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Chemistry

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PREFACE Chemistry is one of the oldest academic discipline and its roots lie in man’s fascination towards study of structure, composition and properties of matter and the reactions by which matter converts from one form to the other.

NEET: Chemistry (Vol. I) not only adds great value towards a progressive society but also contributes greatly to other branches of science like biology, physics, geology, astronomy, biotechnology etc. Thus chemistry serves to be the backbone of all lifesciences.

Target’s “NEET: Chemistry (Vol. I)” has been compiled according to the notified syllabus for NEET-UG & ISEET, which in turn has been framed after reviewing various state syllabi as well as the ones prepared by CBSE, NCERT and COBSE. In the National-Eligibility-cum-Entrance Test (NEET), 25% weightage is given to Chemistry, as there are 45 questions based on Chemistry, out of the total 180 questions.

Target’s “NEET: Chemistry (Vol. I)” comprises of a comprehensive coverage of theoretical concepts & Multiple Choice Questions. In the development of each chapter we have ensured the inclusion of shortcuts & unique points represented as a ‘note’ for the benefit of students. The flow of content & MCQs have been planned keeping in mind the weightage given to a topic as per the NEET-UG & ISEET exam.

MCQs in each chapter comprise of mixture of questions based on theory & numericals and their level of difficulty is at par with that of various competitive examinations like CBSE, AIIMS, CPMT, PMT, JIPMER, IIT, AIEEE, & the likes. This edition of “NEET: Chemistry (Vol. I)” has been conceptualized with a complete focus on the kind of assistance students would require to answer tricky questions, which would give them an edge required to score in this highly competitive exam. Lastly, we are grateful to the publishers of this book for their persistent efforts, commitment to quality & their unending support to bring out this book, without which it would have been difficult for us to partner with students in this journey towards their success.

All the best to all Aspirants! Yours faithfully Authors

No. Topic Name Page No. 1 Some Basic Concepts of Chemistry 1 2 Structure of Atom 48 3 Classification of Elements and Periodicity in Properties 100 4 Chemical Bonding and Molecular Structure 132

5 States of Matter: Gases and Liquids 209

6 Thermodynamics 252 7 Equilibrium 307 8 Redox Reactions 384 9 Hydrogen 424

10 s-Block Elements (Alkali and Alkaline earth metals) 460 11 Some p-Block Elements 512 12 Organic Chemistry – Some Basic Principles and Techniques 565 13 Alkanes 655 14 Alkenes 693 15 Alkynes 727 16 Aromatic hydrocarbons 749 17 Environmental Chemistry 786

TARGET Publications Chemistry (Vol. I)

Some Basic Concepts of Chemistry 1

1.0 General Introduction – Importance and Scope of Chemistry Scientists and their contribution:

Scientists Contribution Joseph Louis Gay-Lussac Gay Lussac’s law of combining volumes of gases Amedeo Avogadro Avogadro’s law Antoine Lavoisier Law of conservation of mass Joseph Proust Law of definite composition John Dalton Law of multiple proportions

Chemistry: “Chemistry is defined as the study of composition, structure and properties of matter and the

reactions by which one form of matter may be converted into another form.” There are five important branches of chemistry such as:

Note: Apart from the above, there are several other branches of chemistry as: i. Medicinal (pharmaceutical) Chemistry: It deals with the application of chemical research

techniques to the synthesis of pharmaceuticals. ii. Environmental Chemistry: It deals with the study of chemistry associated with soil, air and water

and also of the impact of human activities on the natural system. iii. Green Chemistry: It deals with processes and products that eliminate or reduce the use or release of

hazardous substances. iv. Food Chemistry: It deals with the chemical processes associated with all forms of food stuffs. v. Agrochemistry: It deals with the application of chemistry for agricultural production and food

processing. vi. Geo Chemistry: It deals with the study of chemical composition and chemical processes associated

with Earth and the other plants. vii. Astrochemistry: It deals with the study of the compositions and reactions of the chemical elements

and molecules found in the space and the interactions between this matter and radiation. viii. Photochemistry: It deals with the interactions between light and matter.

CHEMISTRY

It deals with the substances which are constituents of living organisms.

It deals with the chemistry of elements other than carbon and their compounds.

It deals with the chemistry of carbon and carbon compounds.

It deals with the separation, extraction, identification and quantitative determination of the composition of different substances.

It 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.

Bio-ChemistryAnalytical Chemistry

Organic Chemistry

Inorganic Chemistry

Physical Chemistry

Some basic concepts of chemistry 0 1

TARGET PublicationsChemistry (Vol. I)

Some Basic Concepts of Chemistry2

ix. Electrochemistry: It deals with the study of chemical reactions in a solution and electron transfer particularly within electrolytic solution.

x. Solid State Chemistry: It deals with the structures, properties and chemical processes that occur in the solid phase.

xi. Polymer Chemistry: It deals with the examination of structure and properties of macromolecules and the study of new ways to synthesize those molecules.

xii. Nuclear Chemistry: It deals with the study of radioactivity, nuclear processes and nuclear properties. xiii. Nano Chemistry: It deals with the production and reactions of nanoparticles and their compounds.

Importance and scope of Chemistry: Chemistry plays a very important role in our everyday lives. Our daily needs of food, clothing,

shelter, potable water, medicines, etc., are in one or the other manner connected with chemical compounds, processes and principles. There is no aspect of life that is not associated with chemistry. In fact, Chemistry is the single branch of science which profoundly influences the existence of human beings, plants, animals as well as their habitat. Thus, mankind owes much to chemistry because it has improved the quality of life.

Some Applications of Chemistry: i. Chemistry in medicines and health care: The chemical substances used for treatment of diseases by destroying the disease causing

agents (antigens) without causing harm to the host tissues are called drugs or medicines. Some of the medicinal compounds are mentioned below:

NAME OF THE COMPOUND

FUNCTION/MEDICINAL PROPERTY EXAMPLES

a. Antipyretics Used to lower the temperature of the body in high fever.

Aspirin (acetylsalicyclic acid), paracetamol, phenacetin

b. Analgesics Used to relieve pain without causing impairment of consciousness. These are of two types:

i. Narcotic drugs: These are sleep inducing.

Naproxen, Ibuprofen

ii. Non-narcotic drugs: These do not induce sleep.

Morphine, Codeine

c. Tranquillizers Used for the treatment of stress, fatigue, mild and severe mental diseases.

Noradrenaline, Iproniazid, Phenelzine (antidepressant drugs)

d. Antiseptics Used to either kill or prevent the growth of micro-organisms. Not harmful and can be applied on living tissues.

Furacin, Soframycin

e. Disinfectant Used to kill micro-organisms, but are harmful to mankind and cannot be applied on living tissues.

Chlorine, Dettol, Bithional, Iodine, Boric acid, Iodoform, Hydrogen peroxide

f. Antimicrobials Used to cure infections caused by micro-organisms.

Salvarsan, prontosil

g. Antibiotics Produced by microbes and are used to inhibit the growth of microbes.

Penicillin, Ampicillin, Streptomycin, Neomycin

h. Antacids Used to neutralize excess acid in the gastric juices and give relief from acid indigestion, acidity and gastric ulcers.

Baking soda (NaHCO3) in water, omeprazole, lansoprazole

i. Antihistamines Used to diminish or abolish the main actions of histamine released in the body, thus prevent the allergic reactions.

Brompheniramine, (Diametapp), Terfenadine (Seldane), Dimithendine (foristal)



j. Anaesthetics Used to produce general or local insensibility to pain and other sensations.

General anaesthetics: Chloroform, Diethyl ethers, Vinyl ethers. Local anaesthetics: Cocaine, Novocaine.

k. Antifertility drugs

Used to control pregnancy. Ethynylestradiol (Novestrol), Mifepristone

ii. Chemistry in food: Many chemicals are added to food for their preservation and enhancing their appeal. These are

called food additives. While antioxidants, preservatives, fat emulsifiers, flour improvers are added to increase the shelf life of the stored food, some additives like dyes, flavours and sweetening agents are added to improve their cosmetic value. Some of these additives are mentioned below:

NAME OF THE FOOD

ADDITIVE FUNCTION EXAMPLES

a. Food preservatives These are added to the food materials to prevent their spoilage and to retain their nutritive value for long periods.

Butylated hydroxyanisole, (BHA), Butylated hydroxy toluene (BHT)

b. Taste enhancers These are used to enhance the taste of food stuffs.

Monosodium glutamate (MSG) – commonly called as aginomoto.

c. Artificial sweetening agents

These give sweetening effect to the food and enhance its odour and flavour.

Aspartame (methylester) Alitame, sucralose.

d. Edible food colours (dyes)

These are used to give an attractive appeal to the food stuffs.

Natural dye like carotene.

iii. Other applications of chemistry: a. Fuel: These are used for transportation and power generation. Petroleum is a rich source

of organic compounds. It is fractionally distilled to obtain various fractions like gasoline, kerosene, diesel and aviation fuel.

b. Dyes: These impart colour to the textiles. A dye should have a suitable colour and capacity to fix to the fibre.

Natural dyes include Indigo, alizarin etc. Synthetic dyes include Azo dyes, pthalocyanin dyes, etc. c. Detergents: These are the substances which possess cleansing properties. E.g. Sodium alkyl sulphates: (C12H25 – OSO3Na: Sodium lauryl sulphate), long chain alkyl benzene sulphonates:

Sodium p-dodecyl benzene sulphonate.

Matter and its constituents: i. The entire universe is made up of matter. ii. Matter is anything which has mass and occupies space. iii. It exists in three physical forms viz., solid, liquid and gas. a. Solids have definite shape and definite volume. b. Liquids have definite volume but no definite shape. c. Gases have neither definite shape nor definite volume. iv. These three states are interconvertible by changing the conditions of temperature and pressure. Solid heat

cool Liquid heat

cool Gas

C12H25 SO3Na



Note: Greek philosopher Democritus had suggested that matter is composed of extremely small atomio.

Classification of matter: Unit and its need: Definition: “The arbitrarily decided and universally accepted standards used in the measurement of physical

quantities are called units.”

Matter

Physical classification

Pure substance It comprises of a single type of particle present in a fixed ratio in which all the constituent particles are same in their chemical nature. Eg. Water, glucose, sodium chloride etc

Homogeneous It comprises of a single phase in which components are completely mixed with each other and its composition is uniform throughout. Eg. Mixture of salt and

water.

Heterogeneous It comprises of two or more phases present in the mixture and its composition is not uniform throughout. Eg. Phenol-water system,

silver chloride-water system etc.

Mixtures It comprises of two or more substances (components) present in any ratio in which the constituent substances retain their separate identities. Eg. Air, Tea, Brass (an alloy of copper and zinc) etc

Elements Pure substances whichare made up of only one component. Eg. Ag, Au, Cu, etc.

Compounds Pure substances which aremade up of two or morecomponents. Eg. Water, ammonia, etc.

Chemical classification

Solids Eg. NaCl

Liquids Eg. H2O

Gases Eg. CH4

Physical methods

Chemical methods

Inorganic Eg. AlCl3

Organic Eg. CH3CHO



Need: i. For calculation of experimental data. ii. For measurement of physical quantities such as mass, pressure, volume, length, time,

temperature, density, etc. iii. Any measured property is expressed as a number along with an appropriate unit associated with

the property as only the number does not give any idea of the property. Various system in which units are expressed: i. Units are expressed in various systems like CGS (centimeter for length, gram for mass and

second for time), FPS (foot, pound, second) and MKS (meter, kilogram, second) systems etc. ii. In 1960, the general conference of weights and measures, proposed a revised metric system,

called International System of units i.e., SI system, abbreviated from its French name Systeme Internationale d′ Units.

Note: NASA’s Mars climate orbiter, the first weather satellite for Mars, was destroyed by heat. The failure

of the mission was due to confusion in estimating the distance between Earth and Mars in miles and kilometers.

Seven fundamental SI units:

No. Fundamental quantity SI unit Symbol 1. Length Meter m 2. Mass Kilogram kg 3. Time Second s 4. Temperature Kelvin K 5. Amount of substance Mole mol 6. Electric current Ampere A 7. Luminous intensity Candela cd

Derived units The units of all physical quantities can be derived from the seven fundamental SI units. These units

are known as derived units. The table given below shows some common derived units.

No. Physical quantity Relationship with fundamental unit Symbol i. Area Length squared m2 ii. Volume Length cubed m3 iii. Density Mass per unit volume kg m−3 iv. Velocity Distance travelled in unit time ms−1 v. Acceleration Velocity change per unit time ms−2 vi. Force Mass × acceleration kg m s−2 (newton, N) vii. Pressure Force per unit area kg m −1 s−2 viii. Electric charge Current × time A s (coulomb, C)

ix. Electric potential or Potential difference Energy per unit charge kg m2s−2A−1 (J A−1 s−1

or volt V or J C−1) x. Energy (work or heat) Force × distance travelled kg m2 s−2 (J s−1) xi. Concentration Mole per cubic metre mol m−3

xii. Heat capacity Cp = dH/dT Cv = dE/dT

kg m2 s−2 K−1 mol−1 (J K−1 mol−1)

xiii. Electrochemical equivalent Z = E/F kg C−1 (kg/coulomb) Note: 1 Litre = 1 dm3



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 10−1 0.1 centi− c 10−2 0.01 milli− m 10−3 0.001 micro− µ 10−6 0.000 001 nano− n 10−9 0.000 000 001 pico− p 10−12 0. 000 000 000 001

femto− f 10−15 0.000 000 000 000 001 1.1 Laws of chemical combination Chemical combination: “The process in which the elements combine with each other chemically, to form compounds, is

called as chemical combination.” 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.”

i. Law of conservation of mass: The law was first stated by Russian scientist Lomonosove in the year 1765. Later in 1774,

French scientist, Antoine Lavoisier also stated the same law independently. Statement: It states that, “Mass is neither created nor destroyed during chemical combination of matter.” Explanation: a. According to Lavoisier, total masses of the reactants before the reaction are found to be

same as that of total masses of the products formed after the reaction. b. Eg. AgNO3 + NaCl ⎯→ AgCl + NaNO3 1.70g 0.555g 1.435g 0.82g ii. Law of definite composition or constant proportions:

This law was first stated by French chemist Joseph Proust in (1799). Statement: It states that, “Any pure compound always contains the same elements in a definite proportion

by weight irrespective of its source or method of preparation.” Explanation:

a. In support of this law, it was experimentally proved that a naturally occurring pure sample of copper carbonate contains 51.35 % copper by weight, 38.91 % carbon by weight and 9.74 % oxygen by weight.

b. Further, a pure sample of copper carbonate was synthesized in laboratory and it was found that the percentage by weight of copper, carbon and oxygen were exactly identical to that of the naturally occurring sample of copper carbonate.

c. French scientist Berthollet opposed Proust’s law of definite proportion by giving examples of the substances containing different proportions of elements.

d. However, Berthollet’s objections were ruled out as the experimental work of analysis mentioned by Berthollet was found to be based on impure samples or incomplete reactions.



iii. Law of multiple proportions: This law was proposed by British scientist John Dalton in 1808. Statement:

It states that, “If two elements, combine chemically with each other forming two or more compounds with different compositions by weight, then the masses of the two interacting elements in the two compounds are in the ratio of small whole numbers.”

Explanation: a. Hydrogen and oxygen combine to form two compounds H2O (water) and H2O2

(Hydrogen peroxide). H2O : 2 parts of Hydrogen, 16 parts of Oxygen H2O2 : 2 parts of Hydrogen, 32 parts of Oxygen. b. The masses of oxygen which combine with same mass of hydrogen in these two

compounds bear a simple ratio 1 : 2. iv. Law of reciprocal proportions: This law was given by Richter in 1794. Statement: It states that, “When two different elements combine separately with the same weight of a third

element, the ratio in which they do so will be the same or some simple multiple of the ratio in which they combine with each other.”

Explanation: a. Definite mass of an element A combines with two other elements B and C to form two

compounds. b. If B and C also combine to form a compound, their combining masses are in same

proportion or bear a simple ratio to the masses of B and C which combine with a constant mass of A.

Eg. Hydrogen combines with sodium and chlorine to form compounds NaH and HCl

respectively. NaH : 23 parts of sodium, 1 part of Hydrogen HCl : 35.5 parts of chlorine, 1 part of Hydrogen Sodium and chlorine also combine to form NaCl in which 23 parts of sodium and 35.5

parts of chlorine are present. These are the same parts which combine with one part of hydrogen in NaH and HCl respectively.

v. Gay-Lussac’s law of combining volumes of gases: This law was enunciated by Gay-Lussac in 1808. Statement:

It 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.”

Explanation: a. Under similar conditions of temperature and pressure, 1 volume of hydrogen reacts with

1 volume of chlorine to give 2 volumes of hydrogen chloride. H2 + Cl2 ⎯→ 2HCl

1 volume 1 volume 2 volumes b. Thus, the volume ratio of hydrogen: chlorine: hydrogen chloride is 1 : 1 : 2. c. This is a simple whole number ratio and is also in agreement with their molar ratios when

they are involved in the reaction.

NaH

H(A)

HCl

Cl (C)NaCl

(B) Na



Note: i. Gay-Lussac’s law of combining volumes is applicable only to gaseous reactions and not to

reactions involving solids and liquids. ii. The volumes of gases in the chemical reaction are not additive, though it appears to be additive.

However in case of hydrogen-oxygen reaction, 2 volumes of hydrogen and 1 volume of oxygen equal to 3 volumes of reactants get converted into 2 volumes of product steam.

iii. Similarly, in case of formation of ammonia, 1 volume of nitrogen reacts with 3 volumes of hydrogen equal to 4 volumes of reactants get converted into 2 volumes of product ammonia.

1.2 Dalton’s atomic theory Dalton’s atomic theory: John Dalton, an English school teacher, proposed the atomic theory in the year 1808. According to him, “Atom is the smallest indivisible particle of a substance.” Postulates/Assumptions: Dalton made the following assumptions in his theory: i. All matters are made up of tiny, indestructible, indivisible unit particles called atoms. ii. Atoms are the smallest particles of an element and molecules are the smallest particles of a

compound. iii. All atoms of the same element have same size, shape, mass and all other properties. iv. Atoms of different elements have different properties. v. Compounds are formed when atoms of different elements combine. vi. The atoms in a compound unite in small whole number ratios like 1:1, 1:2, 1:3, 2:1, 2:3, etc. vii. A chemical reaction involves only the separation, combination or rearrangement of integer

number of atoms. viii. During a chemical reaction, atoms are neither created nor destroyed. Note: The number of atom present in a molecule of a substance is called Atomicity. Avogadro’s Law: Avogadro, in the year 1811, combined Gay -Lussac’s law and Dalton’s theory to propose Avogadro’s

law. Statement : It states that, “Equal volumes of all gases, under identical conditions of temperature and pressure,

contain equal number of molecules.” OR

“At constant pressure and temperature, volume of a gas is directly proportional to the number of molecules.”

∴ V ∝ number of molecules (P, T constant) Explanation: i. If equal volumes of three gases i.e. Hydrogen (H2), Oxygen (O2) and Chlorine (Cl2) are taken in

different flasks of the same capacity under similar conditions of temperature and pressure, all the flasks are found to have the same number of molecules.

ii. However, these molecules may differ in size and mass.

1L of H2 gasat N.T.P

1L of O2 gasat N.T.P

1L of Cl2 gas at N.T.P

Illustration of Avogadro’s hypothesis.



Eg. a. Hydrogen + Chlorine ⎯→ Hydrogen chloride [1 vol ] [1 vol ] [2 vol] b. Applying Avogadro’s hypothesis, assuming that 1 volume contains n molecules, it

follows that Hydrogen + Chlorine ⎯→ Hydrogen chloride n molecules n molecules 2n molecules c. Dividing throughout by 2n, we get

12

molecule + 12

molecule ⎯→ 1 molecule

d. This means that 1 molecule of hydrogen chloride contains 1 / 2 molecule of hydrogen and 1 / 2 molecule of chlorine.

e. Now, 1 / 2 molecule of hydrogen can exist because one molecule of hydrogen contains 2 atoms of hydrogen and therefore 1/ 2 molecule of hydrogen contains one atom of hydrogen.

f. Similarly, 1/2 molecule of chlorine contains an atom of chlorine because chlorine is also a diatomic molecule.

g. Thus, one molecule of hydrogen chloride is formed from one atom of hydrogen and one atom of chlorine.

Note: This generalization is in agreement with Dalton’s atomic theory. 1.3 Concept of Elements, Atoms and Molecules Elements: i. “An element is defined as a substance which cannot be separated into simpler substances by

any chemical process.” Eg. Platinum, Nickel, Cobalt, etc.

ii. A pure substance is made up of only one kind of atoms having the same atomic number. iii. The smallest particle of an element is the atom. iv. Elements can be divided into two groups namely a. naturally occurring and b. artificially synthesized one v. There are about 118 elements out of which about 92 elements are naturally occurring and

nearly 26 elements are synthesized in laboratory. vi. The most abundant element in the earth’s crust is oxygen. vii. Artificially synthesized elements have a very short life as they breakup into more stable lighter

elements. viii. Most of the elements are solids, while eleven of them are gases and only two are liquids. ix. The two liquids are Mercury and Bromine. x. Elements can be broadly divided into four categories. Classification of elements:

Elements Characteristics Example Metals

Generally solid, hard, malleable, ductile, high tensile strength, lustre and good conductors of heat and electricity.

Copper, iron, zinc, etc.

Non-Metals

Generally non-lustrous, brittle, poor conductors of heat and electricity.

Sulphur, phosphorus, nitrogen,etc.

Metalloids Elements that have properties which lie in between those of metals and non- metals.

Arsenic, tin, bismuth, etc.

Noble Gases

Group of six elements that do not combine with other elements and tend to exist by themselves. They are chemically inactive. Hence, are also known as inert elements.

Neon, helium, argon, etc.



Chemical symbols: i. An abbreviated form in which the name of an element is represented is called as a symbol. ii. Chemists represent elements by symbols of one or two letters. iii. The first letter of the symbol is always capital and the second letter if present, is always small. iv. The symbols of most of the elements are derived from the English names of the elements. v. In some cases, Latin names of the elements are used to derive the symbols. Eg. Aluminium − Al , Einsteinium − Es, Gold − Au (Aurum)

Compounds: i. “Compounds are defined as substances of definite compositions which can be decomposed into

two or more substances by a simple chemical process.” Eg. Methane, ammonia, urea, etc.

ii. The properties of all the substances and elements obtained on decomposition of the compounds are completely different.

Eg. Carbon is combustible and oxygen supports combustion, but carbon dioxide is used as a fire extinguisher.

Atoms: i. “The smallest indivisible particle of an element is called atom.” ii. Every atom of an element has a definite mass of the order of 10−26 kg and has a spherical shape

of radius of the order of 10−15 m. iii. The smallest atom of an element is that of hydrogen with mass 1.667 × 10−26 kg. iv. Atoms may or may not exist freely. v. Atoms of almost all the elements can react with one another to form compounds. Note:

Name Number of atoms in a molecule Example Monoatomic Only one atom noble gases, some metals, carbon, silicon, etc. Diatomic Two atoms Hydrogen (H2), Oxygen (O2), Nitrogen (N2), etc. Polyatomic More than two atoms Phosphorus (P4), Sulphur (S8), etc.

Molecules:

i. “A molecule is an aggregate of two or more atoms of definite composition which are held together by chemical bonds.”

OR “The smallest particle of a substance (element or compound) which is capable of independent

existence is called a molecule.” ii. It has all the properties of the original compound. iii. It cannot be divided into its constituent atoms by simple methods. iv. Only under drastic conditions, a molecule can be decomposed into its constituent atoms. v. The properties of the constituent atoms of a compound and the molecule of compound are

completely different. Phlogiston theory: i. The phlogiston theory was proposed by Ernst Stahl (1660 - 1734). ii. Phlogiston was described as a substance in a combustible material which is given off when the

material burns. iii. This theory persisted for about 100 years and was a centre of much controversy. iv. Antoine Lavoisier proved that the flammable air produced by Cavandish was a new gas and

named it as hydrogen gas. v. During the end of the eighteenth century, much work was done with gases, especially by

Joseph Black, Henry Cavendish, Joseph Priestley and Carl Scheele. vi. Priestley was a very conservative scientist. Even after his discovery of oxygen, he still believed

in phlogiston theory.



1.4 Atomic and molecular masses Atomic mass: i. In 1961, the International Union of Chemists selected a new unit for expressing the atomic

masses. ii. They accepted the stable isotope of carbon (12C) with mass number of 12 as the standard for

comparing the atomic and molecular masses of elements and compounds. iii. “Atomic mass is the average relative mass of an atom of an element as compared to the mass of

an atom of carbon (C12) taken as 12”.

Atomic mass = 12

th

Massof an atom1 massof an atomof C

12

Note:

i. 1 a.m.u = 1.66056 × 2410− g , where a.m.u stands for atomic mass unit.

ii. Mass of hydrogen atom = 24

24

1.6736 10 g1.66056 10 g

−

−

××

= 1.00780 a.m.u = 1.0080 a.m.u

iii. Mass of oxygen-16 (16O) = 15.995 a.m.u iv Recently the unit of atomic mass ‘a.m.u.’ is been replaced by ‘u’ known as unified mass. v. “Gram atomic mass is the quantity of an element whose mass in grams is numerically equal to

its atomic mass”. OR

. “Atomic mass of an element expressed in grams is the gram atomic mass or it is also called gram atom”.

Eg. The atomic mass of oxygen = 16 a.m.u Therefore, gram atomic mass of oxygen = 16g. Average atomic mass: i. Majority of elements occur in nature as mixtures of several isotopes. ii. Isotopes are the different atoms of same elements possessing different atomic masses but same

atomic number. iii. The average relative mass depends upon the isotopic composition of that particular element. iv. The best way to define the atomic mass of the elements is to determine the atomic mass of each

isotope separately and then combine them in the ratio of their proportion of occurrence. This is called average atomic mass.

v. Each element has a number of isotopes with different isotopic masses. vi. While calculating the atomic mass of an element, a weighed average of the isotopic masses of

the isotopes of the element is taken, considering the relative quantity of isotopes. vii. Thus, it is the average mass of an atom of the element which is used in calculating the atomic

mass weight of the element. Eg. a. Chlorine has two isotopes, 85Cl and 37Cl, present in 75 % and 25 % proportion

respectively. Hence, the atomic mass of chlorine is the weighed average of these two isotopic masses.

i.e., (35.0 × 0.75) + (37.0 × 0.25) = 35.5. b. Aston’s mass spectrometer proved that neon exists in nature in the form of a mixture of

three isotopes, 1. Neon-20 with atomic mass 19.9924 u with natural abundance 90.92 % 2. Neon-22 with atomic mass 21.9914 u with natural abundance 8.82 % 3. Neon-21 with atomic mass 20.9940 u with natural abundance 0.26 % ∴ Average atomic mass of Ne,



=

20 20 22 22

21 21

(Atomicmassof Ne % of Ne) (Atomicmassof Ne % of Ne)(Atomicmassof Ne % of Ne)

100

× + ×

+ ×

= [(19.9924 u × 90.92) + (21.9914 u × 8.82) + (20.994 u × 0.26)]/100 = 20.1713 u Molecular mass: “Molecular mass of a substance is defined as the ratio of mass of one molecule of a substance to

th

112

of the mass of one atom of 12C.”

It is also the algebraic sum of atomic masses of constituent atoms present in the molecule. Characteristics of molecular mass: i. Like atomic mass, molecular mass is expressed as a.m.u. ii. It is the number that indicates comparative mass of a molecule of a compound with respect to

th

112

of the mass of one atom of 12C.

iii. Gram molecular mass is the molecular mass expressed in grams. iv. 1 gram molecular mass is also known as 1 gram molecule. Eg. Molecular mass of CO2 = 44 a.m.u Therefore gram molecular mass of CO2 = 44 g 1.5 Mole concept and molar mass Mole Concept: i. A mole is defined as the amount of substance that contains the same number of entities (atoms,

molecules, ions or other particles), as present in 12 g (or 0.012 kg) of the 12C isotope. ii. The quantity of a substance equal to its atomic mass or molecular mass in grams is referred as

1 mole of a substance. iii. Avogadro Number (NA):

“The number of atoms, molecules, ions, or electrons etc. present in 1 mole of a substance is found to be equal to 6.023 × 1023, which is called Avogadro Number (NA).”

Thus, NA = 6.023 × 1023 molecules or ions or electrons per mol.

Eg. 1 mole of hydrogen atoms = 6.023 × 1023 hydrogen molecules. 1 mole of sodium ions = 6.023 × 1023 sodium ions. 1 mole of electrons = 6.023 × 1023 electrons. iv. The volume of 1 mole of any pure gas at standard temperature and pressure is always constant

and is equal to 22.414 L or 0.022414 m3. This value is called as Avogadro’s molar volume or molar gas volume at STP.

Eg. 1 mole of chlorine gas = 22.4 L or 0.0224 m3 Molar Mass: i. The mass of one mole of a substance is called its molar mass (M) ii. The units of molar mass are g mol−1 or kg mol−1. iii. The molar mass is equal to atomic mass or molecular mass expressed in grams, depending upon

whether the substance contains atoms or molecules. Note: Mole Triangle: The relationship between the mass of a gas with number of moles, volume of a gas at STP and the

number of molecules is given by the mole triangle.



Note:

a. One mole of atoms = Massof an elementAtomicmass

b. Mass of one atom = 23

Atomic mass6.023 10×

c. One mole of molecules = 6.023 × 1023 molecules = Gram molecular mass of the substance

d. Mass of one molecule = 23

Molecular mass6.023 10×

e. Moles of a compound = Massof the compoundMolecular mass

f. Number of moles (n) = Massof thesubstanceMolar mass of the substance

g. Number of molecules = n × Avogadro’s number h. Volume of gas at S.T.P = n × 22.414 L i. Volume occupied by 1 mole of a gas at N.T.P = 22.4L j. Molecular mass = Vapour density × 2 1.6 Percentage composition and empirical and molecular formula Percentage composition: i. Percentage composition of a compound is the relative mass of each of the constituent element

in 100 parts of it. ii. Percentage composition can be calculated as mass percentage. iii. Mass percentage gives the mass of each element expressed as the percentage of the total mass.

iv. Mass percentage of an element = Massof theelement in1moleof compound 100Molar mass of the compound

×

Eg. Calculate the mass percentage composition of glucose: The formula of glucose = C6H12O6 Molar mass of glucose = (6 × 12) + (12 × 1) + (6 × 16) = 180 The formula of glucose shows that there are 6−C atoms, 12−H atoms and 6−O atoms. Mass of 6−C atoms = (6 × 12) = 72

Mole Triangle

Number offundamental

particles

Multiplied by Avogadro’s number

Divided by Avogadro’s number

Divided by molecular mass

Multiplied by 22.4 dm3

Mass of substance

Multiplied by molecular mass

Divided by 22.4 dm3

Volume occupied by gas at STP

in dm3

Number of moles



Mass % of C = 72180

⎛ ⎞⎜ ⎟⎝ ⎠

× 100 = 40.0 %

Mass of 12−H atoms = 12 × 1 = 12

Mass % of H = 12180

⎛ ⎞⎜ ⎟⎝ ⎠

× 100 = 6.67 %

Mass of 6−O atoms = 6 × 16 = 96

Mass % of O = 96180

⎛ ⎞⎜ ⎟⎝ ⎠

× 100 = 53.33 % Chemical formula: i. Elements and compounds are represented by symbols and formulae respectively. A chemical

formula gives the representation of a molecule of a substance in terms of symbols of various elements present in it.

Eg. Ammonia is represented by the formula NH3, carbon dioxide by CO2, copper sulphate by

CuSO4, etc. The determination of a formula of the substance involves the chemical analysis of a. The constituent elements present. b. The relative amount of elements of each type present in a given mass of the compound. ii. The chemical formula may be of two types: a. 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. Molecular formula of benzene = C6H6 ∴ Empirical formula = CH b. Molecular formula:

“The formula which gives the actual number of each kind of atoms in one molecule of the compound is called the molecular formula of the compound.”

It is an integral multiple of empirical formula. Eg. Molecular formula of benzene = C6H6. Thus, it has six atoms of carbon and six atoms of hydrogen. iii. Empirical and molecular formula of some molecules are given below:

Compound Empirical Formula Molecular Formula Hydrogen peroxide HO H2O2

Benzene CH C6H6

Glucose CH2O C6H12O6

Sucrose C12H22O11 C12H22O11

Naphthalene C5H4 C10H8 iv. Molecular formula and empirical formula are related as: Molecular Formula = n × Empirical formula where ‘n’ is a simple whole number and may have values 1, 2, 3 …..

n = Molecular massEmpiricalformula mass

Eg. The molecular mass of benzene is 78. The empirical formula of benzene is CH and therefore,

its Empirical formula mass is 13. Thus,

n = Molecular massEmpiricalformula mass

= 7813

= 6

Therefore, molecular formula of benzene = 6 × (CH) = C6H6



The various steps involved in determining the empirical formula are: Step 1: Divide the percentage of each element by its atomic mass. This gives the moles of atoms of various

elements in the molecule of the compound.

Moles of atoms = Percentageof an elementAtomic mass of theelement

Step 2: Divide the result obtained in the above step by the smallest value among them to get the simplest ratio

of various atoms. Step 3: Make the values obtained above to the nearest whole number by multiplying if necessary by a

suitable integer. This gives the simplest whole number ratio. Step 4: Write the symbols of the various elements side by side and insert the numerical value at the right

hand lower corner of each symbol. The formula thus obtained represents the empirical formula of the compound.

Steps for determination of the Molecular Formula of a compound: Step 1: Determine the empirical formula as described above. Step 2: Calculate the empirical formula mass by adding the atomic masses of the atoms in the empirical

formula. Step 3: Determine the molecular mass of the compound. Molecular mass can be determined by the following formulae: i. Molecular mass = Vapour Density × 2 ii. Molecular mass of an acid = Equivalent mass × basicity of the acid iii. Molecular mass of a base = Equivalent mass × acidity of the base iv. Molecular mass = Equivalent mass × no. of e− gained or lost. Step 4:

Determine the value of ‘n’ as, n = Molecular massEmpirical formula mass

Change ‘n’ to the nearest whole number. Step 5: Multiply empirical formula by ‘n’ to get the molecular formula. Molecular formula = n × Empirical formula Eg.

A compound with molar mass 159 was found to contain 39.62% copper and 20.13% sulphur. Suggest molecular formula for the compound (Cu = 63, S = 32 and O = 16)

Solution: % copper + % sulphur = 39.62 + 20.13 = 59.75 This is less than 100%. Hence compound contains adequate quantity of oxygen so that total

percentage of elements is 100%. Hence, % of oxygen = 100 − 59.75 = 40.25%



Element % of Element At. Mass Atomic ratio Simplest ratio

Cu 39.62 63 39.62 0.63

63= 0.63 1

0.63=

S 20.13 32

20.13 0.6332

=

0.630.63

= 1

O 40.25 16

40.25 2.5116

= 2.510.63

= 4

Hence, empirical formula is CuSO4 and empirical formula mass = 63 + 32 + 16 × 4 = 159 Molecular mass = empirical formula mass Therefore, molecular formula = empirical formula = CuSO4 Note: Molecular mass is the sum of the atomic masses of all the the atoms as given in the molecular formula of

the substance. 1.7 Chemical reactions, stoichiometry and calculations based on stoichiometry Chemical Reactions: i. “A chemical reaction is a process in which a single substance or many substances interact with

each other to produce one or more substances.” ii. Chemical reactions are represented in terms of chemical equations. iii. Chemical equation is a statement of a chemical reaction in terms of the symbols and formulae

of the species involved in the reaction. iv. The chemical equation may be defined as, “the brief representation of a chemical change in

terms of symbols and formulae of substances involved in it”. v. The chemical reaction when written in the form of chemical equation is always in balanced

form and the masses are always conserved. vi. The substances which react with each other to bring about the chemical changes are known as

reactants. vii. Whereas, the substances which are formed as a result of the chemical change are known as

products. Eg. AgNO3 + NaCl ⎯→ AgCl + NaNO3 Reactants Products Stoichiometry: Stoichiometry means quantitative relationship among the reactants and the products in a reaction. 1N2 (g) + 3H2 (g) ⎯→ 2NH3 (g) 1, 3, and 2 are coefficients of N2, H2 and NH3 respectively. These coefficients of reactants and

products in the balanced chemical reaction are called stoichiometric coefficients. The stoichiometric calculations involve the following steps: i. Write the correct formula of the reacting substances and products. Care must be taken to satisfy

valencies of the atoms of the compound. ii. For writing the balanced chemical equation, following three steps must be followed: Step 1:

Write the names of the reactants with ‘+’ sign separating the reacting substances on the left hand side. Then draw an arrow from left to right and to the right side of the arrow, write the names of all the products with ‘+’ sign separating them. Thus, in case of a reaction involving burning of methane in oxygen, producing carbon dioxide and water, the reaction is written as,

methane + oxygen ⎯→ carbon dioxide + water



Step 2: Rewrite the chemical equation in terms of chemical formula of each substance as shown. CH4 (g) + O2 (g) ⎯→ CO2 (g) + H2O(l) Step 3: Balance the mass of the chemical reaction by selecting the proper whole number coefficients

for each reactant and product as shown. CH4(g) + 2O2 (g) ⎯→ CO2 (g) + 2H2O(l) This is the balanced chemical equation. Note: Some Common Compounds:

No. Compound Formula No. Compound Formula 1. Phosphoric acid H3PO4 11. Cupric chloride CuCl2 2. Sodium phosphate Na3PO4 12. Ferrous chloride FeCl2 3. Ferric phosphate FePO4 13. Ferric chloride FeCl3 4. Aluminium phosphate AlPO4 14. Stannous chloride SnCl2 5. Copper phosphate Cu3(PO4)2 15. Stannic chloride SnCl4 6. Ferrous phosphate Fe3(PO4)2 16. Hydrogen sulphate H2SO4 7. Hydrogen chloride HCl 17. Sodium sulphate Na2SO4 8. Potassium chloride KCl 18. Copper sulphate CuSO4 9. Sodium chloride NaCl 19. Ferrous sulphate FeSO4 10. Cuprous chloride Cu2Cl2 20. Ferric sulphate Fe2(SO4)3

Mass relationship: i. A balanced chemical reaction may be used to establish the weight relationships of reactants and

products. ii. This is based on the law of conservation of mass, which states that, total mass of reactants is

always equal to total mass of the products. Atomic masses: (Na = 23, Cl = 35.5) 2Na(s) + Cl2(g) ⎯→ 2 NaCl(s) (2 atoms) (1 molecule) ⎯→ (2 molecules) [2 × 23] [1 × 35.5 × 2] ⎯→ [2 (23 + 35.5)] [46 g ] [71g] ⎯→ [117g] 117g ⎯→ 117g Limiting reactants: “It is the reactant that reacts completely but limits further progress of the reaction.” Excess reactant: “It is the reactant which is taken in excess than the limiting reactant.” Eg. 2H2(g) + O2(g) ⎯→ 2H2O(l)

3g 2g xg Limiting Excess Water reactant reactant



Calculations based on stoichiometry: Solving of stoichiometric problems is very important. It requires grasp and application of mole

concept, balancing of chemical equations and care in the conversion of units.

The problems based upon chemical equations may be classified as:

i. Mole to mole relationships: In these problems, the moles of one of the reactants/products is to be calculated if that of other

reactants/products are given. ii. Mass-mass relationships:

In these problems, the mass of one of the reactants/products is to be calculated if that of the other reactants/products are given.

iii. Mass-volume relationship:

In these problems, mass or volume of one of the reactants or products is calculated from the mass or volume of other substances.

iv. Volume-volume relationship:

In these problems, the volume of one of the reactants/products is given and that of the other is to be calculated.

The main steps for solving such problems are: a. Write down the balanced chemical equation.

b. Write down the moles or gram atomic or gram molecular masses of the substances whose quantities are given or have to be calculated. In case, there are two or more atoms or molecules of a substance, multiply the mole or gram atomic mass or molecular mass by the number of atoms or molecules.

c. Write down the actual quantities of the substances given. For the substances whose masses or volumes have to be calculated, write the symbol of interrogation (?).

d. Calculate the result by a unitary method.

Eg. Calculate, how many moles of methane are required to produce 22g of CO2(g) after combustion.

(Atomic masses, C = 12u, H = 1u, O = 16u)

Solution: The balanced combustion chemical reaction is,

CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O(g)

1 mol 1 mol

44g

? 22 g

44 g of CO2(g) are produced from 1 mole of CH4

22 g of CO2(g) will be produced by burning 1 2244× moles of CH4.

= 0.5 mole of CH4.



1.0 General Introduction – Importance and

Scope of Chemistry 1. The branch of chemistry which deals with

carbon compounds is called _____ chemistry. (A) Organic (B) Inorganic (C) Carbon (D) Bio 2. The branch of chemistry which deals with the

separation, identification and quantitative determination of the composition of different substances is called _____ chemistry.

(A) Organic (B) Inorganic (C) Analytical (D) Bio 3. _____ chemistry deals with the chemistry of

elements other than carbon and of their compounds.

(A) Organic (B) Physical (C) Inorganic (D) Bio 4. 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 is called _____ chemistry.

(A) Inorganic (B) Organic (C) Analytical (D) Physical 5. _____ chemistry is the chemistry of the

substances consisting of living organisms. (A) Organic (B) Physical (C) Inorganic (D) Bio 6. Which of the following branch of chemistry

deals with the application of chemical research techniques to the synthesis of Pharmaceuticals ?

(A) Nano chemistry (B) Polymer chemistry (C) Medicinal chemistry (D) Green chemistry 7. Nuclear chemistry deals with (A) Chemical processes associated with

food stuffs. (B) Structure, properties and chemical

processes occuring in the solid phase. (C) Production and reactions of

nanoparticles. (D) Study of radioactivity, nuclear,

processes and nuclear properties.

8. Which of the following drugs is used as a tranquilizer?

(A) Naproxen (B) Furacin (C) Noradrenaline (D) Ampicillin 9. Aspirin is used as an (A) Antipyretic (B) Antibiotic (C) Antihistamine (D) Tranquilizer 10. Aspartame is used as a food additive to (A) Impart attractive colour to the food stuff. (B) give sweetening effect to the food and

enhance its odour and flavour. (C) To prevent food spoilage (D) To retain the food nutritive value for a

longer time. 11. Indigo is used as a/an (A) Natural dye (B) Detergent (C) Artificial dye (D) Fuel 12. Greek philosopher _____ had suggested that

matter is composed of extremely small atomio.

(A) Dalton (B) Aristotle (C) Ptolemy (D) Democritus 13. Which of the following is CORRECT? (A) 1 L = 1 dm3 (B) 1 L = 10 dm3

(C) 10 L = 1 dm3 (D) 1 L = 1 m3

14. The prefix femto is used for expressing (A) 109 (B) 10−12 (C) 10−15 (D) 105

15. 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 16. If two or more phases are present in a mixture

then it is called a _____ mixture. (A) heterogeneous (B) homogeneous (C) homologous (D) heterologous 17. Mixture of all gases constitute _____ system. (A) homogeneous (B) heterogeneous (C) homologous (D) heterologous 18. Mixture of liquids constitute _____ system. (A) homogeneous (B) heterogeneous (C) either (A) or (B) (D) neither (A) nor (B)

Multiple Choice Questions



19. Phenol−water system is a/an (A) element. (B) compound. (C) homogeneous system. (D) heterogeneous system. 20. Which out of the following is NOT a

homogeneous mixture? (A) Air (B) Solution of salt in water (C) Solution of sugar in water (D) Smoke 21. Which one of the following is NOT a mixture? (A) Iodized table salt (B) Gasoline (C) Liquefied Petroleum Gas (LPG) (D) Distilled water 1.1 Laws of chemical combination 22. Law of conservation of mass was first stated

by (A) Lomonosove (B) Antoine Lavoisier (C) Joseph Proust (D) Dalton 23. After a chemical reaction, the total mass of

reactants and products [MP PMT 1989] (A) Is always increased (B) Is always decreased (C) Is not changed (D) Is always less or more 24. 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 25 The law of conservation of mass holds good

for all of the following except (A) All chemical reactions. (B) Nuclear reactions. (C) Endothermic reactions. (D) Exothermic reactions. 26. 1.5 g of hydrocarbon on combustion in excess

of oxygen produces 4.4 g of CO2 and 2.7 g of H2O, The data illustrates

(A) Law of conservation of mass (B) Law of multiple proportion (C) Law of constant composition (D) Law of reciprocal proportion

27. Which of the following is the best example of law of conservation of mass?

[NCERT 1975] (A) 12 g of carbon combines with 32 g of

oxygen to form 44 g of CO2 (B) When 12 g of carbon is heated in a

vacuum there is no change in mass. (C) A sample of air increases in volume

when heated at constant pressure but its mass remains unaltered.

(D) The weight of a piece of platinum is the same before and after heating in air.

28. ‘n’ g of substance X reacts with ‘m’ g of substance Y to form ‘p’ g of substance R and ‘q’ g of substance S. This reaction can be represented as, X + Y = R + S. The relation which can be established in the amounts of the reactants and the products will be

(A) n − m = p − q (B) n + m = p + q (C) n = m (D) p = q 29. If law of conservation of mass was to hold

true, then 20.8 g of BaCl2 on reaction with 9.8 g of H2SO4 will produce 7.3 g of HCl and BaSO4 equal to

(A) 11.65 g (B) 23.3 g (C) 25.5 g (D) 30.6 g 30. The law of definite composition was proposed

by (A) Lomonosove (B) Antoine Lavoisier (C) Joseph Proust (D) Dalton 31 If water samples are taken from sea, rivers,

clouds, lake or snow, they will be found to contain H2 and O2 in the fixed ratio of 1 : 8. This indicates the law of

(A) Multiple proportion (B) Definite proportion (C) Reciprocal proportion (D) None of these 32. The percentage of copper and oxygen in

samples of CuO obtained by different methods were found to be the same. This illustrates the law of [AMU 1982, 92]

(A) Constant proportion (B) Conservation of mass (C) Multiple proportion (D) Reciprocal proportion



33. Irrespective of the source, pure sample of water always yields 88.89 % mass of oxygen and 11.11 % mass of hydrogen. This is explained by the law of [Kerala CEE 2002]

(A) Conservation of mass (B) Constant composition (C) Multiple proportions (D) Constant volume 34. A sample of pure carbon dioxide, irrespective

of its source contains 27.27 % carbon and 2.73 % oxygen. The data supports

[AIIMS 1992] (A) Law of constant composition (B) Law of conservation of mass (C) Law of reciprocal proportion (D) Law of multiple proportion 35. Zinc sulphate contains 22.65 % of zinc and

43.9 % of water of crystallization. If the law of constant proportions is true, then the weight of zinc required to produce 20 g of the crystals will be

(A) 45.3 g (B) 4.53 g (C) 0.453 g (D) 453 g 36. 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 37. The law of definite proportion is NOT

applicable to nitrogen oxide because [EAMCET 1981]

(A) Nitrogen atomic weight is not constant (B) Nitrogen molecular weight is variable (C) Nitrogen equivalent weight is variable (D) Oxygen atomic weight is variable 38. The law of multiple proportions was given by (A) Proust (B) Dalton (C) Avogadro (D) Lavoisier 39. In SO2 and SO3, the ratio of the masses of

oxygen which combine with a fixed mass of sulphur is 2 : 3. This is an example of the law of

(A) Constant proportion (B) Multiple proportion (C) Reciprocal proportion (D) Gay Lussac

40. Different proportions of oxygen in the various oxides of nitrogen proves the law of

[MP PMT 1985] (A) Equivalent proportion (B) Multiple proportion (C) Constant proportion (D) Conservation of mass 41. Hydrogen and oxygen combine to form H2O2

and H2O containing 5.93 % and 11.2 % Hydrogen respectively. The data illustrates

(A) Law of conservation of mass (B) Law of constant proportion (C) Law of reciprocal proportion (D) Law of multiple proportion 42. Which one of the following pairs of

compounds illustrates the law of multiple proportion? [EAMCET 1989]

(A) H2O, Na2O (B) MgO, Na2O (C) Na2O, BaO (D) SnCl2, SnCl4 43. Which of the following pairs of substances

illustrates the law of multiple proportions ? [CPMT 1972, 78]

(A) CO and CO2

(B) H2O and D2O (C) NaCl and NaBr (D) MgO and Mg(OH)2 44. Two samples of lead oxide were separately

reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustrates

[AMU 1983] (A) Law of reciprocal proportions (B) Law of constant proportions (C) Law of multiple proportions (D) Law of equivalent proportions 45. 1.0 g of an oxide of A contains 0.5 g of A.

4.0 g of another oxide of A contains 1.6 g of A. The data indicates the law of

(A) Reciprocal proportion (B) Constant proportion (C) Conservation of energy (D) Multiple proportion



46. In compound A, 1.00 g of nitrogen unites with 0.57 g of oxygen. In compound B, 2.00 g of nitrogen combines with 2.24 g of oxygen. In compound C, 3.00 g of nitrogen combines with 5.11 g of oxygen. These results obey the

[CPMT 1971] (A) Law of constant proportion (B) Law of multiple proportion (C) Law of reciprocal proportion (D) Dalton’s law of partial pressure 47. Two elements, X (Atomic mass 16) and Y

(Atomic mass 14) combine to form compounds A, B and C. The ratio of different masses of Y which combines with fixed mass of X in A, B and C is 1:3:5. If 32 parts by mass of X combines with 84 parts by mass of Y in B, then in C, 16 parts by mass of X will combine with

(A) 14 parts by mass of Y (B) 42 parts by mass of Y (C) 70 parts by mass of Y (D) 82 parts by mass of Y 48. Two elements X and Y have atomic masses of

14 and 16 respectively. They form a series of compounds A, B, C, D and E in which for the same amount of element; X, Y is present in the ratio 1 : 2 : 3 : 4 : 5. If the compound A has 28 parts by mass of X and 16 parts by mass of Y, then the compound of C will have 28 parts by mass of X and

[NCERT 1971] (A) 32 parts by mass of Y (B) 48 parts by mass of Y (C) 64 parts by mass of Y (D) 80 parts by mass of Y 49. One part of an element A combines with two

parts of an element B. Six parts of the element C combines with four parts of element B. If A and C combine together the ratio of their weights will be governed by

(A) Law of definite proportion (B) Law of multiple proportion (C) Law of reciprocal proportion (D) Law of conservation of mass 50. 2 g of hydrogen combines with 16 g of oxygen

to form water and with 6 g of carbon to form methane. In carbon dioxide, 12 g of carbon is combined with 32 g of oxygen. These figures illustrate the law of

(A) Multiple proportion (B) Constant proportion

(C) Reciprocal proportion (D) Conservation of mass 51. 6 g of carbon combines with 32 g of sulphur to

form CS2. 12 g of C also combines with 32 g of oxygen to form carbondioxide. 10 g of sulphur combines with 10 g of oxygen to form sulphur dioxide. Which law is illustrated by them?

(A) Law of multiple proportion (B) Law of constant composition (C) Law of Reciprocal proportion (D) Gay Lussac’s law 52. ______ law of combining volumes states that

“Whenever gases combine, they do so in simple ratio by volumes”.

(A) Avogadro’s (B) Gay Lussac’s (C) Dalton’s (D) Thomson’s 53. 1 L of N2 combines with 3 L of H2 to form 2 L

of NH3 under the same conditions. This illustrates the

(A) Law of constant composition (B) Law of multiple proportions (C) Law of reciprocal proportions (D) Gay-Lussac’s law of gaseous volumes 54. The balancing of chemical equation is based

upon_____ (A) Law of combining volumes (B) Law of multiple proportions (C) Law of constant volume (D) Law of definite proportions 55. Which of the following reactions has the ratio

of volumes of reacting gases and the product as 1 : 2 : 2 ?

(A) 2CO(g) + O2(g) ⎯→ 2CO2(g) (B) O2(g) + 2H2(g) ⎯→ 2H2O(g) (C) H2(g) + F2(g) ⎯→ 2HF(g) (D) N2(g) + 3H2(g) ⎯→ 2NH3(g) 56. How many litres of ammonia will be formed

when 2 L of N2 and 2 L of H2 are allowed to react?

(A) 0.665 (B) 1.0 (C) 4.00 (D) 1.33 57. What volume of ammonia would be formed

when 0.36 dm3 of nitrogen reacts with sufficient amount of hydrogen? (all volumes are measured under same conditions of temperature and pressure)

(A) 0.36 dm3 (B) 0.72 dm3

(C) 0.18 dm3 (D) 0.12 dm3



58. The volume of oxygen required for the complete combustion of 0.25 cm3 of CH4 at S.T.P is

(A) 0.25 cm3 (B) 0.5 cm3 (C) 0.75 cm3 (D) 1 cm3

1.2 Dalton’s atomic theory 59. Who proposed the atomic theory? (A) Democritus (B) Newton (C) Thompson (D) Dalton 60. According to Dalton’s atomic theory, an atom

can (A) be created. (B) be destroyed. (C) neither be created nor destroyed. (D) be created and destroyed. 61. Dalton assumed that _____ are the smallest

particles of a compound. (A) atoms (B) molecules (C) ions (D) elements 62. On the basis of his assumptions, Dalton

explained (A) Law of conservation of mass (B) Law of multiple proportion (C) Both (A) and (B) (D) Neither (A) nor (B) 63. The number of atoms present in a molecule of

a substance is called ________ (A) Atomicity (B) Volume (C) Density (D) Mass 64. Atomicity of mercury vapour is_____ (A) 1 (B) 2 (C) 3 (D) 4 65. Atomicity of ammonium phosphate molecule

is_____ (A) 5 (B) 10 (C) 15 (D) 20 66. Avogadro’s law distinguishes between (A) cations and anions. (B) atoms and molecules. (C) atoms and ions. (D) molecules and ions. 67. Two containers of the same size are filled

separately with H2 gas and CO2 gas. Both the containers under the same temperature and pressure will contain the same

(A) number of atoms (B) mass of gas (C) number of molecules (D) number of electrons

1.3 Concepts of Elements, Atoms and Molecules

68. Substances which CANNOT be decomposed

into two different substances by chemical process are called

(A) Elements (B) Molecules (C) Compounds (D) All of these 69. _____ elements exist naturally. (A) 92 (B) 35 (C) 118 (D) 105 70. The most abundant element in the earth’s crust

is (A) Iron (B) Aluminium (C) Oxygen (D) Nitrogen 71. Which of the following is indivisible by

chemical methods? (A) atom (B) molecule (C) compound (D) mixture 72. Atom is the smallest particle of a/an (A) compound (B) substance (C) mixture (D) element 73. Atoms have a mass of the order (A) 10−26 kg (B) 10−15 kg (C) 10−26 g (D) 10−15 g 74. 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 75. Atoms have a radius of the order (A) 10−26 m (B) 10−15 µm (C) 10−15 mm (D) 10−15 m 76. The number of atoms in 5.4 g of NH3 is

approximately (A) 8 × 1023 (B) 2 × 1023 (C) 4 × 1023 (D) 6 × 1023

77. The number of atoms of oxygen present in

11.2 L of ozone at N.T.P. are (A) 3.01 × 1022 (B) 6.02 × 1023 (C) 9.03 × 1023 (D) 1.20 × 1019

78. _____ is a substance of definite composition

which can be decomposed into two or more substances by a simple chemical process.

(A) Compound (B) Element (C) Solution (D) Mixture



79. The phlogiston theory was suggested for (A) neutralisation reaction. (B) oxidation reaction. (C) reduction reaction. (D) combustion reaction. 80. Antoine Lavoisier proved that the flammable

air produced by Cavendish was a new gas and named it as

(A) Oxygen (B) Hydrogen (C) Methane (D) Nitrogen 1.4 Atomic and Molecular masses 81. The modern atomic mass scale is based on (A) 12C (B) 16O (C) 1H (D) 13C 82. Recently the unit of atomic mass amu is

replaced by (A) u (B) mol (C) g (D) kg 83. 1 amu is equal to

(A) 112

of C − 12 (B) 114

of O − 16

(C) 1 g of H2 (D) 1.66 × 10−23 kg 84. The number of atoms in 6 amu of He is (A) 18 (B) 18 × 6.022 × 1023 (C) 54 (D) 54 × 6.023 × 1023

85. The element whose atom has mass of

10.86 × 10−26 kg is (A) Boron (B) Calcium (C) Silver (D) Zinc 86. An atom of an element weighs 1.792 × 10–22 g,

atomic mass of the element is (A) 108 (B) 17.92 (C) 1.192 (D) 64 87. The number of gram atoms of oxygen present

in 0.3 gram mole of (COOH)2.2H2O is (A) 0.6 (B) 1.8 (C) 1.2 (D) 3.6 88. The sulphate of a metal M contains 9.87 % of

M. This sulphate is isomorphous with ZnSO4.7H2O. The atomic mass of M is

[IIT 1991] (A) 40.3 u (B) 36.3 u (C) 24.3 u (D) 11.3 u

89. _____ mass of a substance is defined as the ratio of mass of one molecule of a substance

to th1

12of mass of one 12C atom.

(A) Chemical (B) Molecular (C) Molar (D) Gram molar 90. The molecular mass of carbon dioxide is 44.

What is the unit of molecular mass? (A) g (B) mol (C) a.m.u (D) mol g−1

91. Vapour density of a gas is 22. What is its

molecular mass? [AFMC 2000] (A) 33 (B) 22 (C) 44 (D) 11 92. The vapour density of gas A is four times that

of B. If molecular mass of B is M, then molecular mass of A is

(A) M (B) 4M (C) 3M (D) 2M 93. Boron has two stable isotopes, 10B (19 %) and

11B (81 %). The atomic mass that should appear for boron in the periodic table is

[CBSE PMT 1990] (A) 10.8 amu (B) 10.2 amu (C) 11.2 amu (D) 10.0 amu 1.5 Mole concept and molar mass 94. Avogadro’s number is (A) number of atoms in one gram of

element. (B) number of millilitres which one mole of

a gaseous substance occupies at N.T.P. (C) number of molecules present in a gram

molecular mass of a substance. (D) All of these. 95. Avogadro number is the number of particles

present in (A) 1 molecule (B) 1 atom (C) 1 kg (D) 1 mole 96. NA = _________ atoms mol−1. (A) 6.021 × 1021 (B) 6.024 × 1024

(C) 6.051 × 1015 (D) 6.023 × 1023

97. One _____ is the collection of 6.023 × 1023

atoms /molecules/ions. (A) kg (B) g (C) mole (D) cm



98. The number of molecules in 22.4 cm3 of nitrogen gas at STP is

(A) 6.023 × 1020 (B) 6.023 × 1023 (C) 22.4 × 1020 (D) 22.4 × 1023

99. Number of molecules in 0.4 g of He is (A) 6.023 × 1023 (B) 6.023 × 1022 (C) 3.011 × 1023 (D) 3.011 × 1022

100. If NA is the Avogadro’s number then number

of valence electrons in 4.2 g of nitride ions N3− is

(A) 2.4 NA (B) 4.2 NA (C) 1.6 NA (D) 3.2 NA 101. 11.2 cm3 of hydrogen gas at STP,

contains____ (A) 0.0005 mol (B) 0.01 mol (C) 0.029 mol (D) 3.011 × 1023 molecules 102. The number of molecules present in 0.032 mg

of methane is (A) 12.046 × 1017 (B) 1.2044 × 1017 (C) 12.044 × 107 (D) 2 × 10−6

103. What is the mass of 0.5 mole of ozone

molecule? (A) 8 g (B) 16 g (C) 24 g (D) 48 g 104. At STP, 2 g of helium gas (molar mass = 4)

occupies a volume of (A) 22.4 dm3 (B) 11.2 dm3 (C) 5.6 dm3 (D) 2 dm3

105. The number of molecules in 16 g of oxygen is (A) 6.023 × 1023 (B) 3.011 × 1023

(C) 3.011 × 1022 (D) 1.5 × 1023

106. The number of sulphur atoms present in

0.2 moles of S8 molecules is (A) 4.82 × 1023

(B) 9.63 × 1022

(C) 9.63 × 1023 (D) 1.20 × 1023

107. 19.7 kg of gold was recovered from a

smuggler. How many atoms of gold were recovered? (Au = 197)

[Pb. CET 1985] (A) 100 (B) 6.023 × 1023 (C) 6.023 × 1024 (D) 6.023 × 1025

108. How many moles of electrons weigh one kilogram?

(A) 6.023 × 1023

(B) 1

9.108 × 1031

(C) 6.0239.108

× 1054

(D) 19.108 6.023×

× 108

109. The number of molecule at NTP in 1 mL of an

ideal gas will be (A) 6 × 1023 (B) 2.69 × 1019 (C) 2.69 × 1023 (D) 2.69 × 1034

110. 4.4 g of an unknown gas occupies 2.24 L of

volume under NTP conditions. The gas may be [MP PMT 1995]

(A) CO2 (B) CO (C) O2 (D) SO2 111. One mole of CO2 contains (A) 6.023 × 1023 atoms of C (B) 6.023 × 1023 atoms of O (C) 18.1 × 1023 molecules of CO2 (D) 3 g atoms of CO2 112. How many grams are contained in 1 g atom of

Na? (A) 13 g (B) 23 g

(C) 1 g (D) 123

g 113. One mole of oxygen weighs______. (A) 8 g (B) 32 g (C) 1 g (D) 64 g 114. 1 mol of CH4 contains (A) 6.02 × 1023 atoms of H (B) 4 g atom of Hydrogen (C) 1.81 × 1023 molecules of CH4 (D) 3.0 g of carbon 115. The mass of carbon present in 0.5 mole of

K4[Fe(CN)6] is (A) 1.8 g (B) 18 g (C) 3.6 g (D) 36 g 116. How many molecules are present in one gram

of hydrogen? [AIIMS 1982] (A) 6.023 × 1023 (B) 3.012 × 1023 (C) 2.512 × 1023 (D) 1.512 × 1023 117. The number of moles of sodium oxide in

620 g is [BHU 1992] (A) 1 mole (B) 10 moles (C) 18 moles (D) 100 moles



118. How many atoms are contained in one mole of sucrose (C12H22O11)? [Pb. PMT 2002]

(A) 45 × 6.023 × 1023 atoms/mole (B) 5 × 6.623 × 1023 atoms/mole (C) 5 × 6.023 × 1023 atoms/mole (D) 40 × 6.023 × 1023 atoms/mole 119. One mole of P4 molecule contains (A) 1 molecule (B) 4 molecules

(C) 14

× 6.022 × 1023 atoms

(D) 24.092 × 1023 atoms 120. Total number of atoms in 44 g of CO2 is (A) 6.023 × 1023 (B) 6.023 × 1024 (C) 1.807 × 1024 (D) 18.06 × 1022

121. The mass of 1 atom of hydrogen is (A) 1 g (B) 0.5 g (C) 1.66 × 10−24 g (D) 3.2 × 10−24 g 122. The mass of 2.01 × 1023 molecules of CO is (A) 9.3 g (B) 7.2 g (C) 1.2 g (D) 3 g 123. How many moles of Helium gas occupy 22.4 L at 0 °C at 1 atm pressure? (A) 0.11 (B) 0.90 (C) 1.0 (D) 1.11 124. The mass of a molecule of water is

[Bihar CEE 1995] (A) 3 × 10–26 kg (B) 3 × 10–25 kg (C) 1.5 × 10–26 kg (D) 2.5 × 10–26 kg 125. The mass of 1 molecule of N2 is (A) 3.24 × 10−22 g (B) 3.45 × 10−25 g (C) 4.65 × 10−23 g (D) 4.56 × 1022 g 126. The mass of a molecule of the compound

C60H122 is (A) 1.4 × 10−21 g (B) 1.09 × 10−21 g (C) 5.025 × 1023 g (D) 16.023 × 1023 g 127. The number of moles of oxygen in 1 L of air

containing 21 % oxygen by volume, in standard conditions, is

[CBSE PMT 1995; Pb. PMT 2004] (A) 0.186 mol (B) 0.21 mol (C) 2.10 mol (D) 0.0093 mol

128. 2 moles of H2 at NTP occupy a volume of (A) 11.2 litre (B) 44.8 litre (C) 2 litre (D) 22.4 litre 129. If the density of water is 1 g/cm3, then the

volume occupied by one molecule of water is approximately [Pb. PMT 2004]

(A) 18 cm3 (B) 22400 cm3 (C) 6.02 × 10−23 cm3 (D) 3.0 × 10−23 cm3

130. The number of molecules in 8.96 L of a gas at

0°C and 1 atmospheric pressure is approximately [BHU 1993]

(A) 6.023 × 1023 (B) 12.04 × 1023 (C) 18.06 × 1023 (D) 24.09 × 1022

131. Number of g of oxygen in 32.2 g of

Na2SO4.10H2O is [Haryana PMT 2000] (A) 20.8 (B) 22.4 (C) 2.24 (D) 2.08 132. Number of moles of water in 488 gm of

BaCl2.2H2O are (Ba = 137) (A) 2 moles (B) 4 moles (C) 3 moles (D) 5 moles 133. The mass of 1 × 1022 molecules of

CuSO4.5H2O is [CBSE PMT 1999; MH CET 2003]

(A) 41.51 g (B) 415.1 g (C) 4.151 g (D) 4151 g 134. Mass of H2O in 1000 kg CuSO4.5H2O is

(Cu = 63.5) (A) 3.607 kg (B) 36.07 kg (C) 360.7 kg (D) 3607 kg 135. The number of molecules in 16 g of methane is (A) 3.0 × 1023 (B) 6.023 × 1023

(C) 166.02

× 1023 (D) 163.0

× 1023

136. The number of water molecules in 1 litre of

water is [EAMCET 1990] (A) 18 (B) 18 × 1000 (C) NA (D) 55.55 NA 137. The numbers of moles of BaCO3 which

contain 1.5 moles of oxygen atoms is [EAMCET 1991]

(A) 0.5 (B) 1 (C) 3 (D) 6.02 × 1023

138. 1.24 g of P is present in 2.2 g of (A) P4S3 (B) P2S2 (C) PS2 (D) P2S4



139. 2 g of oxygen contains number of atoms equal to that in [BHU 1992]

(A) 0.5 g of hydrogen (B) 4 g of sulphur (C) 7 g of nitrogen (D) 2.3 g of sodium 140. If 1021 molecules are removed from 200 mg of

CO2, then the number of moles of CO2 left are [IIT 1983]

(A) 2.89 × 10−3 (B) 28.8 × 10−3 (C) 0.288 × 10−3 (D) 1.68 × 10−2

141. Mole triangle is the relationship between the

mass of a gas, the number of moles, the volume 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. 1.6 Percentage composition and empirical and

molecular formula 142. _____ of a compound is the chemical formula

indicating the relative number of atoms in the simplest ratio.

(A) Empirical formula (B) Molecular formula (C) Empirical mass (D) Molecular mass 143. _____ indicates the actual number of

constituent atoms in a molecule. (A) Empirical formula (B) Molecular formula (C) Empirical mass (D) Molecular mass 144. The mass percentage of each constituent

element present in 100 g of compound is called its

(A) Molecular composition (B) Atomic composition (C) Percentage composition (D) Mass composition 145. If two compounds have the same empirical

formula but different molecular formulae, they must have [MP PMT 1986]

(A) Different percentage composition (B) Different molecular mass (C) Same viscosity (D) Same vapour density 146. The percentage of oxygen in NaOH is

[CPMT 1979] (A) 40 (B) 60 (C) 8 (D) 10

147. Percentage of nitrogen in urea is about (A) 46 % (B) 85 % (C) 18 % (D) 28 % 148. The percentage composition of carbon in urea,

[CO(NH2)2] is (A) 40 % (B) 50 % (C) 20 % (D) 80 % 149. What is the % of H2O in Fe(CNS)3.3H2O? (A) 45 (B) 30 (C) 19 (D) 25 150. The percentage of P2O5 in diammonium

hydrogen phosphate (NH4)2HPO4 is [CPMT 1992]

(A) 23.48 (B) 46.96 (C) 53.79 (D) 71.00 151. A 400 mg iron capsule contains 100 mg of

ferrous fumarate (CHCOO)2Fe. The percentage of iron present in it is approximately

(A) 33 % (B) 25 % (C) 14 % (D) 8 % 152. Which pair of species have same percentage

of carbon? (A) CH3COOH and C6H12O6 (B) CH3COOH and C2H5OH (C) HCOOCH3 and C12H22O11 (D) C6H12O6 and C12H22O11 153. Empirical formula of glucose is (A) C6H12O6 (B) C6H11O6 (C) CHO (D) CH2O 154. The empirical formula of C2H2 is _____. (A) C2H4 (B) CH (C) CH4 (D) all of these 155. A compound (80 g) on analysis gave C = 24 g,

H = 4 g, O = 32 g. Its empirical formula is [CPMT 1981]

(A) C2H2O2 (B) C2H2O (C) CH2O2 (D) CH2O 156. Which of the following has same molecular

formula and empirical formula? (A) CO2 (B) C6H12O6 (C) C2H4 (D) all of these 157. The molecular mass of an organic compound

is 78. Its empirical formula is CH. The molecular formula is

(A) C2H4 (B) C2H2 (C) C6H6 (D) C4H4



158. The empirical formula of an acid is CH2O2, the probable molecular formula of acid may be

[AFMC 2000] (A) CH2O (B) CH2O2 (C) C2H4O2 (D) C3H6O4 159. The empirical formula of a compound is

CH2O. If 0.0835 mole of the compound contains 1.0 g of hydrogen, then the molecular formula of the compound is

(A) C6H12O6 (B) C5H10O5

(C) C4H8O8 (D) C3H6O3 160. On analysis, a certain compound was found to

have 254 g of iodine (At. mass 127) and 80 g oxygen (At. mass 16). What is the molecular formula of the compound?

(A) IO (B) I2O (C) I5O3 (D) I2O5 161. The molecular formula of the compound with

molecular mass 159, containing 39.62 % Cu and 20.13 % S is

(A) Cu2S (B) CuS (C) CuSO3 (D) CuSO4 162. A compound made of two elements A and B is

found to contain 25 % A (At. mass 12.5) and 75 % B (At. mass 37.5). The simplest formula of the compound is

(A) AB (B) AB2 (C) AB3 (D) A3B 163. Two elements X (At. mass 75) and Y (At.

mass 16) combine to give a compound having 75.8 % X. The formula of the compound is

(A) XY (B) XY2

(C) X2Y2 (D) X2Y3 164. An oxide of a metal (M) contains 40 % by

mass of oxygen. Metal (M) has atomic mass of 24. The empirical formula of the oxide is

(A) M2O (B) MO (C) M2O3 (D) M2O4 165. Two oxides of metal contain 27.6 % and 30 %

oxygen respectively. If the formula of first oxide is M3O4 then formula of second oxide is

(A) MO (B) M2O (C) M2O3 (D) MO2

166. 14 g of element X combines with 16 g of oxygen. On the basis of this information, which of the following is a CORRECT statement?

(A) The element X could have an atomic mass of 7 and its oxide formula is XO

(B) The element X could have an atomic mass of 14 and its oxide formula is X2O

(C) The element X could have an atomic mass of 7 and its oxide formula is X2O

(D) The element X could have an atomic mass of 14 and its oxide formula is XO2

1.7 Chemical reactions, stoichiometry and calculations based on stoichiometry

167. ________ is the quantitative relationship

between the reactants and products in a balanced chemical equation.

(A) Stoichiometry (B) Complexometry (C) Chemistry (D) Reactions 168. The starting material which takes part in

chemical reaction is called (A) product. (B) reactant. (C) catalyst. (D) starter. 169. _____ reactant is the reactant that reacts

completely but limits further progress of the reaction.

(A) Oxidizing (B) Reducing (C) Limiting (D) Excess 170. 3 g of H2 reacts with 29 g of O2 to yield water.

Which is the limiting reactant? (A) H2 (B) O2 (C) H2O (D) none of these 171. _____ reactant is the reactant which is taken in

excess than the limiting reactant. (A) Oxidizing (B) Reducing (C) Limiting (D) Excess 172. A _____ chemical reaction may be used to

establish the weight relationships of reactants and products.

(A) thermal (B) molecular (C) balanced (D) molar 173. The set of numerical coefficient that balances

the equation K2CrO4 + HCl ⎯→ K2Cr2O7 + KCl + H2O is

[Kerala CEE 2001] (A) 1, 1, 2, 2, 1 (B) 2, 2, 1, 1, 1 (C) 2, 1, 1, 2, 1 (D) 2, 2, 1, 2, 1



174. One mole of calcium phosphide on reaction with excess of water gives [IIT 1999]

(A) One mole of phosphine (B) Two mole of phosphoric acid (C) Two moles of phosphine (D) One mole of phosphorus pentoxide 175. For the reaction : A + 2B ⎯→ C 5 moles of A and 8 moles of B will produce (A) 5 moles of C (B) 4 moles of C (C) 8 moles of C (D) 13 moles of C 176. The moles of O2 required for reacting with

6.8 g of ammonia in the following reaction (..... NH3 +..... O2 ⎯→ ..... NO + ..... H2O) is (A) 5 (B) 2.5 (C) 1 (D) 0.5 177. One mole of potassium dichromate completely

oxidises ______ number of moles of ferrous sulphate in acidic medium.

(A) 1 (B) 3 (C) 5 (D) 6 178. 27 g of Al (At. mass = 27) will react with

oxygen equal to (A) 24 g (B) 8 g (C) 40 g (D) 10 g 179. 1.2 g of Mg (At. mass 24) will produce MgO

equal to (A) 0.05 mol (B) 0.03 mol (C) 0.01 mol (D) 0.02 mol 180. 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) 0.1 181. If one mole of ethanol (C2H5OH) completely

burns to form carbon dioxide and water, the mass of carbon dioxide formed is about

(A) 22 gm (B) 45 gm (C) 66 gm (D) 88 gm 182. Complete combustion of 0.858 g of compound

X gives 2.63 g of CO2 and 1.28 g of H2O. The lowest molecular mass X can have is

(A) 43 g (B) 86 g (C) 129 g (D) 172 g 183. 1.12 mL of a gas is produced at STP by the

action of 4.12 mg of alcohol (ROH) with methyl magnesium iodide. The molecular mass of alcohol is [Roorkee 1992; IIT 1993]

(A) 16.0 (B) 41.2 (C) 82.4 (D) 156.0

184. Hydrogen reacts with nitrogen to form ammonia as: N2(g) + 3H2(g) ⎯→ 2NH3(g)

The amount of ammonia that would be produced if 200 g of H2 reacts with N2 is

(A) 1032.2 g (B) 11332 g (C) 1133.3 g (D) 8692.6 g 185. What volume of Hydrogen gas, at 273 K and 1

atm pressure will be consumed in obtaining 21.6 g of elemental boron (At. mass = 10.8) from the reduction of boron trichloride by Hydrogen ? [AIEEE 2003]

(A) 22.4 L (B) 89.6 L (C) 67.2 L (D) 44.8 L 186. 12 g of Mg (At. mass 24) will react

completely with acid to give [MNR 1985] (A) One mole of H2

(B) 12

mole of H2

(C) 23

mole of O2

(D) Both 12

mol of H2 and 12

mol of O2

187. H2 evolved at STP on complete reaction of 27

g of Aluminium with excess of aqueous NaOH would be [CPMT 1991]

(A) 22.4 litres (B) 44.8 litres (C) 67.2 litres (D) 33.6 litres 188. What mass of CaO will be obtained by heating

3 mole of CaCO3? [At. mass of Ca = 40] (A) 150 g (B) 168 g (C) 16.8 g (D) 15 g 189. How much of NaOH is required to neutralise

1500 cm3 of 0.1N HCl? (Na = 23) [KCET 2001]

(A) 40 g (B) 4 g (C) 6 g (D) 60 g 190. How many grams of caustic potash is required

to completely neutralize 12.6 g HNO3? (A) 22.4 g of KOH (B) 1.01 g of KOH (C) 6.02 g of KOH (D) 11.2 g of KOH 191. The mass of CaCO3 produced when carbon

dioxide is passed in excess through 500 mL of 0.5 M Ca(OH)2 will be

(A) 10 g (B) 20 g (C) 50 g (D) 25 g



192. What mass of calcium chloride in grams would be enough to produce 14.35 g of AgCl?

(At. Mass of Ca = 40, Ag = 108) (A) 5.50 g (B) 8.295 g (C) 16.59 g (D) 11.19 g 193. The amount of sulphur required to produce

100 moles of H2SO4 is (A) 3.2 × 103 g (B) 32.65 g (C) 32 g (D) 3.2 g 194. What should be the weight of 50 % HCl which

reacts with 100 g of limestone ? (A) 50 % pure (B) 25 % pure (C) 10 % pure (D) 8 % pure 195. The conversion of oxygen to ozone occurs to

the extent of 15 % only. The mass of ozone that can be prepared from 67.2 L of oxygen at STP will be

(A) 144 g (B) 96 g (C) 640 g (D) 64 g Miscellaneous 196. An element forms two oxides containing

respectively 53.33 and 36.36 percent of oxygen. These figures illustrate the law of

(A) Conservation of mass (B) Constant proportion (C) Reciprocal proportion (D) Multiple proportion 197. One sample of atmospheric air is found to

have 0.03 % of carbon dioxide and another sample 0.04 %, this is an evidence that

(A) The law of constant composition is not always true

(B) The law of multiple proportions is true (C) Air is a compound (D) Air is a mixture 198. Among the following pairs of compounds, the

one that illustrates the law of multiple proportions is

(A) NH3 and NCl3 (B) H2S and SO2 (C) CuO and Cu2O (D) CS2 and FeSO4 199. An example of a chemical change is (A) the melting of an ice cube (B) the boiling of gasoline (C) the frying of an egg (D) all of these 200. ________ is the sum of the atomic masses of

all the atoms as given in the molecular formula of the substance.

(A) Molecular mass (B) Empirical mass (C) Percentage mass (D) Equivalent mass

201. Molecular mass = vapour density × 2, is valid for

(A) metals. (B) non-metals. (C) solids. (D) gases. 202. One mole of H2O corresponds to (A) 22.4 litres at 1 atm and 25 °C (B) 6.02 × 1023 atoms of hydrogen and

6.02 × 1023 atoms of oxygen (C) 18 g (D) 1 g 203 4.48 litres of methane at N.T.P. corresponds to (A) 1.2 × 1022 molecules of methane (B) 0.5 mole of methane (C) 3.2 g of methane (D) 0.1 mole of methane 204. 6.023 × 1023 electrons make an electrical

charge of one (A) electron volt (B) avogadro (C) coulomb (D) faraday 205. The mass of a substance that displaces

22.4 litre air at NTP is (A) Molecular mass (B) Empirical mass (C) Equivalent mass (D) All of these 206. The number of atoms present in 0.1 mole of

P4 are (A) 2.4 × 1023 atoms (B) 6.02 × 1022 atoms (C) same as in 0.2 mole of S8 (D) same as in 3.1 g of phosphorus 207. Number of moles of water in 1 L of water with

density 1 g/cc are (A) 55.56 (B) 45.56 (C) 56.55 (D) 46.55 208. Which one of the following pairs of gases

contain the same number of molecules? (A) 16 g of O2 and 14 g of N2 (B) 8 g of O2 and 22 g of CO2 (C) 28 g of N2 and 22 g of CO2 (D) 32 g of O2 and 32 g of N2 209. 54 grams of aluminium (atomic mass = 27)

will react with how many grams of oxygen ? (A) 16 g (B) 48 g (C) 40 g (D) 15 g 210. The largest number of molecules is in

[BHU 1997] (A) 34 g of water (B) 28 g of CO2 (C) 46 g of CH3OH (D) 54 g of N2O5



211. Which of the following has least mass? (A) 2 g atom of nitrogen (B) 3 × 1023 atoms of C (C) 1 mole of S (D) 7.0 g of Ag 212. Which of the following contains maximum

number of atoms? [JIPMER 2000] (A) 6.023 × 1021 molecules of CO2 (B) 22.4 L of CO2 at S.T.P. (C) 0.44 g of CO2 (D) None of these 213. 4.0 g of caustic soda (molar mass 40) contains

same number of sodium ions as are present in (A) 10.6 g of Na2CO3 (molar mass 106) (B) 58.5 g of NaCl (Formula mass 58.5) (C) 100 mL of 0.5 M Na2SO4 (Formula

mass 142) (D) 1 mol of NaNO3 (molar mass 85) 214. Four containers of 2L capacity contain

dinitrogen as described below. Which one contains maximum number of molecules under similar conditions ?

(A) 2.5 g of N2 molecules (B) 4 g of N atoms (C) 40 g of N atoms (D) 84 g of dinitrogen 215. Which of the following contains the largest

number of atoms? (A) 11 g of CO2 (B) 4 g of H2 (C) 5 g of NH3

(D) 8 g of SO2 216. 4.4 g of CO2 and 2.24 litre of H2 at STP are

mixed in a container. The total number of molecules present in the container will be

(A) 6.023 × 1023

(B) 1.2046 × 1023 (C) 2 moles (D) 6.023 × 1024 217. Number of moles of KMnO4 required to

oxidize one mole of Fe(C2O4) in acidic medium is [Haryana CEE 1996]

(A) 0.6 (B) 0.167 (C) 0.2 (D) 0.4

218. 14 g of nitrogen represents (A) 6.02 × 1023 N2 molecules (B) 22.4 L of N2 at N.T.P. (C) 11.2 L of N2 at N.T.P. (D) 28 gm of nitrogen. 219. In the reaction, 2Al(s) + 6HCl(aq) ⎯→ 3

(aq)2Al + + (aq)6Cl− + 3H2(g),

(A) 6 L HCl(aq) is consumed for every 3L H2(g) produced.

(B) 33.6 L H2(g) is produced regardless of temperature and pressure for every mole of Al that reacts.

(C) 67.2 L H2(g) at STP is produced for every mole of Al that reacts.

(D) 11.2 L H2(g) at STP is produced for every mole of HCl(aq) consumed.

220. Number of water molecules in a drop of water,

if 1 mL of water has 20 drops and A is Avogadro number, is

(A) 0.5 A / 18 (B) 0.05 A (C) 0.5 A (D) 0.05 A / 18 221. M is the molecular mass of KMnO4. The

equivalent mass of KMnO4 when it is converted into K2MnO4 is

(A) M (B) M3

(C) M5

(D) M7

222. Volume of a gas at N.T.P. is 1.12 × 10−7 cc.

Calculate the number of molecules in it. (A) 3.01 × 1020 (B) 3.01 × 1012 (C) 3.01 × 1023

(D) 3.01 × 1024

223. Under similar conditions, oxygen and nitrogen

are taken in the same mass. The ratio of their volume will be_______.

(A) 7 : 8 (B) 3 : 5 (C) 6 : 5 (D) 9 : 2



1. (A) 2. (C) 3. (C) 4. (D) 5. (D) 6. (C) 7. (D) 8. (C) 9. (A) 10. (B) 11. (A) 12. (D) 13. (A) 14. (C) 15. (B) 16. (A) 17. (A) 18. (C) 19. (D) 20. (D) 21. (D) 22. (A) 23. (C) 24. (C) 25. (B) 26. (A) 27. (A) 28. (B) 29. (B) 30. (C) 31. (B) 32. (A) 33. (B) 34. (A) 35. (B) 36. (C) 37. (C) 38. (B) 39. (B) 40. (B) 41. (D) 42. (D) 43. (A) 44. (C) 45. (D) 46. (B) 47. (C) 48. (B) 49. (C) 50. (C) 51. (C) 52. (B) 53. (D) 54. (C) 55. (B) 56. (C) 57. (B) 58. (B) 59. (D) 60. (C) 61. (B) 62. (B) 63. (A) 64. (A) 65. (D) 66. (B) 67. (C) 68. (A) 69. (A) 70. (C) 71. (A) 72. (D) 73. (A) 74. (B) 75. (D) 76. (B) 77. (C) 78. (A) 79. (D) 80. (B) 81. (A) 82. (A) 83. (A) 84. (A) 85. (D) 86. (A) 87. (B) 88. (C) 89. (B) 90. (C) 91. (C) 92. (B) 93. (A) 94. (C) 95. (D) 96. (D) 97. (C) 98. (B) 99. (B) 100. (A) 101. (A) 102. (A) 103. (C) 104. (B) 105. (B) 106. (C) 107. (D) 108. (D) 109. (B) 110. (A) 111. (A) 112. (B) 113. (B) 114. (B) 115. (D) 116. (B) 117. (B) 118. (A) 119. (D) 120. (C) 121. (C) 122. (A) 123. (C) 124. (A) 125. (C) 126. (A) 127. (D) 128. (B) 129. (D) 130. (D) 131. (B) 132. (B) 133. (C) 134. (C) 135. (B) 136. (D) 137. (A) 138. (A) 139. (B) 140. (A) 141. (B) 142. (A) 143. (B) 144. (C) 145. (B) 146. (A) 147. (A) 148. (C) 149. (C) 150. (C) 151. (D) 152. (A) 153. (D) 154. (B) 155. (D) 156. (A) 157. (C) 158. (B) 159. (A) 160. (D) 161. (D) 162. (A) 163. (B) 164. (B) 165. (C) 166. (C) 167. (A) 168. (B) 169. (C) 170. (A) 171. (D) 172. (C) 173. (D) 174. (C) 175. (B) 176. (D) 177. (D) 178. (A) 179. (A) 180. (D) 181. (D) 182. (A) 183. (C) 184. (C) 185. (C) 186. (B) 187. (D) 188. (B) 189. (C) 190. (D) 191. (D) 192. (A) 193. (A) 194. (A) 195. (A) 196. (D) 197. (D) 198. (C) 199. (C) 200. (A) 201. (D) 202. (C) 203. (C) 204. (D) 205. (A) 206. (A) 207. (A) 208. (A) 209. (B) 210. (A) 211. (B) 212. (B) 213. (C) 214. (D) 215. (B) 216. (B) 217. (A) 218. (C) 219. (D) 220. (D) 221. (A) 222. (B) 223. (A)

29. BaCl2 + H2SO4 ⎯→ HCl + BaSO4

∴ 20.8 + 9.8 = 7.3 + x ∴ x = 23.3 g 35. 100 g zinc sulphate crystals are obtained from 22.65 g of zinc.

∴ 1 g of zinc sulphate crystal will be obtained from = 22.65100

g of zinc.

∴ 20 g of zinc sulphate crystals will be obtained from = 22.65100

× 20 = 4.53 g of zinc. 36. 100 g of CaCO3 is obtained from 40 g of Ca

∴ 1 g of CaCO3 will be obtained from 40100

g of Ca

∴ weight of Ca in 4 g of a sample of calcium carbonate from another source will be = 40100

× 4 = 1.6 g 47. In compound B, 32 parts of X react with 84 parts of Y. ∴ In compound B, 16 parts of X react with 42 parts of Y. In compound C, 16 parts of X react with x parts of Y. The ratio of masses of Y which combines with fixed mass of X in compounds B and C is 3:5

B 42 3 C x 5

∴ x = 42 5

3×

= 70

Answers to Multiple Choice Questions

Hints to Multiple Choice Questions



48. A 16 1 C x 3 x = 16 × 3 x = 48 57. N2 + 3 H2 ⎯→ 2 NH3 (1 vol.) (3 vol.) (2 vol.) 1 : 3 : 2 0.36 : 3 × 0.36 : x

x = 2 1.083

× = 0.72 dm3

58. CH4 + 2O2 ⎯→ CO2 + 2H2O (1 vol.) (2 vol.) (1 vol.) (2 vol.) 1 cm3 of CH4 requires 2 cm3 of O2 ∴ 0.25 cm3 of CH4 requires 0.5 cm3 of O2 64. In vapour state, atomicity of mercury is 1. 65. Ammonium phosphate [(NH4)3PO4] contains 3N, 12H, 1P and 4O, atoms, hence total number of atoms are

20, hence atomicity is 20. 76. Molecular mass of NH3 = 17 Amount of NH3 = 5.4 g

∴ Number of moles = 5.417

= 0.32 mole

∴ 1 mole of NH3 = 6.023 × 1023 atoms

∴ 0.32 mole of NH3 = 236.023 10 0.32

1× × atoms

≈ 2 × 1023 atoms 77. 22.4 L of ozone ≡ 6.023 × 1023 atoms

∴ 11.2 L of ozone ≡ 236.023 10 11.2

22.4× ×

= 3.011 × 1023 atoms Hence, number of oxygen atoms in ozone (O3) = 3 × 3.011 × 1023 = 9.03 × 1023 atoms

84. 1 atom of He ≡ 13

amu

∴ x atom of He ≡ 6 amu

∴ 1 × 6 = 3x

∴ x = 6 × 3 = 18 atoms 85. Atomic mass of the given element = 6.023 × 1023 × 10.86 × 10–26 kg = 65.4 × 10–3 kg = 65.4 g ∴ The element whose atom has mass of 10.86 × 10–26 kg is Zinc.



86. Atomic mass of an element = 1.792 × 10−22 × 6.023 × 1023 ≈ 108 87. ∵ 1mole (COOH)2.2H2O has 96 g of oxygen ∴ 0.3 mole (COOH)2.2H2O has 96 × 0.3 = 28.8 g

∴ Number of gram atoms of oxygen = 28.816

= 1.8 88. As the given sulphate is isomorphous with ZnSO4.7H2O, its formula would be MSO4.7H2O; Let m be the

atomic mass of M, molecular mass of MSO4.7H2O = m + 32 + 64 + 126 = m + 222

Hence % of M = mm 222+

× 100

= 9.87 (given) ∴ 100 m = 9.87 m + 222 × 9.87 90.13 m = 222 × 9.87

∴ m = 222 9.8790.13

× = 24.3 u 91. Molecular mass = Vapour density × 2 = 22 × 2 = 44 92. Molecular mass = Vapour density × 2

Vapour density of A is 4 × M2

= 2M

∵ (Vapour density of A is four times that of B) ∴ Molecular mass of A = 2M × 2 = 4M 93. Contribution of 10B = 10.0 × 0.19 = 1.9 amu ….(i) Contribution of 11B = 11.0 × 0.81 = 8.91 amu ….(ii) Adding (i) and (ii) = 1.9 + 8.91 = 10.81 amu

Thus, the average atomic mass of boron is 10.81 amu. 99. Molecular mass of He = 4 Amount of He = 0.4 g

∴ Number of moles of He = 0.44

= 0.1 mole

∵ 1 mole of He = 6.023 × 1023 molecules

∴ 0.1 mole of He = 230.1 6.023 10

1× ×

= 6.023 × 1022 molecules 100. 14 g N3– ions have 8NA valence electrons ∴ 4.2 g of N3– ions have valence electrons

= A8N 4.214

×

= 2.4NA



101. 1 cm3 = 0.001 L ∴ 11.2 cm3 = 0.001 × 11.2 = 0.0112 L

Number of moles in 11.2 cm3 of H2 is = 0.011222.4

= 0.0005 mol 102. Molecular Mass of CH4 = 12 + 4 = 16 g mol−1 Amount of CH4 = 0.032 mg = 3.2 × 10−5g

Number of moles of CH4 = 53.2 10

16

−× = 2 × 10−6 mol

1 mole of CH4 contains 6.023 × 1023 molecules of CH4 ∴ 2 × 10−6 moles of CH4 contains = 2 × 10−6 × 6.023 × 1023 = 12.046 × 1017 mol 103. Ozone has molecular structure as O3 ∵ 1 mole of ozone = 48 g

∴ 0.5 mole of ozone = 0.5 481× = 24 g

104. 4 g of Helium gas = 22.4 dm3 of volume

∴ 2 g of Helium gas = 22.4 24

× = 11.2 dm3

105. Molecular mass of oxygen = 32 g/mol Amount of oxygen = 16 g

∴ Number of moles of oxygen = 1632

= 0.5 mole

∵ 1 mole of oxygen = 6.023 × 1023 molecules

∴ 0.5 moles of oxygen = 236.023 10 0.5

1× × = 3.011 × 1023 molecules

106. Number of S atoms = 6.023 × 1023 × 0.2 × 8 ≈ 9.63 × 1023

107. Gram atomic mass of Au = 197

Number of moles in 19700 g = 19700197

= 100

Number of Au atoms = 100 × 6.023 × 1023 = 6.023 × 1025

108. 1 mole ≡ 6.023 × 1023 electrons

Mass of 1 electron is 9.108 × 10−31 kg ∴ 6.023 × 1023 electrons mass is 6.023 × 1023 × 9.108 × 10−31 kg

Now, one kg of electron contains = 31 23

19.108 10 6.023 10−× × ×

mole

= 19.108 6.023×

× 108 mole 109. ∵ 22400 mL at NTP has 6.023 × 1023 molecule

∴ 1 mL at NTP has = 236.023 10

22400× = 0.0002688 × 1023 = 2.69 × 1019 mol



110. ∵ 2.24 L of gas has mass = 4.4 g

∴ 22.4 L of gas has mass = 4.42.24

× 22.4 = 44 g

So, given gas is CO2 because CO2 has molecular mass = 44 g 114. 1 mole of CH4 contains 4 mole of hydrogen atom i.e., 4 g atom of hydrogen. 115. 1 mole of K4[Fe(CN)6] ≡ 12 × 6 g of carbon

∴ 0.5 mole of K4[Fe(CN)6] ≡0.5 12 6

1× × g ≡ 36 g of carbon

116. ∵ 2 g of hydrogen = 6.023 × 1023 molecules

∴ 1 g of hydrogen = 236.023 10

2× = 3.012 × 1023 molecule.

117. Sodium oxide = Na2O Molecular mass = 46 +16 = 62 ∵ 62 g of Na2O = 1 mole

∴ 620 g of Na2O = 10 moles 118. 1 mole of sucrose contains 6.023 × 1023 molecules. ∵ 1 molecule of sucrose has 45 atoms

∴ 6.023 × 1023 molecule of sucrose has 45 × 6.023 × 1023 atoms/mole 119. 1 mole of P ≡ 6.023 × 1023 molecules ∵ 1 molecule of P4 ≡ 4 atoms

∴ 6.023 × 1023 molecules ≡ 4 × 6.023 × 1023 ≡ 24.092 × 1023 atoms 120. 1 mole of CO2

contains 6.023 × 1023 molecules ∵ 1 molecule of CO2 ≡ 3 atoms

∴ 6.023 × 1023 molecules of CO2 ≡ 3 × 6.023 × 1023 atoms ≡ 1.807 × 1024 atoms

121. Mass of 1 atom of hydrogen = 23

16.023 10×

= 1.66 × 10−24 g

122. 6.023 × 1023 molecules ≡ 28 g of CO

∴ 2.01 × 1023 molecules ≡23

23

28 2.01 106.023 10× ×

×g of CO ≡ 9.3 g of CO

123. 4 g of He ≡ 22.4 L of He at 0 °C and at 1 atm pressure

∴ Number of moles = Weight in g of HeMolecular mass of He

= 44

= 1 mole 124. 6.023 × 1023 molecules = 18 g of water

∴ 1 molecule of water = 23

18 16.023 10

××

= 3 × 10–23g = 3×10–26 kg 125. 6.023 × 1023 molecules of N2 ≡ 28 g of N2

∴ 1 molecule of N2 ≡ 23

28 16.023 10

××

≡ 4.65 × 10–23 g of N2



126. Mass of Molecule = 23

Molar mass in gram6.023 10×

= Molar mass in amu ×1.66 × 10−24 g = 842 × 1.66 × 10−24 (∵ Molar mass of C60H122 = 842)

= 1.4 × 10−21

127. 1 L of air = 210 cc of O2 ∵ 22400 cc = 1 mole

∴ 210 cc = 122400

× 210 = 0.0093 mol 128. 1 mole of H2 ≡ 22.4 L ∴ 2 mole of H2 ≡ 44.8 L

129. d = MV

; 1 = MV

or M = V; 18 g = 18 mL

∵ 6.023 × 1023 molecule of water has volume = 18 cc or cm3

∴ 1 molecule of water has volume = 23

186.023 10×

= 2.988 × 10–23 ≈ 3× 10–23 cm3

130. ∵ 22.4 L of a gas at STP has number of molecules = 6.023 × 1023

∴ 8.96 L of a gas at STP has number of molecules = 236.023 10 8.96

22.4× × = 2.409 × 1023 = 24.09 × 1022 mol

131. Na2SO4.10H2O = 2 × 23 + 32 + 4×16 + 10 × 18 = 46 + 32 + 64 + 180 = 322 g ∵ 322g Na2SO4.10H2O contains 224g of oxygen

∴ 32.2 g Na2SO4.10H2O contains 32.2 224322

× = 22.4 g of oxygen 132. Molecular mass of BaCl2.2H2O ≡ 137 + 35.5 × 2 + 2 × 18 ≡ 244 g ∵ 244 g of BaCl2.2H2O ≡ 2 mole of water

∴ 488 g of BaCl2.2H2O ≡ 488 2244

× ≡ 4 moles of water 133. Gram−molecular mass of CuSO4.5H2O = 250 g Number of molecules in 250 g (one mole) of CuSO4.5H2O = 6.023 × 1023

Let the mass of 1 × 1022 molecule of CuSO4.5H2O = x g

So, 22

23

1 106.023 10

××

= 250

x or

x = 22

23

250 1 106.023 10

× ××

x = 4.151 g 134. Molecular mass of CuSO4.5H2O ≡ 249.5 g ∵ 249.5 g of CuSO4.5H2O ≡ 90 g of H2O

∴ 1000×103 g of CuSO4.5H2O ≡610 90

249.5× g of H2O ≡ 360721 g of H2O ≡ 360.7 kg of H2O

135. 16 g of CH4 = 1mole = 6.023 × 1023 molecules



136. d = MV

(d = density, M = mass, V = volume)

Since d = 1 So, M = V 18 g = 18 mL ∵ 18 mL or 18 g of water = NA molecules (NA = avogadro’s number)

∴ 1000 mL = AN18

× 1000 = 55.55 NA mol 137. ∵ 3 moles of oxygen is present in 1 mole of BaCO3

∴ 1.5 moles of oxygen is present in BaCO3 = 13

× 1.5 = 12

= 0.5 mol 138. ∵ (31 × 4 = 124)g of P is present in 220g of P4S3

∴ 1.24 g P is present in = 220124

× 1.24 = 2.2 g of P4S3

139. 2 g of oxygen contains atom = 2 116 8

= mole

Similarly, 4 g of sulphur = 4 132 8

= mole 140. 200 mg of CO2 = 200 × 10–3 = 0.2 g ∵ 44 g of CO2 = 6.023 × 1023 molecules

∴ 0.2 g of CO2 = 236.023 10

44×

× 0.2 = 0.0274 × 1023 = 2.74 × 1021 molecules

Now 1021 molecules are removed, So, remaining molecules = 2.74 × 1021 – 1021 = 1021(2.74 – 1) =1.74×1021 molecules

Now, 6.023 × 1023 molecules = 1mole

1.74 × 1021 molecules = 21

23

1 1.74 106.023 10× ×

× = 0.289 × 10–2 = 2.89 × 10–3

moles 146. ∵ 40 g of NaOH contains 16 g of oxygen

∴ 100 g of NaOH contains = 1640

× 100 = 40 % oxygen 147. Urea- (NH2 – CO – NH2) ∵ 60 g of urea contains 28 g of nitrogen

∴ 100 g of urea contains 2860

× 100 = 46.66 %

≈ 46 % 148. 60g of urea contains 12g of carbon

100g of urea contains 1260 × 100 = 20%

149. In Fe(CNS)3.3H2O

% of H2O = 3 18284× × 100 = 19 %



150. 2(NH4)2HPO4 ≡ P2O5 2(36 + 1 + 31 + 64) = 62 + 80 264 = 142 (Molecular mass (Molecular mass of of the salt) P2O5)

% of P2O5 = 2 5Molecular mass of P OMolecular mass of salt

× 100 = 142264

× 100 = 53.79 % 151. Molecular mass of (CHCOO)2Fe = 170

∴ Fe present in 100 mg of (CHCOO)2Fe = 56170

× 100 mg = 32.9 mg

32.9 mg of Fe is present in 400 mg of capsule

∴ % of Fe in the capsule = 32.9400

× 100 = 8.225 ≈ 8 % 152. Molecular mass of CH3COOH ≡ 60 ∵ 60 g of CH3COOH ≡ 12 × 2 g of carbon

∴ Percentage of carbon ≡ 12 260× × 100 ≡ 40 %

Similarly, Molecular mass of C6H12O6 ≡ 180 ∵ 180 g of C6H12O6 ≡ 12 × 6 g of carbon

∴ Percentage of carbon ≡ 12 6180

× × 100 ≡ 40 %

Therefore, both the compounds have same percentage of carbon. 153. Molecular formula = n(Empirical formula) = 6(CH2O) = C6H12O6 (Glucose) ∴ Empirical formula = CH2O 155. C = 24 g, H = 4 g, O = 32 g Hence number of moles is

2412

= 2; 41

= 4; 3216

= 2

So, Molecular formula = C2H4O2 So, Empirical formula = CH2O (Simplest formula) 156. CO2 has same molecular and empirical formula. 157. n = Molecular mass 78

Emperical formula mass 13= = 6

Molecular formula = n (Empirical formula) = 6 (CH)= C6H6 158. Empirical formula of the acid is CH2O2 (Empirical formula)n = Molecular formula

n = whole number multiple i.e. 1,2,3,4...... If n = 1; then the molecular formula will be CH2O2



159. ∵ 0.0835 mole of compound contains 1 g of hydrogen

∴ 1 mole of compound contains = 10.0835

= 11.97 ≈ 12 g of hydrogen.

∵ 12 g of hydrogen is present only in C6H12O6 ∴ Molecular formula = C6H12O6 160.

Element Amount (g) Atomic ratio Simplest ratio

I 254 254127 = 2

22 = 1 × 2 = 2

O 80 8016 = 5

52 = 2.5 × 2 = 5

∴ Molecular formula of compund is I2O5 161. As percentage of Cu = 39.62 % S = 20.13 % ∴ Percentage of O = 100 – (39.62 + 20.13) = 40.25 %

Element % of Element At. Mass Atomic ratio Simplest ratio

Cu 39.62 63 39.62 0.63

63= 0.63 1

0.63=

S 20.13 32

20.13 0.6332

=0.630.63

= 1

O 40.25 16

40.25 2.5116

= 2.510.63

= 4

∴ Empirical formula = CuSO4

∵ n = Molecular massEmpiricalformula mass

= 159159

= 1

∴ Molecular formula = (CuSO4) × n = (CuSO4) × 1 = CuSO4 162.

Elements % Composition Atomic Mass Atomic ratio Simplest ratio

A 25 12.5 25

12.5= 2

22

= 1

B 75 37.5 75

37.5= 2

22

= 1

Hence, the simplest formula of the compound is AB (1 : 1). 163. % of X = 75.8 ∴ % of Y = 100 – 75.8 = 24.2

Thus, empirical formula of the compound is XY2

Hence, the Molecular formula of the compound = XY2

Element % Composition Atomic ratio Simplest ratio

X 75.8 75.875

= 1.011 1.0111.011

= 1

Y 24.2 24.216

= 1.513 1.5131.011

= 1.5 ≈ 2



164. % of oxygen ≡ 40 % % of metal (M) ≡ 100 – 40 ≡ 60 %

Element % Composition Atomic ratio Simplest ratio

Oxygen 40 4016 = 2.5

2.52.5 = 1

Metal (M) 60 6024 = 2.5

2.52.5 = 1

∴ Empirical formula of the oxide is MO (1 : 1) 165. Percentage of oxygen in M3O4 = 27.6 % ∴ Percentage of M in M3O4 = 100 – 27.6 = 72.4 %

Now, to find out the formula of second oxide, we have to calculate the atomic mass of metal (M). Let the atomic mass of metal (M) be x ∴ Molecular mass of M3O4 = x + 16 × 4 = x + 64

∴ Percentage of M = 64+

xx

× 100 = 72.4 (calculated)

∴ 72.4 = 64+

xx

× 100

∴ 100x = 72.4x + 64 × 72.4 ∴ 27.6x = 64 × 72.4

∴ x = 64 72.427.6× = 168

As, number of metal (M) in M3O4 = 3

∴ Atomic mass of M = 1683

= 56

Now, we know that percentage of oxygen in second oxide is 30 % ∴ Percentage of metal (M) = 100 – 30 = 70 % ∴ Formula of second oxide is given as follows:

Elements % Composition Atomic ratio Simplest ratio

M 70 7056

= 1.25 1.251.25

= 1×2 = 2

O 30 3016

= 1.875 1.8751.25

=1.5×2=3 ∴ Simplest formula of the second oxide is M2O3 173. 2K2CrO4 + 2HCl ⎯→ K2Cr2O7 + 2KCl + H2O 174. Ca3P2 + 6H2O ⎯→ 2PH3 + 3Ca(OH)2

175. A + 2B ⎯→ C ∵ 2 mole of B ≡ 1 mole of C

∴ 8 mole of B ≡ 8 12× ≡ 4 mole of C

Calcium Phosphide

Phosphine



176. 2NH3 + 52

O2 ⎯→ 2NO + 3H2O

∵ (2 × 17) g of NH3 ≡ 52

mole of O2

∴ 6.8 g of NH3 ≡ 6.8 52 17 2

×× ×

mole of O2 ≡ 0.5 mole of O2

177. K2Cr2O7 + 6FeSO4 + 7H2SO4 ⎯→ 3Fe2(SO4)3 + Cr2(SO4)3 + 7H2O + K2SO4 ∴ 1 mole of K2Cr2O7 oxidizes 6 mole of FeSO4 completely. 178. 4Al + 3O2 ⎯→ 2Al2O3 (108 g) (96 g) (204 g) ∵ 108 g Al reacts with 96 g of O2

∴ 27 g Al will react with = 96 27108× = 24 g of O2

179. 24 g of Mg ≡ 1 mol of MgO

∴ 1.2 g of Mg forms ≡ 1.2 124

× ≡ 0.05 mol of MgO 180. 3BaCl2 + 2Na3PO4 ⎯→ Ba3(PO4)2 + 6NaCl ∵ 2 mol of Na3PO4 ≡ 1 mol of Ba3(PO4)2

∴ 0.2 mol of Na3PO4 ≡ 0.2 12×

≡ 0.1 mol 181. 1 mole of ethanol (C2H5OH) completely burns to form 2 mole of carbon dioxide and 3 mole of water ∵ 1 mole of carbon dioxide ≡ 44 g

∴ 2 mole of carbon dioxide ≡ 88 g

182. % C = CO2W12

44 W× × 100 = 12 2.63

44 0.858× × 100 = 83.6%

% H = H O2W2 100

18 W× × = 2 1.28

18 0.858× × 100 = 16.6 %

Element % (A) At.wt. (B) AB

Ratio Simplest ratio

C 83.6 12 6.97 1 × 3 ≈ 3 H 16.6 1 16.6 2.38 × 3 ≈ 7

∴ Molecular formula is C3H7

∴ C3H7 = (12 × 3) + (7 × 1) = 36 + 7 = 43 g 183. 1.12 mL is obtained from 4.12 mg ∴ 22400 mL will be obtained from

4.121.12

× 22400 = 82400 mg = 82.4 g

ROH + CH3MgI ⎯→ CH4 + Mg OR

I1mol 1mol = 22400mL



184. N2 + 3H2 ⎯→ 2NH3 (28 g) (6 g) (34 g) 6 g of H2 produces 34 g of NH3

∴ 200 g of H2 produces = 34 200

6×

= 1133.3 g

185. BCl3 + 32

H2 ⎯→ B + 3HCl;

(Number of moles of B) = 21.610.8

= 2 mole

B ≡ 32

H2

For 1 mole of B ≡ 32

mole of H2

∴ For 2 mole of B ≡ 2 32× ≡ 3 mole of H2

∴ V = 3 × 22.4 L = 67.2 L 186. Mg+2 ≡ H2 Mg2+ + 2HCl ⎯→ MgCl2 + H2↑ ∵ 24 g of Mg = 1 mole of H2

∴ 12 g of Mg = 1224

= 12

mole of H2

187. Hence, volume of hydrogen evolved is

32

× 22.4 = 33.6 L

188. CaCO3 ∆⎯⎯→ CaO + CO2 (94) (56) (44) 1 mol of CaCO3 ≡ 56 g of CaO

∴ 3 mole of CaCO3 ≡ 3 561

× ≡ 168 g of CaO

189. N = W(g) 1000V Eq.wt.

××

0.1 = W(g) 10001500 40

××

∴ W = 0.1 1500 401000

× ×

∴ W = 6 g 190. HNO3 + KOH ⎯→ KNO3 + H2O

12.663

= 0.2 mole; HNO3 ≡ KOH

∵ 0.2 mole of HNO3 ≡ 0.2 mole of KOH ∴ 0.2 × Molecular mass (KOH) ≡ 0.2 × 56 ≡ 11.2 g of KOH

H2O + Al + NaOH ⎯→ NaAlO2 + 32

H227 g



191. N = (g)W 1000V Molecular mass

×

×

W(g) = 0.5 500 1001000

× × = 25 g 192. CaCl2 + Ag+ ⎯→ 2AgCl + Ca2+ (110) (108) (2 × 143.5) (40) ∴ 110 g of CaCl2 ≡ 2 × 143.5 g of AgCl ∴ For 14.35 g of AgCl,

Mass of CaCl2 required ≡ 110 14.352 143.5

××

≡ 5.5 g 193. 1 mole of H2SO4 ≡ 32 g of S

∴ 100 moles of H2SO4 ≡ 100 321×

≡ 3.2 × 103 g of S 194. 50 % HCl itself means 50 g HCl reacts with 100 g sample

∴ % Purity = 50100

× 100 = 50 % 195. ∵ 22.4 L of O2 ≡ 48 g of O3

∴ 67.2 L of O2 ≡ 48 67.222.4×

≡ 144 g of O3 203. 22.4 L ≡ 16 g of CH4

∴ 4.48 L ≡ 16 4.4822.4×

≡ 3.2 g of CH4

206. 1 mole of P4 = 6.023 × 1023 atoms ∴ 0.1 mole of P4 = 6.023 × 1023 × 0.1 atoms = 6.023 × 1022 atoms ∴ 0.1 mole of P4 = 6.023 × 1022 × 4 atoms = 2.4 × 1023 atoms 207. n = Mass

Molar mass=

3 1

1 kg18×10 kg mol− −

= 55.56 mol

208. 16 g of O2 has number of moles = 16 132 2

=

14 g of N2 has number of moles = 14 128 2

=

Number of moles are same, so number of molecules are same. 209. 4Al + 3O2 ⎯→ 2Al2O3 (108 g) (96 g) (204 g)

∵ 54 g Al will react with 96 54108

× = 48 g of O2

210. (A) 34 g of water : ∵ 18 g H2O = 6.023 × 1023 molecules

∴ 34 g H2O = 236.023 10

18×

× 34

= 11.38 × 1023 molecules



(B) 28 g of CO2 : ∵ 44 g CO2 = 6.023 × 1023 molecules

∴ 28 g CO2 = 236.023 10

44×

× 28 = 3.83 × 1023 molecules (C) 46 g of CH3OH : ∵ 32 g CH3OH = 6.023 × 1023 molecules

∴ 46 g CH3OH = 236.023 10

32×

× 46 = 8.658 × 1023 molecules (D) 54 g of N2O5 : ∵ 108 g of N2O5 = 6.023 × 1023 molecules

∴ 54 g of N2O5 = 236.023 10

108×

× 54 = 3.01 × 1023 molecules. 211. (A) 2 g atom of nitrogen = 28 g (B) 6.023 × 1023 atoms of C has mass = 12 g

∴ 3 × 1023 atoms of C has mass = 23

23

12 3 106.023 10

× ××

= 6 g

(C) 1 mole of S has mass = 32 g (D) 7.0 g of Ag So, lowest mass = 6 g of C 212. (A) 6.023 × 1021 molecules of CO2 : Number of atoms = 3 × 6.023 × 1021 = 18.069 × 1021 atoms (B) 22.4 L of CO2 : Number of atoms = 6.023 × 1023 × 3 = 18.069 × 1023 atoms (C) 0.44 g of CO2 :

Number of moles = 0.4444

×6.023×1023moles

= 6.023 × 1021 moles = 3×6.023× 1021 atoms = 18.069 × 1021 atoms 213. ∵ 40 g of caustic soda (NaOH) contains 6.023 × 1023 ions of Na

∴ 4 g of NaOH = 234 6.023 10

40× × = 6.023 × 1022 ions

Similarly, for 100 mL of 0.5 M Na2SO4 ∵ 1000 mL of 1M Na2SO4 = 2 × 6.023 × 1023 ions of Na

∴ 100 mL of 0.5M Na2SO4 = 23100 0.5 2 6.023 10

1000 1× × × ×

× = 6.023 × 1022 ions

214. (A) 2.5 g of N2 molecules : ∵ 28 g of N2 ≡ 6.023 × 1023 molecules

∴ 2.5 g of N2 ≡ 236.023 10 2.5

28× × ≡ 5.4 × 1022 molecules



(B). 4 g of N atoms : ∵ 14 g atom of N ≡ 6.023 × 1023 molecules

∴ 4 g atom of N ≡ 236.023 10 4

14× ×

≡ 17.2 × 1022 molecules

(C) 40 g of N atoms : 14 g atom of N ≡ 6.023 × 1023 molecules

40 g of N atoms ≡ 236.023 10 40

14× ×

≡ 172.1 × 1022 molecules

(D) 84 g of dinitrogen : ∵ 28 g of dinitrogen ≡ 6.023×1023 molecules

∴ 84 g of dinitrogen ≡ 236.023 10 84

28× × ≡ 180.7 × 1022 molecules

∴ Maximum number of molecules are present in 84 g of dinitrogen. 215. (A) 11 g of CO2 : 44 g of CO2 ≡ 6.023 × 1023 atoms

11 g of CO2 = 236.023 10 11

44× × = 1.505 × 1023 atoms

(B) 4 g of H2: 1.008 g of H2 ≡ 6.023 × 1023 atoms

4 g of H2 = 236.023 10 4

1.008× × = 23.900 × 1023 atoms

(C) 5 g of NH3 : 17 g of NH3 ≡ 6.023 × 1023 atoms

5 g of NH3 = 236.023 10 5

17× × = 1.77 × 1023 atoms

(D) 8 g of SO2 : 48 g of SO2 ≡ 6.023 × 1023 atoms

8 g of SO2 = 236.023 10 8

48× × = 1.0 × 1023 atoms

216. 44 g of CO2 ≡ 6.023 × 1023 molecules

∴ 4.4 g of CO2 ≡ 236.023 10 4.4

44× × ≡ 0.6023 × 1023 molecules

Similarly, 22.4 L of H2 ≡ 6.023 × 1023 molecules

∴ 2.24 L of H2 ≡ 236.023 10 2.24

22.4× ×

≡ 0.6023 × 1023 molecules

∴ Total number of molecules present in the container is = 0.6023 × 1023 + 0.6023 × 1023 = (0.6023 + 0.6023) × 1023 = 1.2046 × 1023 molecules 217. Equivalent of KMnO4 = Equivalent of Fe(C2O4) x × 5 = 1 × 3 x = 0.6 mol



218. ∵ 28 g of N2 ≡ 22.4 L of N2

∴ 14 g of nitrogen ≡ 22.4 1428

× ≡ 11.2 L of N2

220. Number of molecules = n × A

n = MassMolar Mass

Mass of 1 mL of water = 1g

∴ n = 118

∴ Number of molecules in 1 mL (20 drops) of water = 118

× A

∴ Number of molecules in 1 drop of water = 118 20×

× A = 0.05A18

mol 221. Equivalent mass = Molecular mass (M)

Number of e gain or lost−

When KMnO4 is converted in K2MnO4 transfer of 1 electron takes place.

∴ Equivalent mass = M1

= M 222. n =

A

Number of moleculesN

Also, n = Volumeof gasat N.T.P.22.4L

∴ A

Number of moleculesN

= Volumeof gasat N.T.P.22.4L

∴ 236.023 10×x =

101.12 10 L22.4L

−×

∴ x = 10 231.12 10 6.023 1022.4

−× × × = 3.01 × 1012 mol 223. Molecular mass of nitrogen = 14 Molecular mass of oxygen = 16 ∴ Ratio = 14 : 16 = 7 : 8

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