Unit: The standard of reference chosen to measure any physical quantity is called unit. For example kilogram is the unit used to express the mass. Kilogram is defined as The mass of Pt-Ir cylinder stored in air-tight glass jar at International Bureau of weights and measures in France. SI unit: The international system of units popularly known as S. I. units was adopted by the General Conference of Weights and Measures in 1960.Physical quantities: All quantities which can be measured either directly or indirectly are called physical quantities.

Or Properties such as mass, length, time, temperature, volume etc. are expressed in numerals with suitable units are generally called physical quantities. Derived units: A large number of physical quantities can always be expressed as some combination of the fundamental units. Any unit obtained by multiplying or dividing by another fundamental unit is known as Derived units. Significant Figures: The total number of digits in a number in the last digit whose value is uncertain is called the number of significant figures. Precision: Precision is a measure of how closely individual measurements agree with the correct value or true value. Accuracy: Accuracy refers to how closely individual measurements agree with the correct value or true value. OR Accuracy means the closeness of a single measurement to the correct value of the quantity measured. Law of chemical combination:- In the seventeenth century, scientists had been trying to find out methods for converting one substance into another. During their quantitative studies of chemical changes they made certain generalizations. These generalizations are known as laws of chemical combination. Law of conservation of mass: It states that, during any physical or chemical change the total mass of the products is equal to the total mass of the reactants. In other words, matter is neither created nor destroyed during any physical or chemical change. Law of constant composition: It states that a pure chemical compound always contains same elements combined together in the same definite proportion by weight. OR The sample of a pure compound whatever its source, always consists of the same elements combined in the same proportions by mass. Law of multiple proportion: It states that when two elements combine to form two or more than two compounds, the weights of one of the element which combines with a fixed weight of the other, bear a simple whole number ratio Law of Reciprocal Proportion: (Law of combining weights) It states that, when two different elements combine separately with the same weight of the 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. OR If masses of two elements which separately react chemically with identical masses of a third are also the masses, which react with each other or simple multiple of them. Atom: The simplest particle of an element which may or may not have independent existence and posses its entire characteristic is called an element. Molecule: The smallest- particle of a substance which is capable of independent existence is called a molecule. Atomic Mass:

BASIC CONCEPT OF CHEMISTRY

1

The atomic mass of an element is the number of times an atom at that element is heavier than an atom of carbon taken as 12 amu. One atomic mass unit (a.m.u.): It is defined as the quantity of mass equal to 1/12 of the mass of an atom of carbon (C12). Molecular mass: It is defined as the average relative mass of its molecule as compared to the mass of an atom carbon (C12) taken as 12 amu. Mole: The amount of substance which contains the same number of elementary particles as the number of atoms present in 12 gm. of carbon (C12). . Chemical Stoichiometry: In order to study chemical compounds in the laboratory it is necessary to have knowledge of the quantitative relationship that exists among the, amount of the substance that take part in the chemical reaction. Stoichiometry is derived from Greek words stoichion = element and metrou = measure is the term we use to refer all the quantitative aspects of chemical compounds and reactions. Normality: Number of gram equivalent weight of solute dissolved in one litre solution at normal temperature is called normality of the solution. One normal solution: If 1 gram equivalent of a solute substance is dissolved in one litre of a solution at normal temperature is called a normal solution. Molarity: The number of moles of the solute dissolved in one liter of solution at normal temperature is called molarity of the solution. What is Chemistry?

Chemistry is an important branch of science. Chemistry deals with the matter that form our environment and the transformation that the matter

undergoes. It is a science of atoms and molecules. Chemistry is a study of composition , structure and properties of the matter.

Organic, inorganic, physical, analytical, industrial, and bio-chemistry are various branches of chemistry. Q. What are fundamental and derived units? Explain derived units with example.

All quantities which can be measured either directly or indirectly are called physical quantities can always be expressed as some combination of the units which are known as fundamental units.

Any units obtained by multiplying or dividing by another fundamental unit are known as derived units. Volume = length x breadth x height

Now the S.I unit for length is metre. Volume = metre metre metre = metre3 = m3 Thus m3 is S.I unit of volume and it is derived unit.

= =

Q. Give the list of some non S.I units which are still widely used in chemistry. (1) Angstrom (A0) for length (2) atmosphere atm for pressure (3)Litre for volume (4) Celcius C0 for temperature.

Q. Write a note on SI units. Earlier there were several units used for expressing the same quantity. For example mile, furlong, foot etc.

were used to measure a distance, pound sher, chatank etc. for a weight. This created difficulty in everyday affair as there was no uniformity.

To overcome such problem in 1971 the French Academy of Sciences devised a simple system which is now known as the metric system. This system of measurement immediately becomes popular among the scientific community throughout the world.

This system is still in use in practice but as this system has several units to express the same quantity. Therefore in 1960, the General conference of weights and measures, adopted the international system of units or SI units.

This system has seven SI base units.

In SI, large and small quantities are expressed by using an appropriate prefix with the base units.e.g. The meter (m) is the SI base unit of length. By combining it with one of the SI prefixes we can get a unit of appropriate size. For instance, national highway distances are given in kilometers (km.) means 1000 or 103

metres(m). The second (s) is the SI base unit of time. Combining this unit with prefixes such as milli, micro, nano, etc.

we can create units of time for measuring very fast changes. In S I unit system other units are derived with the help of seven base units for example unit for volume can

be derived as follows. Volume = length x breadth x height = metre x metre x metre = metre3 = m3 A modern high-speed computer takes roughly a nanosecond in adding two 10- digit numbers.

What are significant figures? Explain the rules to determine them.

The digits in a properly recorded measurement are known as significant figures. The greater the number of significant figures in a reported result, smaller is the uncertainty and greater the

precision. Significant figures are those figures or digits in a measured number that include all certain digits plus a final

one that is some uncertain. The last digits with some uncertainty in its called the least significant digit. Rules for Determining significant figures:- All the digits in a number not containing zero are significant figure. e.g. the number of significant figures in

46, 4.6, 4.62 and 4.635 are respectively 2,2,3 and 4. In any number the digit zero occurring between two digits is always considered as a significant figure. e.g.

the significant figures in the numbers 2.03 , 204 and 6.053 are 3,3 and 4 respectively. In any number, zero located between the figure al1.d the decimal point or zeros located before the decimal

point are not considered for significant figures e.g. the significant figures in the numbers 0.15, 0.015, 0.0015 is 2 in all the cases.

Zeros placed after a decimal point in an integer are considered in fixing significant figures. They indicate precision of the measurement and instrument used e.g. the significant figures in numbers 23.0, 23.00 and 23.000 are respectively 3,4 and 5.

Explain units and dimensional analysis with example. Sometimes we are required to convert one set of units into other units that may or may not be SI units e.g.

distance reported in miles, bond length reported in Angstrom (AO) and weight reported in pounds are non-SI units and a convenient way of their conversions in S.I unit is the unit factor method.

Here we write the units with every number and carry the units through the calculation, treating them as algebraic quantities. For inter conversion of the units of times(minutes and seconds) we know the basic relationship

1 min = 60 s. which may be written as, 60s 1 min11 min 60s

= =

These are the unit factors to convert min into second or second into min. Such equalities are known as unit factor because the overall effect of multiplication by such factor is equals

to multiplication by one. Thus multiplication of an original quantity by this factor does not change the magnitude of the quantity. Only numeric value and units are changed in such process. e.g. to find the

number of seconds in 5.0 min 60s

5 min 5.0 min 300s1 min

= =

Physical quantity Name of the unit Symbol Mass Kilogram kg Length Metre m Time Second s Temperature Kelvin K Electric current Ampere A Luminous intensity Candela Cd Amount of substance Mole Mol

By using this method we can inter convert Angstrom into meters. e.g. we can convert 5 A0 into meters as

under.

0 101 10A m=

10 0

0 10

10 m 1Aunit factor1

1A 10 m

= =

0 10

0 10

0

5 A 10 m5 A 5 10 m

1A

= =

Explain what are element, mixture and compound. An element is the simplest form of a pure substance. It may be defined as the simplest form of a pure

substance which can neither be prepared form nor be decomposed into simpler form of matter by ordinary physical or chemical methods. An element is consists of only one kind of atoms.

There are 109 elements known at present. Out of these 96 occur naturally and remaining have been prepared by or detected in nuclear reactions. Some of common elements are oxygen, silicon, aluminum, carbon, nitrogen etc.

A compound may be defined as a substance which is formed by the combination of two or more than two elements in a definite proportion by mass and into which it may be decomposed by suitable chemical methods. e.g. water always contains hydrogen and oxygen in the ratio of 1:8 by mass. Similarly CO2 always contains carbon and oxygen in the ratio of 3:8.

A mixture is a system in which two or more elements or compounds or both are simply mixed together in any proportion. For example air is a mixture of gases like oxygen, nitrogen, CO2, water vapour, argon and dust particles etc. Similarly, gun powder is a mixture of carbon, sulphur and potassium nitrate.

Thus in a mixture, each of the constituents retain its characteristic property e.g.sugar retains its characteristic sweetness in aqueous solution mixture. A mixture may be (1) homogeneous or (2) heterogeneous.

Homogeneous mixtures have same compositions throughout. Sugar forms homogeneous mixture with water. On the contrary the mixture of grains of sand and salt is heterogeneous.

Law of constant composition: (Proust 1779) A pure compound has the same composition and its every molecule contains the same number of atoms of

each kind. Thus a chemical compound always consists of the same elements combined together in the same proportion by mass.

This law states that a sample of a pure compound whatever its source, always consists of the same elements combined in the same proportions by mass.

For example one molecule of water always contains 2.016 g of hydrogen and 16.00 g. of oxygen. According to the theory of atoms, molecules are made up of atoms and the relative number of atoms of its

constituent elements for a given compound is same in each of its sample. As the mass of each atom of a given element is same, the mass of its each molecule will always be same. Thus a definite composition of a compound provides an experimental base for the atomic theory. Limitations:

Physical Classification Chemical Classification

Classification of Matter

soilid gas liquid

Mixture Elements Compound

Homogeneous Heterogeneous

This law is not true if same compound is obtained from isotopes of an element because mass of each

isotope is different. This law also not applicable to non stoichiometry compounds because they have variable composition. Law of multiple proportions: ( Dalton 1803) If two elements combine to form more than one compound then for a given mass of an element, the

masses of the other element in two or more compounds are in the ratio of small integers. For example nitrogen and oxygen can combine to form different compounds. i.e. oxygen as 8: 16 : 24 : 32 :

40 i.e. 1 : 2 : 3 : 4 : 5

Law of combining weights: ( Richter 1792) "If masses of two elements which separately react chemically with identical masses of a third are also the

masses which react with each other or simple multiple of them." Equivalent wt. of an element is either equal to its atomic wt. or simple multiples of it. e.g. the equivalent

masses of hydrogen, oxygen and chlorine are 1.008, 8.00 and 35.5 g m respectively. According to the law of combining weights 35.5 gm of chlorine reacts with 1.008 gm of hydrogen and

8.000gm of oxygen respectively. If three elements A, B and C combine to form three compounds AB, BC and CA, the ratio of the masses of B

and C which combine with a fixed mass of A is either same or simple multiple of the masses of B and C when they combine with each other to form BC.

For example H, O and S combine to form three compounds water (H2O), sulphur dioxide (SO2) and hydrogen sulphide (H2S).

Copmpound Weight of H Weight of S Weight of O H2O 2g 16g Ratio of S:O with 2 g of H

2:1 H2S 2g 32g SO2 32g 16g Ratio of S:O in SO2 1:1

Explain Mole concept.

In a sample of every element there are innumerable atoms. As atoms are very small particles therefore single atom or molecule cannot be considered as unit so chemist have adopted mole as the chemical counting unit.

The number of molecules, atoms or ions in one mole will be 6.022 1023. One mole of oxygen gas contains 6.022 1023 molecules of oxygen and 2 6.022 1023 atoms of oxygen as it

is di-atomic gas. As one mole contains a very large number of entities (atoms molecules ions etc.) chemist have preferred a

mole as a convenient unit for expressing the amount of a substance. For example the number of molecules of oxygen present in 32 g of oxygen is considered as one mole as a

standard. This number is known as Avogadro constant. In order to determine this number precisely atom of carbon 12 is considered as reference sample and its

mass was determined by a mass spectrometer and it is 1.992648 10-23 g. knowing that 1 mole of carbon weights 12g.

No.of atoms in 12g C12 12

2323 12

12g / molC6.022137 10

1.992648 10 g / C atom= =

atoms / mol

= 6.022 1023 atoms / mol

Oxide of nitrogen

Mass of nitrogen Mass of oxygen Fixed mass of nitrogen Fixed mass of oxygen

N2O 28 16 14 8 NO 14 16 14 16 N2O3 28 48 14 24 N2O4 28 64 14 32 N2O5 28 80 14 40

weight of element .

atomic weightno.of mole = weight of sub.

molar massno.of mole =

23no.of particles of substance

6.022 10no.of mole

=

vol of gaseous substance at S.T.P.22.4 lit

no.of mole =

Explain atomic mass, molecular mass and molar mass. Dalton gave the idea of atomic mass in relative terms, that is, the average mass of one atom relative to the

average mass of the other. The relative atomic masses were referred to as the atomic weights. On the present atomic mass scale, the carbon- 12 isotope is chosen as the standard and arbitrarily assigned

a mass of exactly12 atomic mass unit. One atomic mass unit (amu) is therefore, a mass equal to exactly 1/12 the mass of carbon-12 atom.

The average relative atomic mass depends upon the isotopic composition of that particular element. For example there exists two isotopes of chlorine in nature with atomic masses 35 and 37, their relative abundance being,3 : 1

The average atomic mass of chlorine is = ( 35 3 ) ( 35 1 )3 1

+

+ = 35.5 amu.

Average atomic mass = fractional abundance Mass

fractional abundance = % abundance 1

100

Molecular mass is calculated relative to the mass of carbon 12. The atomic weight unit is thus 1/12 of carbon-12. Thus one molecule of carbon dioxide is 44 times heavier than the weight of one atom of carbon-12. So its molecular wt. is 44.

"Molecular mass is the sum of the atomic masses of all atoms in it." The molar mass means the formula mass of a compound is the sum of the atomic masses of all the atoms in

formula unit of the compound. e.g. molar mass of H2O is (2+16)= 18g. Similarly molar masses of H2S04, HCl and KCl are 98, 36.5 and 74.5g.

The average atomic mass of chlorine is 35.5 Why? The average relative atomic mass depends upon the isotopic composition of that particular element.

Chlorine has two isotopes with atomic masses 35 and 37. Their relative abundance being 3 : 1 ,So the average atomic mass of chlorine will be =35.5 amu.

Gay Lussacs Law of Combining Volumes: French Chemist Gay-Lussac experimented with reactions of gases and proposed the following law: When

gases react with each other they do so in volumes which bear simple whole number ratio to one another and to the volumes of products. If there are also gases, provided all volumes are measured under similar conditions of temperature and pressure. Gay-Lussacs discovery of integral ratio in volume relationship is actually the law of definite proportions by volumes.

+ ( vapour)hydrogen oxygen water2 vol 1vol 2 vol

Scientist Berzelius tried to correlate Daltons atomic theory with Gay Lussacs observation and gave his hypothesis.

Equal volume of different gases contains same number of atoms under identical condition of temperature and pressure.

+ ( vapour)hydrogen oxygen water2 vol 1vol 2 vol2natom natom 2natom2atom 1 2atom 1atom

Thus according to above calculation there is atom of oxygen in 1 compound atom of water which is not appropriate as per Daltons atomic theory. This problem was solved by Avogadro he made clear distinction between two ultimate particles of matter. (1) atom (2) molecule.

According to Avogadro atom is the ultimate particle of an element which takes part in the chemical reaction but it may or may not be capable of independent existence. Molecule is the ultimate particle of element or compound which is capable of independent existence.

Equal volume of different gases contains same number of molecules under identical condition of temperature and pressure.

+ ( vapour)hydrogen oxygen water

2 vol 1vol 2 vol2nmolecule nmolecule 2nmolecule2molecule 1 2 molecule 1molecule

Thus 1 molecule of water contain molecule of oxygen i.e. 1 atom of oxygen. Avogadros hypothesis has following applications.

(i) Provides a method to determine the atomic weight of gaseous elements. (ii) Provides a relationship between vapour density (V.D.) and molecular masses of substances

Molecular mass vapour density= 2 (iii) It helps in the determination of mass of fixed volume of a particular gas.

Mass of 1 ml gas = V.D. 0.0000897 gm. (iv) It also helps in the determination of molar volume at N.T.P.

V.D. 0.0000897 gm. gas has volume = 1 ml 2 V.D.(i.e., molecular mass) gm. has

Molar mass of a gas or its 1 mole occupies 22.4 L volume at S.T.P. (v) It helps in determination of molecular formulae of gases and is very useful in gas analysis. By knowing

the molecular volumes of reactants and products of reaction, molecular composition can be determined easily. Equivalent Mass (E)

The Number of parts by weight of the substance which combines with (or) displaces 1.008 parts by weight of hydrogen or 8 parts by weight of oxygen or 35.5 parts by weight of chlorine is called equivalent weight.

The equivalent mass of different substances can be calculated as summarized below Substances Equivalent Mass Elements

Atomic Mass

Valency

Acids

Molecular MassBasicity* of acid

Bases

Molecular MassAcidity** of base

Salts ( ) ( )

Formula MassChange on metal atom Number of metal atoms

Reductant

Formula MassNo. of e it gains

Oxidant

Formula MassNo. of e it loses

*Basicity is the number of replacable H atoms per molecule of acid. **Acidity is the number of OH ions furnished by each molecule of the base.

(i) Weight to weight percent Wt. of solute 100

Wt. of solutio% /

nw w =

Example : 10% Na2CO3 solution w/w means 10 g of Na2CO3 is dissolved in 100g of the solution. (It means 10 g Na2CO3 is dissolved in 90 g of H2O)

(ii) Weight to volume percent Wt. of solute 100

Volume of solutio% /

nw v =

Example : 10% Na2CO3 (w/v) means 10g Na2CO3 is dissolved in 100 cc of solution.

(iii) Volume to volume percent Vol. of solute%V / V 100

Vol. of solution=

Example : 10% ethanol (v/v) means 10 cc of ethanol dissolved in 100 cc of solution.

(iv) Volume to weight percent Vol. of solute 100

Wt. of solutio% /

nv w =

Example : 10% ethanol (v/w) means 10 cc of ethanol dissolved in 100g of solution. Normality (N) : It is defined as the number of gram equivalents (equivalent weight in grams) of a solute present per litre of the solution. Unit of normality is gram equivalents litre1. Normality changes with temperature since it involves volume. When a solution is diluted x times, its normality also decreases by x times. Solutions in term of normality generally expressed as, N= Normal solution; 5N = Penta normal, 10 N= Deca normal; = semi normal N/10 = Deci normal; N/5 = Penti normal N/100 or 0.01 = centinormal, N/1000 or 0.001= millinormal Mathematically normality can be calculated by following formulas,

(i) Number of g .eq. of solute

Normality (N )Volume of solution ( l )

= (ii) Weight of solute in g.

Ng. eq. weight of solute Volume of solution (l )

=

(iii) Wt. of solute per litre of solution N

g eq. wt. of solute= (iv)

Wt. of solute 1000N

g.eq. wt. of solute Vol. of solution in ml=

(v) Percent of solute 10N

g eq. wt. of solute

= , (vi) -1Strength in g l of solutionN

g eq. wt. of solute= (vii)

Wt% density 10N

Eq. wt .

=

(viii) If volume V1 and normality N1 is so changed that new normality and volume N2 and V2 then, N1V1 = N2V2 (Normality equation) (ix) When two solutions of the same solute are mixed then normality of mixture (N) is 1 1 2 2

1 2

N V N VN

V V+

=+

(x) Vol. of water to be added i.e., (V2V1) to get a solution of normality N2 from V1ml of normality N1

1 22 1 1

2

N NV V V

N

=

(xi) If Wg of an acid is completely neutralised by V ml of base of normality N Wt. of acid VN

g eq. wt. of acid 1000=

Similarly, Wt. of base Vol. of acid N of acid

g eq. wt. of base 1000

=

(xii) When Va ml of acid of normality Na is mixed with Vb ml of base of normality Nb NaVa = NbVb (a) If NaVa = NbVb (Solution neutral) (b) If NaVa > NbVb (Solution is acidic) (c) If NaVa < NbVb (Solution is basic)

(xiii) Normality of the acidic mixture a a b bmixa b

V N V NN

(V V )+

=+

(xiv) Normality of the basic mixture

b b a amix

a b

V N V NN

(V V )+

=+

(xv) No. of meq of soluteNVol. of solution in ml

= (* 1 equivalent = 1000 milliequivalent or meq.)

(5) Molarity (M) : Molarity of a solution is the number of moles of the solute per litre of solution (or number of millimoles per

ml. of solution). Unit of molarity is mol/litre or mol/dm3 For example, a molar )1( M solution of sugar means a solution containing 1 mole of sugar (i.e., 342 g or 231002.6 molecules of it) per litre of the solution. Solutions in term of molarity generally expressed as,

1 M = Molar solution, 2 M = Molarity is two,

or 0.5 M = Semimolar solution,

or 0.1 M = Decimolar solution,

or 0.01 M = Centimolar solution

or 0.001 M = Millimolar solution

Molarity is most common way of representing the concentration of solution.

Molarity is depend on temperature as, T1M

When a solution is diluted (x times), its molarity also decreases (by x times) Mathematically molarity can be calculated by following formulas,

(i) No. of moles of solute (n)MVol. of solution in litres

= , (ii) Wt. of solute (in gm) per litre of solutionM

Mol. wt. of solute=

(iii) Wt. of solute (in gm) 1000M

Mol. wt. of solute Vol. of solution in ml .= (iv)

No. of millimoles of soluteMVol. of solution in ml

=

(v) Percent of solute 10M

Mol. wt. of solute

= (vi) -Strength in gl of solutionM

Mol. wt. of solute

1

=

(vii) 10 Sp. gr. of the solution Wt. % of the soluteM

Mol. wt. of the solute

=

(viii) If molarity and volume of solution are changed from M1V1 to M2V2. Then, M V M V1 1 2 2= (Molarity equation)

(ix) In balanced chemical equation, if n1 moles of reactant one react with n2 moles of reactant two. Then, M V M V

n n1 1 2 2

1 2

=

(x) If two solutions of the same solute are mixed then molarity (M) of resulting solution. M V M V

MV V

1 1 2 2

1 2( )+

=+

(xi) Volume of water added to get a solution of molarity M2 from V1 ml of molarity M1 is M M

V V VM

1 22 1 1

2

=

Relation between molarity and normality Molecular massEquivalent a

m s

s

Normality of solution molarity=

Normality equivalent mass = molarity molecular mass

For an acid, Molecular massEquivalent mass

= Basicity So, Normality of acid = molarity basicity.

For a base, Molecular massEquivalent mass

= Acidity

So, Normality of base = Molarity Acidity. (6) Molality (m) : It is the number of moles or gram molecules of the solute per 1000 g of the solvent. Unit of molality is kgmol / . For example, a 2.0 molal )2.0( m solution of glucose means a solution obtained by dissolving 0.2 mole of glucose in gm1000 of water. Molality (m) does not depend on temperature since it involves measurement of weight of liquids. Molal solutions are less concentrated than molar solution. Mathematically molality can be calculated by following formulas,

(i) Number of moles of the solute Strength per 1000 grams of solventmWeight of the solvent in grams Molecular mass of solute

1000= =

(ii) Strength per 1000 grams of solvent m

Molecular mass of solute= (iii)

No. of gm moles of solutemWt. of solvent in kg

=

(iv) Wt. of solutem

Mol. wt. of solute Wt. of solvent in g1000

= (v) No. of millimoles of solutem

Wt. of solvent in g=

(vi) 10 solubilitym

Mol. wt. of solute

= (vii) 1000 wt. % of solute (x )m

(100 x ) mol. wt. of solute

=

(viii) 1000 Molaritym

(1000 sp. gravity) (Molarity Mol. wt. of solute)

=

Relation between molarity (M) and molality (m)

Molality Molaritym Molarity molecular massDensity

1000

=

Molarity Molality densityM Molality molecular mass1

1000

=

+

Chemical Classification of Matter 1. Which of the following is not a homogeneous mixture?

(A) A mixture of O2and N2 (B) Brass (C) Solution of sugar in water (D) Milk. 2. Which of the following is not an element?

(A) Diamond (B) Graphite (C) Ozone (D) Silica. 3. Which of the following is a chemical change?

(A) Dissolution of sugar in water (B) Dissolution of zinc in HCI (C) Dissolution of O2in water (D) Dissolution of ethanol in water.

4. 6 g of carbon combines with 32 g of sulphur to form CS2. 12 g of C also combine with 32 g of oxygen to form carbon dioxide. 10 g of sulphur combines with 10 g of oxygen to form sulphur dioxide. Which law is illustrated by them? (A) Law of multiple proportions (B) Law of constant composition (C) Law of Reciprocal proportions (D) Gay Lussac's law.

5. Which of the following data illustrates the law of conservation of mass? (A) 56 g of CO reacts with 32 g of oxygen to produce 44 g of CO2

(B) 1.70 g of AgNO3 reacts with 100 mL of 0.1 MHCI to produce 1.435 g of AgCI and 0.63 g of HNO3. (C) 12 g of C is heated in vacuum and on cooling there is no change in mass. (D) None of the above.

6. If law of conservation of mass was to hold true, then 20.8 g of BaCl2on reaction with 9.8 g of H2SO4 will produce 7.3 g of HCl and BaSO4 equal to (A)11.65g (B)23.3g (C) 25.5 g (D) 30.6 g.

7. 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 proportions (C) Law of constant composition (D) Law of reciprocal proportions.

8. The law of multiple proportions is illustrated by (A) Carbon monoxide and carbon dioxide (B) potassium bromide and potassium chloride (C) Water and heavy water. (D) Calcium hydroxide and barium hydroxide.

9. 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 proportions (C)Law of reciprocal proportions (D) Law of multiple proportions.

10. Two elements X (at. mass 16) and Y (at. mass 14) combine to form compounds A, Band C. The ratio of different masses of Y which combine with a fixed mass of X in A, Band 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 16parts by mass of X will combine with (A) 4 parts by mass of Y (B) 42 parts by mass of Y (C) 70 parts by mass of Y (D)84 parts by mass of Y.

11. Oxygen combines with two isotopes of carbon 12C and 14C to form two samples of carbon dioxide. The data illustrates (A) Law of conservation of mass (B) Law of multiple proportions (C) Law of reciprocal proportions (D) none of these.

12. 4.4 g of an oxide of nitrogen gives 2.24 L of nitrogen and 60 g of another oxide of nitrogen gives 22.4 L of nitrogen at S.T.P. The data illustrates the (A) Law of conservation of mass (B) Law of constant proportions (C)Law of multiple proportions (D) Law of reciprocal proportions.

13. Which one of the followingpair of substances illustrates law of multiple proportions? (A) CO, CO2 (B) NaCI, NaBr (C) H2O, D2O (D) MgO, Mg(OH)2.

14. One of the following combinations which illustrates the law of reciprocal proportions? (A) N2O3, N2O4, N2O5 (B) NaCI, NaBr, NaI (C)CS2, CO2, SO2 (D)PH3, P2O3 ,P2O5

15. Two elements X and Y combine in gaseous state to form XY fn the ratio 1 : 35 .5 by mass. The mass of Y that will be required to react with 2 g of X is. (A)7.1 g (B)3.55g (C) 71 g (D) 35.5 g.

16. Two samples of NaCI are produced when Na combines separately with two isotopes of chlorineCl35 and Cl37. Which law is illustrated? (A) Law of multiple proportions (B) Law of reciprocal proportions (C)Law of constant volume (D) none of the above.

17. The balancing of chemical equations is based upon (A)Law of combining volumes (B) Law of multiple proportions (C)Law of conservation of mass (D) Law of definite proportions.

18. Two compounds A and B have same percentage composition. The compounds A and B (A) are identical (B) are isomers (C) are neither identical nor isomers (D)All the three are correct.

Mole Concept 19. 1 g atom of nitrogen represents

(A) 6.02 1023 N2 molecules (B) 22.4 L of N2 at S.T.P. (C})11.2 10 f N2 at S.T.P. (D) 28 g of nitrogen.

20. 5.6 L of a gas at S.T.P. weighs equal to 8 g. The vapour density of gas is (A) 32 (B) 16 (C) 8 (D) 40

21. The maximum volume at S.T.P. is occupied by (A) 12.8 g of SO2 (B) 6.02 10

22molecules of CH4

(C) 0.5 mol of NO2 (D)1 gm molecule of CO2

22. A sample of Na2CO3 contains 6.02 1023 Na+ ions. The mass of the sample is (at. mass Na = 23, C =12,O = 16) (A) 53 g (B) 106 g (C) 165 g (D) 212 g.

23. What is correct for 10 g of CaCO3 ? (A) It contains 1 g atom of carbon (B) It contains 0.3 g atoms of oxygen (C)It contains 12 g of calcium (D) It refers to 0.1 g equivalent of CaCO3.

24. Which of the following is not correct regarding 14 gram of carbon monoxide? (A)It corresponds to 0.05 mole of CO (B)It occupies2.24 litres at S.T.P. (C)It corresponds to 3.01 1023molecules of CO (D)It corresponds to same number of moles of CO2 and nitrogen (I) oxide gases.

25. 4 .48 litre of methane at S.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.

26. If the density of water is 1 g cm 3, then the volume occupied by one molecule of water is approximately (A) 18 cm3 (B) 22400 cm3 (C) 6.02 10 23cm3 (D) 3.0 10 23cm3

27. Which of the following has the highest mass? (A) 20 of sulphur (B) 4 mol of carbon dioxide (C) 12 1024 atoms of hydrogen (D) 11.2L of helium at N.T.P.

28. If NA is Avogadro's number, then the number of oxygen atom in one g equivalent of oxygen is (A)NA (B) NA/2 (C) NA/4 (D) 2NA.

29. Which of the following represents 1 g molecule of the substance? (A)6.02 1024molecules of NH3 (B) 4 g of helium (C) 40 g of CaO (D) 127 g of iodine.

30. One atom of an element weighs 1.8 10 22g, its atomic mass is (A) 29.9 (B) 18 (C)108.36 (D) 154.

31. The number of sodium atoms in 2 moles of sodium ferrocyanide is (A) 2 (B) 6.02 1023 (C) 8 6.02 1023 (D) 4 6.02 1023.

32. Four containers of 2 L capacity contains dinitrogen as described below. Which one contains maximum number of molecules? (A)2.5 g molecule of N2 (B)4 g atom of nitrogen (C)3.01 X 10

24N atoms (D)82 g of dinitrogen. 33. The number of oxygen atoms present in 14.6 g of magnesium bicarbonate is

(A) 6No (B) 0.6 No (C) No (D) N0/2. 34. If isotopic distribution of C 12 and C 14 is98% and 2% respectively, then the number of C 14 atoms in 12 g of

carbon is (A) 1.032 1022 (B) 3.01 1022 (C) 5.88 1023 (D) 6.02 1023.

35. Out of 1.0 g dioxygen, 1.0 g (atomic) oxygen and 1.0 g ozone, the maximum number of oxygen atoms are contained in (A)1.0 g of atomic oxygen (B)1.0 g of ozone (C)1.0 g of oxygen gas (D)All contain same number of atoms.

36. 4.0 g of caustic soda (Mol. mass 40) contains same number of sodium ions as are present in (A) 10.6 g of Na2CO3 (Mol. mass 106) (B) 58.5 g of NaCI (Formula mass 58.5) (B) 100 mL of 0.5 M Na2SO4 (Formula mass 142)(C) 1g equivalent ofNaNO3 (Mol. mass 85).

37. If H2SO4 ionizes as H2SO4 + 2H2O 2H3O+ + SO42 then total number of ions produced by 0.1 M H2SO4 will be (A) 9.03 1021 (B)3.01 1022 (C) 6.02 1022 (D) 1.8 x 1023.

38. Total number of atoms in 44 g of CO2is (A) 6.02 1023 (B) 6.02 1024 (C) 1.806 1024 (D) 18.06 1022.

39. The flask A and B of equal size contain 2 g of H2 and 2 g of N2respectively at the same temperature. The number of molecules in flask A is (A) same as those in flask B (B) less than those in flask B (C) greater than those in flask B (D) exactly half than those in flask B.

40. A person adds 1. 71 gram of sugar (C12H22O11) in order to sweeten his tea. The number of carbon atoms added are (mol. mass of sugar =342) (A) 3.6 1022 (B)7.2 1021 (C) 0.05 (D) 6.6 1022.

41. Which of the following contains maximum number of atoms?

(A) 2.0 mol of S8 (B) 6.0 mol of S (C) 5.5 mol of SO2 (D)44.8 L of CO2 at S.T.P.

42. A sample of AlF3 contains 3.0 x 1024F ions. The number of formula units of this sample are (A) 9.0 1024 (B) 3.0 1024 (C)0.75 1024 (D) 1.0 1024.

43. The number of atoms present in 0.1 mole of P4(at. mass =31) are (A)2.4 1024atoms (B) same as in 0.05 mol of S8 (C)6.02 1022atoms (D) same as in 3.1 g of phosphorus.

44. Out of the following the largest number of atoms are contained in (A) 11 g of CO2 (B) 4 g of H2 (C) 5 g of NH3 (D) 8 g of SO2.

45. Which of the following will not have a mass of 10 g? (A)0.1 mol CaCO3 (B) 1.51 10

23Ca+2ions (C)0.16 mol of ions (D) 7.525 1022 Br atom.

46. The number of neutrons in 1.8 g of water will approximately be (a)4.216 x 1023 (B)8.432 1023 (C)4.816 1023 (D)4.216 1024.

47. X L of N2 at STP contains 3 x 1022molecules. The number of molecules in X/2 L of ozone at STP will be (A) 3 1022 (B) 1.5 .1022 (C) 1.5 1021 (D) 1.5 1011

Stoichiometry 48. The moles of O2 required for reacting with 6.8 g ammonia , ( ..NH3 +.. O2 ,...NO + H2O)is

(A) 5 (B)2.5 (C) 1 (D)0.5. 49. If 3.01 x 1020molecules are removed from 98 mg of H2SO4 then the number of moles of H2SO4 left are

(A) 0.1 10 3 (B) 0.5 10 3 (C) 1.66 10 3 (D) 9.95 10 2 50. X g of Ag was dissolved in HNO3 and the solution was treated with excess of .NaCI when 2.87 g of AgCI

was precipitated. The value of X is (A) 1.08 g (B) 2.16 g (C)2.70g (D) 1.62g.

51. The mass of CaO that shall be obtained by heating 20 kg of 90% pure lime stone (CaCO3) is (A) 11.2 kg (B) 8.4 kg (C)10.08kg (D) 16.8kg

52. The mass of CaCO3 produced when carbon dioxide is bubbled through 500 mL of 0.5 M Ca(OH)2 will be (A) 10 g (B) 20 g (C) 50 g (D) 25 g.

53. What mass of calcium chloride in grams would be enough to produce 14.35 g of AgCl? (At. Mass:Ca =40 ; Ag =108) (A) 5 .55 g (B) 8.295 g (C) 16.59 g (D) 11.19 g.

54. If LPG cylinder contains mixture of butane and iso butane, then the amount of oxygen that would be required for combustion of 1 kg of it will be (A)1.8kg (B)2.7kg (C)4.5kg (D)3.58kg.

55. The mass of 70% H2SO4 by mass required for neutralization of 1 mol of NaOH is (A) 49 g (B) 98 g (C) 70 g (D) 34.3 g.

56. If potassium chlorate is 80% pure, then 48 g of oxygen would be produced from (atomic mass of K =39) ; (A) 153. 12 g of KClO3 (B) 122.5 gofKClO3 (C) 245 g of KClO3 (D) 98.0 g of KClO3.

57. A sample of hard water is found to contain 40 mg of Ca2+ ions per litre. The amount of washing soda (Na2CO3) required to soften 5.0 L of the sample would be (A) 1.06 g (B) 5.3 g (C) 53 mg (D) 530 mg.

58. The mass of oxygen that would be required to produce enough CO which completely reduces 1.6 kg Fe2O3 (at. mass of Fe =56) is (A) 240 g (B) 480 g (C) 720 g (D) 960 g.

59. NO reacts with O2 to form NO2. When 10 g of NO2 is formed during the reaction, the mass of O2 consumed is (A) 1.90 g (B) 5.0 g (C)3.48g (D)13.9g.

60. The mass of 60% HCI by mass required for the neutralisation of 10 L of 0.1 M NaOH is (A) 60.8 g (B) 21.9 g (C) 100 g (D) 219 g.

61. 100 tons of Fe2O3 containing 20% impurities will give iron by reduction with H2 equal to (A) 112 tons (B) 80 tons (C) 160 tons (D) 56 tons.

62. In a given sample of H2O2each 100 mL of solution contains 68 mg of H2O2. The molarity of the solution is (A)0.02M (B)0.2M (C)2M (D) 20M.

63. A mixture containing 2.0 mol each of H2 and O2 is ignited so that water is formed. The amount of water formed is (A) 18.0 g (B) 36.0 g (C) 1.80g (D)3.60g

64. 1.2 g of Mg (At. mass 24) will produce MgO equal to (A) 0.05 mol (B) 40 g (C) 40 mg (D) 4 g.

65. A solution of 0.1 M of a metal chloride MClx requires 500 mL of 0.6 M AgN03 solution for complete

precipitation. The value of x is (A) 1 (B) 2 (C) 4 (D) 3.

Empirical and Molecular Formula, Molecular Mass 66. 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) 4 M (C)

(D) 2 M.

67. A compound made of two elements A and Bare 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.

68. On analysis a certain compound was found to contain iodine and oxygen in the ratio of 254 g of iodine (at. mass 127) and 80 g oxygen (at. Mass 16). What is the formula of the compound? (A) IO (B) I2O (C) I5O3 (D) I2O5

69. A container of volume V, contains 0.28 g of N2 gas. If same volume of an unknown gas under similar conditions of temperature and pressure weighs 0.44 g, the molecular mass of the gas is (A) 22 (B) 44 (C) 66 (D) 88.

70. Insulin contains 3. 4% sulphur. What will be the minimum molecular weight of insulin? (A) 94.117 (B) 1884 (C) 941.76 (D) 976.

71. B1 g of an element gives B2 g of its chloride, the equivalent mass of the element is (A)

35.5 (B)

35.5 (C)

35.5 (D)

35.5.

72. The haemoglobin from red blood corpuscles of most mammals contain approximately 0.33% of iron by mass. The molecular mass of haemoglobin is 67200. The number of iron atoms in each molecule of haemoglobin is (A) 3 (B) 4 (C) 2 (D) 6.

73. Two oxides of a metal contain 50% and 40% metal (M) respectively. If the formula of first oxide is MO2 the formula of second oxide will be (A) MO2 (B) MO3 (C) M2O (D) M2O5

74. 60 g of a compound on analysis gave 24 g C, 4 g H and 32 g O. The empirical formula of the compound is (A) C2H4O2 (B) C2H2O2 (C) CH2O2 (D) CH2O.

75. 400 mg of capsule contains 100 mg of ferrous fumerate. The percentage of iron present in the capsule is approximately (A) 8.2 % (B) 25% (C) 16% (D) unpredictable.

76. An unsaturated, hydrocarbon weighing 1.68 g has a volume of 488 mL at S.T.P. If it contains 14% of hydrogen. Then the 'family to which hydrocarbon belongs is (A) Alkane (B) Alkane (C) Alkyne (D) Benzene.

77. 0 .5 mol of potassium ferrocyanide contains carbon equal to (A) 1.5 mol (B) 36 g (C) 18g (D)3.6g.

78. Caffine contains 28.9% by mass of nitrogen. If molecular mass of Caffine is 194, then the number of N atoms present in one molecule of Caffine is (A) 3 (B) 4 (C) 5 (D) 6.

79. 0.16 g of dibasic acid requires 25 mL decimolar NaOH solution for complete neutralisation. The molecular mass of the acid is (A) 256 (B) 64 (C) 32 (D) 128.

Miscellaneous Problems 80. In a certain reaction ferrous oxalate is oxidised to ferric sulphate and CO2 by acidified potassium

permanganate, the equivalent mass of ferrous oxalate is

(A) .

(B)

.

(C) .

(D)

.. .

81. If 224 mL of a triatomic gas has a mass of 1 g at 273 K and 1 atm pressure, then the mass of one atom is (A) 8.30 10 23 g (B) 2.08 10 23 g (C) 5.53 10 23 g (D) 6.24 10 23 g.

82. Potassium chromate is isomorphous to potassium sulphate (K2SO4) It is found to contain 26.78% chromium. Calculate the atomic mass. of chromium (K =39.10) (A) 58 (B) 52 (C) 48 (D) 49.

83. Which one of the following properties of an element is not variable? (A) Valency (B) Equivalent mass (C) Atomic mass (D) All the three.

84. Divide a piece of ice into half. Devide it further and keep on dividing it many times. The smallest piece of

ice that you can get by this division is (A) an atom (B) a molecule (C) a particle (D) a crystal.

85. The notations of symbols for elements that we use today was started by (A) Boyle (B) Dalton (C) Berzelius (D) Lavoisier.

86. Atomicity of mercury vapour is (A) 1 (B) 2 (C) 3 (D) 4.

87. If atomic mass of oxygen were taken as 100, the molecular mass of water would be approximately (A) 6.25 (B) 112.5 (C) 102 (D) 106.25.

1.( D) 2.( D) 3( B) 4.( C) 5.(B ) 6.(B ) 7.( A) 8.(A ) 9.(D) 10.(C) 11.(D) 12. (C ) 13. (A) 14. (C) 15. (C) 16. (D) 17. (C) 18. (D) 19. (C) 20. (B) 21.(D) 22. (A) 23. (B) 24. (B) 25. (C) 2 6.(D) 27. (B) 28. (C) 29. (B) 30. (C) 31.(C) 32. (D) 33. (B ) 34. (A) 35. (D) 36. (C) 37. (D) 38. (C) 39. (C) 40. (A) 41.(C) 42. (D) 43. (B) 44. (B) 45. (C) 46. (C) 47. (B) 48. (D) 49. (B) 50. (B) 51.(C) 52. (D) 53. (A) 54. (D) 55. (C) 56. (A) 57. (D) 58. (B) 59. (C) 60. (A) 61.(D) 62. (A) 63. (B) 64. (A) 65. (D) 66. (B) 67. (A) 68. (D) 69. (B) 70. (C) 71.(A) 72. (B) 73. (B) 74. (D) 75. (A) 76. (B) 77. (B) 78. (B) 79. (D) 80. (A) 81.(C) 82. (B) 83. (C) 84. (B) 85. (C) 86. (A) 87. (B)