67 kg Hydrogen

Answers to Multiple Choice Question

1. (B) 5. (C) 9. (C) 13. (D) 17. (B) 21. (A) 25. (C) 29. (B)

2. 6. 10. 14. 18. 22. 26. 30.

(A) (B) (C) (B) (D) (C) (B) (C)

3. 7. 11. 15. 19. 23. 27. 31.

(A) (D) (B) (B) (A) (D) (A) (D)

4. 8. 12. 16. 20. 24. 28. 32.

(D) (A) (D) (A) (A) (D) (A) (D) 37