TOPIC : MOLE CONCEPT - Microsoft · NTSE_WORKSHOP Chemistry_Mole Concept (d) Gay Lussac’s Law of gaseous volume ... (By Avogadro's Law) 2 ... Mole in Terms of volume :

A Pre-Foundation Program

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NTSE_WORKSHOP Chemistry_Mole Concept

ATOMSAll the matter is made up of atoms.An atom is the smallest particle of an element that can take part in achemical reaction.Atoms of most of the elements are very reactive and do not exist in the free state (assingle atom). Theyexist in combination with the atoms of the same element or another element.Atoms are very, verysmall in size. The size of an atom is indicated by its radius which is called "atomicradius" (radius of an atom).Atomic radius is measured in "nanometres"(nm).1 metre = 109 nanometre or 1nm = 10–9 m.Atoms are so small that we cannot see them under the most powerful optical microscope.

(a) Symbols of Elements : Symbol is a small abbreviation to represent a full and lengthy name of the

element .

Dalton was the first to use symbols to represent elements in a short waybut Dalton's symbols for elementwere difficult to draw and inconvenient to use, so Dalton's symbols are only of historical importance.They are not used at all.

It was J.J. Berzelius who proposed the modern system of representing an element. Symbols have been

derived :

(i) Either by taking the first alphabet of the name of the element which is capitalized.

O–Oxygen N–Nitrogen F–Fluorine

C–Carbon

P–Phosphorus

H–Hydrogen

S–Sulphur

U–Uranium

I–Iodine

(ii) Or by taking the first alphabet and one more alphabet from the name of the element. The firstalphabet is capitalized.

Ca–Calcium Ni–Nickel Al–AluminiumMg–Magnesium

Cl–Chlorine

Co–Cobalt

Br–Bromine

Bi–Bismuth

Ba–Barium

(iii) Or from names of the elements in other languages such is Latin, German etc.

Symbol Element Other Language Name

NaCu

SodiumCopper

Natrium (Latin)Cuprum(Latin)

Fe

Pb

K

W

Ag

Au

Hg

Hg

Iron

Lead

Potassium

Tungsten

Silver

Gold

Mercury

Mercury

Ferrum (Latin)

Plumbum (Latin)

Kalium (Latin)

Wolfram (German)

Argentum (Latin)

Aurum(Latin)

Hydragyrum (Latin)

Hydragyrum (Latin)


CHEMISTRYWORKSHOP

NTSE

TOPIC : MOLE CONCEPT


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(b) Significance of The Symbol of an Element :

(i) Symbol represents name of the element.

(ii) Symbol represents one atom of the element.

(iii) Symbol also represents one mole of the element. That is, symbol also represent 6.023 × 1023

atoms of the element.

(iv) Symbol represent a definite mass of the element i.e. atomic mass of the element.Example :(i) Symbol H represents hydrogen element.

(ii) Symbol H also represents one atom of hydrogen element.

(iii) Symbol H also represents one mole of hydrogen atom.

(iv) SymbolHalsorepresentsonegramhydrogenatom.IONS

An ion is a positively or negatively charged atom or group of atoms.

Every atom contains equal number of electrons (negatively charged) and protons (positively charged).

Both charges balance each other, hence atom is electricallyneutral.

Classification of ions

(a) On the basis of charge

(i) Cation :

If an atom has less electrons than a neutral atom, then it gets positivelycharged and a positivelycharged

ion is known as cation.

e.g. Sodium ion (Na+), Magnesium ion (Mg2+) etc.

A cation bears that much units of positive charge as there are the number of electrons lost bythe neutral

atom to form that cation.

e.g.An aluminium atom loses 3 electrons to form aluminium ion, so aluminium ion bears 3 units of

positivechargeand it is represented asAl3+.

(ii) Anion :

If an atom has more number of electrons than that of neutral atom, then it gets negativelycharged and a

negativelycharged ion is known as anion.

e.g. Chloride ion (Cl¯), oxide ion (O2-) etc.

An anion bears that much units of negative charge as there are the number of electrons gained by the

neutral atom to form that anion.

e.g.Anitrogen atom gains 3 electrons to form nitride ion, so nitride ion bears 3 units of negative charge

and it is represented as N3-.

Note :

Size of a cation is always smaller and anion is always greater than that of the correspondingneutral atom.

(b) On the basis of number of atoms

(i) Monoatomic ions : Those ions which are formed from single atoms are called monoatomic

ions or simple ions.

e.g. Na+, Mg2+ etc.

(ii) Polyatomic ions : Those ions which are formed from group of atoms joined together are called

polyatomic ions or compound ions.

e.g.Ammonium ion (NH4+) , hydroxide ion (OH–) etc. which are formed by the joining of two

typesofatoms,nitrogenandhydrogenin the first caseandoxygenandhydrogenin thesecond.


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VALENCY OF IONSThe valency of an ion is same as the charge present on the ion.If an ion has 1 unit of positive charge, its valencyis 1 and it is known asa monovalent cation. If an ion has2 units of negative charge, its valency is 2 and it is known as a divalent anion.

LIST OFCOMMON ELECTROVALENT POSITIVE RADICALS

LIST OFCOMMON ELECTROVALENT NEGATIVE RADICALS

Note :Cation contains less no. of electrons and anion contains more no. of electrons than the no. of protonspresent in them.

LAWS OFCHEMICALCOMBINATION

The laws of chemical combination are the experimental laws which led to the idea of

atoms being the smallest unit of matter.The laws of chemical combination played a significant role in the

development of Dalton’s atomic theoryof matter.

There are two important laws of chemical combination. These are:

(a) Law of conservation of mass

(b) Law of constant proportions


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(a) Law of Conservation of Mass or Matter :

This law was given byAntonie L. Lavoisier (Father of chemistry) in 1774 .

According to this law, “ In a chemical reaction, the total mass of reactants. is equal to the total mass of

the products.”

Suppose we carry out a chemical reaction between A and B and if the products formed are C and D

then,

A + B C + D

Suppose 'a' g of A and 'b' g of B react to produce 'c' g of C and 'd' g of D. Then, according to the law

of conservation of mass, we have,

a + b = c + d

Example: CaO + CO2

56g 44g

56g + 44g =100g

(100 gm)

Heat

(b) Law of Constant Composition or Definte Proportion

This law was given by a french chemist proust in 1799.

According to this law, “The composition of a compound always remains fixed and it is independent of

the source from which the compound is obtained.”

This law cannot be applied to

(i) The compound obtained byusing different isotopes of the elements, selectively.

(ii) Those non-stoichiometric compounds, for which the exact formula can not be given

Example:

Water is a compound of hydrogen and oxygen. It can be obtained from various sources (like river, sea,

well etc.) or even synthesized in the laboratory. From whatever source we may get it, 9 parts by weight

of water is always found to contain 1 part byweight of hydrogen and 8 parts byweight of oxygen. Thus,

in water, this proportion of hydrogen and oxygen always remains constant.

(c) Law of Multiple Proportion

The law of multiple proportion was given byDalton in 1808.

According to this law, “ If two elements combine to form more than one compound, then for the fixed

mass of one element, the mass of other element combined will be in simple ratio.”

Example:

Carbon and oxygen when combine, can form two oxides that are CO (carbon monoxide), CO2(carbon

dioxide).

In CO,12 g carbon combined with 16 g of oxygen.

In CO2,12 g carbon combined with 32 g of oxygen.

Thus, we can see the mass of oxygen which combine with a constant mass of carbon (12 g) bear simple

ratio of 16 : 32 or 1 : 2


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(d) Gay Lussac’s Law of gaseous volume

This law is applicable to only those chemical reactions in which at least two reaction components are

gaseous.

According to this law, “The volumes of all gaseous reactants reacted and the volumes of all gaseous

products formed, measured at the same pressure and temperature, bear a simple ratio.”

Example:

Combination between hydrogen and chlorine to form hydrogen chloride gas. One volume of hydrogen

and one volume of chlorine always combine to form two volumes of hydrogen chloride gas.

H2(g) + Cl

2(g) 2HCl (g)

1vol. 1 vol. 2 vol.

The ratio between the volume of the reactants and the product in this reaction is simple, i.e., 1 : 1 : 2.

Hence it illustrates the Law of combining volumes.

(e) Avogadro’s Hypothesis :

According to this, “Equal volumes of all gases under similar conditions of temperature and pressure

contain equal number of molecules.”

This hypothesishas been found toexplain elegantlyall thegaseous reactionsandisnowwidelyrecognized

as a law or a principle known asAvogadro’s Law orAvogadro’s principle.The reaction between hydrogen and chlorine can be explainedon thebasisofAvogadro’sLawas follows

:

Hydrogen + Chlorine Hydrogen chloride gas1 vol. (By experiment)1 vol. 2 vol.

n molecules. n molecules. 2n molecules.(By Avogadro's Law)

21 molecules. molecules. 1 molecules. (By dividing throughout by 2n)

1 Atom 1 Atom 1 Molecule (Applying Avogadro's hypothesis)

21

It implies that one molecule of hydrogen chloride gas is made up of 1 atom of hydrogen and 1 atom of

chlorine.

Applications ofAvogadro’s hypothesis :

(a) In the calculation of atomicity of elementary gases.

(b) To find the relationship between molecular mass and vapour density of a gas.Molecular mass = 2 × Vapour density

ATOMIC MASS UNITThe atomic mass unit (amu) is equal to one-twelth (1/12) of the mass of atom of carbon-12. The mass ofan atom of carbon-12 isotope was given the atomic mass of 12 units, i.e., 12 amu of 12 u.

1 amu = atom12-Canofmass12

1

The actual mass of one atom of C-12 = 1.9926 × 10–26 kg

Hence, 1 amu =12

109926.1 26kg

= 1.66 × 10–27 Kg = 1.66 × 10–24 gm =AN

1gm.

Presently, amu has been replaced byunified mass (u).


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RELATIVEATOMIC MASSRelative Atomic mass of an element can be defined as the number which indicates how many times the

mass of one atom of the element is heavier in comparison to12

1th part of the mass of one atom of

Carbon-12.

RelativeAtomicmass =12-carbonofatomanofMass

12

1

elementtheofatomanofMass

=

amu1

amuinatom,anofMass

GRAM ATOMIC MASS

The gram atomic mass can be defined as the mass of 1 mole atoms of an element, in gm. In other words,it is the atomic mass of the element expressed in gm.

e.g., Mass of one oxygen atom = 16 amu =AN

16gm.

Mass of NA

oxygen atom = A

A

N.N

16= 16 gm

Hence, the g-atomic mass of oxygen is 16 gm.

Note : Be clear in the difference between 1 amu and 1 gm.

AtomicNumber Element Symbol

Atomicmass

1 Hydrogen H 12 Helium He 43 Lithium Li 74 Beryllium Be 95 Boron B 116 Carbon C 127 Nitrogen N 148 Oxygen O 169 Fluorine F 19

10 Neon Ne 2011 Sodium Na 2312 Magnesium Mg 2413 Aluminium Al 2714 Silicon Si 2815 Phosphorus P 3116 Sulphur S 3217 Chlorine Cl 35.518 Argon Ar 4019 Potassium K 3920 Calcium Ca 40

RELATIVE MOLECULAR MASS

Relative molecular mass of a substance can be defined as the number which indicates how many times

the mass of one molecule of a substance is heavier in comparison to th12

1part of the mass of one atom

of C-12.

Relativemolecular mass

=12-CofatomanofMass

12

1

substancetheofmoleculeoneofMass

=

amu1

amuinsubstance,theofmoleculeoneofMass


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GRAM MOLECULAR MASS

Gram molecular mass can be defined as the mass of 1 mole of molecules.

e.g., Mass of one molecule of O2= 32 amu = gram

N

32

A.

Mass of NA

molecules of O2= gmN

N

32A

A

= 32 gm

Ex.1 Calculate the molecular mass of water (Atomic masses : H = 1u, O = 16u)

Sol. The molecular formula of water is H2O

Molecular mass of water = (2 × atomic mass of H) + (1 × atomic mass of O) = 2 × 1 + 1 × 16 = 18

i.e., molecular mass of water = 18u.

Ex.2 Find out the molecular mass of sulphuric acid.

(Atomic masses : H = 1u, O = 16u, S = 32u)

Sol. The molecular formula of sulphuric acid is H2SO

4.

Molecular mass of H2SO

4

= (2 × atomic mass of H) + (1 × atomic mass of S) + (4 × atomic mass of O)

= (2 × 1) + (1 × 32) + (4 × 16) = 2 + 32 + 64 = 98

i.e., Molecular mass of H2SO

4= 98u.

Ex.3 (a) What is the mass of one molecule of HNO3?

(b) What is the molecular mass of HNO3?

(c) What is the gram molecular mass of HNO3?

Sol. (a) Mass of one molecule of HNO3= (1 + 14 + 3 × 16) amu = 63 amu.

(b) Molecular mass of HNO3= 63

amu1

amu63

(c) Gram molecular mass of HNO3= Mass of 1-molecule of HNO

3× N

A

= 63 amu × NA

=AN

63gm × N

A= 63 gm

FORMULA MASS

The term ‘formula mass’ is used for ionic compounds and other where discrete molecules do not exist.

e.g., sodium chloride, which is best represented as (Na+Cl–)n, but, for reasons of simplicityas NaCl or

Na+Cl–. Here, formula mass means the sum of the masses of all the species in the formula.

Thus, the formula mass of sodium chloride = (atomic mass of sodium) + (atomic mass of chlorine)

= 23 + 35.5

= 58.5 amu.

MOLE CONCEPT

A mole is the amount of substance that contains as many species (Atoms, molecules, ions or otherparticles) as there are atoms in exactly 12 gm of C-12.

species106.022mole1 23


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The number of species in 1 mole is known as avogadro's number orAvogadro's constant (NA).

NA = 6.022 × 1023

1 mole of atoms = 6.022 × 1023 atoms = 1 g -atom

1 mole of molecules = 6.022 × 1023 molecules = 1 g - molecule

1 mole of ions = 6.022 × 1023 ions = 1 g - ion

1 moleof formula units = 6.022 × 1023 formula units = 1 g - formula unit

Note : The term mole was introduced by Ostwald in 1896.

SOME IMPORTANT RELATIONS AND FORMULAE

(i) 1 mole of atoms = 6.023 × 1023 atoms = NA

atoms

(ii) Mass of 1 mole of atoms = Gram atomic mass = mass of 6.023 × 1023 atoms = mass of NA

atoms

(iii) 1 mole of molecules = 6.023 × 1023 molecules = NA

molecules

(iv) mass of 1 mole of molecules = Gram moleculer mass = mass of 6.023 × 1023 molecules = mass of

NA

molecules

(v) Number of moles of atoms =elementofmasstomicaGram

gramsinelementofMass

(vi) Number of moles of molecules = substanceofMassmolecuarGram

gramsinsubstanceofMass

(vii) Number of moles of a molecules =numberAvogadro

elementofmoleculesofNo.=

AN

N

Mole in Terms of volume :

Volume occupied by1 grams Molecular Massof 1 mole ofa gasunder standardconditionsof temperature

and pressure (273 K and 1 atm pressure) is called Gram Molecular Volume./ Its value is 22.4 litres for

each gas.

Volume of a mole of gas = 22.4 lite (at STP)

S.T.P.: 1 bar pressure and 273 K.

22.4

litre)273K(inand1atmatgasofvolumen

Some Important Relations

Units of Pressure :

1 atm = 760 mm of Hg = 76 cm of Hg = 760 torr = 1.01325 bar = 1.01325 × 105 pa.

Units of temperature :

273CK

Value of R :

R = 0.0821 litre-atm/mole.K

= 8.314 J/mole.K = 1.987 2 cal/mole.K


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Units of volume :

ml10m10litre1cm10dm1 333333

1m3 = 103 litre

Mole(gm-atom)

(gm-molecule)

×22.4

lt

÷22.4

lt

×N

A÷ NA

× gm mol. mass

× gm atomic mass

÷ gm mol. mass

÷ gm atomic mass

Volume of gasat S.T.P.

Number of atomsor molecules

mass of substance

PROBLEMS BASED ON THE MOLE CONCEPT

Ex.4 A piece of Cu contains 6.022 × 1024 atoms. How many mole of Cu atoms does it contain?

(A) 1 (B) 10 (C) 100 (D) 0.1

Ans. (B)

Sol. No. of mole of atoms =A

N

atomsof.No= 23

24

10022.6

10022.6

= 10

Ex.5 5 mole of CO2are present in a gaseous sample. How manymolecules of CO

2are present in the sample?

(A) 3.011 × 1023 (B) 3.011 × 1024 (C) 3.011 × 1022 (D) 15

Ans. (B)

Sol. No. of mole =A

N

moleculesof.No

No. of molecules = no. of mole × NA

= 5 × 6.022 × 1023

= 3.011 × 1024

Ex.6 You have 7 gm of a sample of nitrogen gas. Moles of nitrogen atoms and nitrogen molecules present in

the sample are respectively ?

(A) 0.5, 0.5 (B) 0.5, 0.25 (C) 0.25, 0.25 (D) 0.25, 0.50

Ans. (B)

Sol. Moles of atom =massatomicg

gm)(inmass

=

14

7= 0.5

Moles of molecules =massmolecular-g

gm)(inmass=

28

7= 0.25


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Ex.7 What will be the mass of 5 mole of SO2?

(A) 320 gm (B) 32 gm (C) 640 gm (D) 12.8 gmAns. (A)Sol. Molecular mass of SO

2= 64 gm

Now, n =M

m

m = n × m = 5 × 64 = 320 gm

Ex.8 A sample of He gas occupies 5.6 litre volume at 1 atm and 273 K. How many mole of He are present inthe sample?

(A) 0.25 (B) 0.5 (C) 0.125 (D) 5.6

Ans. (A)

Sol. No. of mole =4.22

K273andatm1at)(v = 25.0

4

1

4.22

6.5

AVERAGEATOMICAND MOLECULAR MASS

(a) Average atomic mass :

Average atomic mass = atomsofmoletotal

masstotal

Let a sample contains n1 mole of atoms with atomic massA1 and n2 mole of atoms with atomic massA2.then,

21

2211av nn

AnAnA

(b) Average molecular mass :

Average molecular mass =moleculerofmolestotal

masstotal

Let a sample contains n1 mole of molecules with molecular mass M1 and n2 mole of molecules with

molecular molecular mass M2 , then

21

2211av nn

MnMnM

Ex.9 Find the average atomic mass of a mixture containing 25% bymole Cl37 and 75% bymole Cl35 atoms?

(A) 35.5 (B) 36.0 (C) 36.5 (D) 35.75

Ans. (A)

Sol. n1 = 25 M1 = 37

n2 = 75 M2 = 35

Mav =21

2211

nn

AnAn

= 5.35

7525

35753725


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PERCENTAGE COMPOSITION

The percentage composition of elements in a compound is calculated from the molecular formula of the

compounds.

The molecular mass of the compound is calculated from the atomic masses of the various elements

present in the compound. The percentage bymass of each element is then computed with the help of the

followingrelations

Percentage mass of the element in the compound = massMolecular

elementtheofmassTotal× 100

Ex.10 What si the percentage of calcium in calcium carbonate (CaCO3) ?

Sol. Molecular mass of CaCO3= 40 + 12 + 3 × 16 = 100 amu.

Mass of calcium in 1 mol of CaCO3= 40g.

Percentage of calcium =100

10040= 40%

Ex.11 What are the percentage compositions of hydrogen and oxygen in water (H2O)?

Sol. Molecular mass of water, H2O = 2 + 16 = 18 amu.

So, total mass of hydrogen in H2O = 2 amu.

Percentage of H =18

1002= 11.11%

similarly,

Percentage of oxygen =18

10016= 88.89%

STOICHIOMETRYAND STOICHIOMETRIC CALCULATION

The word 'stoichiometry' is derived from two greek words – stoicheion (meaning element) and metron

(meaning measure). Hence, stoichiometry deals with the calculations of amounts of the reactants ad

products involved in a chemical reaction.

Significance of Chemical Equations

A chemical equation describes the chemical process both qualitatively and quantitatively.

The stoichiometric coefficients in the chemical equationgive the quantitative informationof thechemical

process. These coefficients represent the relative number of molecules or moles of the reactants and

products For example,

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

2 molecules 2 molecules 3 molecules

or, 2 N molecules 2 N molecules 3 N molecules

or, 2 moles 2 moles 3 moles

Again,Avogadro’s hypothesis states that under the same conditions of temperature and pressure,

equal volumes of gases contain the equal number of molecules. Thus, for homogeneous gaseous

reactions, the stoichiometric coefficients of the chemical equation also signify the relative volumes of

each reactant and product under the same conditions of temperature and pressure, For example.


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H2(g) + I2(g) 2 HI(g)

1 molecule 1 molecule 2 molecule

or 1 mole 1 mole 2 mole

or 1volume 1volume 2 volume (T & P constant)

or 1 pressure 1 pressure 2 pressure (T & V constant)

LIMITING REAGENT

The reactant which gives least amount of product on being completelyconsumed is known as limiting

reagent. It may also be defined as the reactant that is completely consumed when a reaction goes to

completion. It comes into the picture when reaction involves two or more reactants. For solving such

reactions, first step is to decide the Limiting Reagent.

Determination of Limiting Reagent :

Methode–I Bycalculating the required amount ofother reactant, bytheequation and comparing itwith given

amount.

[Useful when only two reactant are there]

Methode–II Bycalculating amount of anyone product obtained taking each reactant one byone, irrespective

of other reactants. Theonegiving least product is limiting reagent.

Methode–III Divide given moles of each reactantbytheir stoichiometric coefficient.The one with least ratio is

limiting reagent. [Useful when number of reactants are more than two.]

Ex.12 If 20 gm of CaCO3 is treated with 20 gm of HCl, how manygrams of CO2 can be generated according

to followingreaction :

CaCO3(g) + 2HCl(aq) CaCl2(aq) + H2O() + CO2(g)

(A) 8.8 gm (B) 12.054 gn (C) 20 gm (D) 6.027 gm

Ans. (A)

Sol. Mole of CaCO3 = 2.0100

20

Mole of HCl =5.36

20= 0.548

tcoefficientricStoichiome

Molefor CaCO3 = 2.0

1

2.0

tcoefficientricStoichiome

Molefor HCl = 2747.0

2

548.0 > 0.2

So, CaCO3 is limiting reagent

According to reaction :

100 gm of CaCO3 gives 44gm of CO2

20 gm CaCO3 will give 20100

44 =8.8 gm CO2


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EXERCISE

Q.1 Calculate the number of moles in 5.75g of sodium. (Atomic mass of sodium = 23u)

(1) 0.25 mole (2) 0.50 mole (3) 2 mole (4) 1 mole

Q.2 What is the mass in grams of a single atoms of chlorine : (Atomic mass of chlorine = 35.5 u)

(1) 5.9 × 10–27g. (2) 2.5 × 10–24g. (3) 5.9 × 10–23g. (4) 9.5 × 10–23g.

Q.3 The densityof mercury is 13.6 g cm–3. How many molesofmercuryare there in1 litreof themetal ?

(Atomic mass of Hg = 200 u).

(1) 50 (2) 75 (3) 100 (4) 68

Q.4 The mass of a single atom of an element M is 3.15× 10–23 g . What is its atomic mass ?

(1) 19 (2) 25 (3) 30 (4) 38

Q.5 An atom of neon has a mass of 3.35 × 10–23 g. How many atoms of neon are there in 20 g of the

gas ?

(1) 9.57 × 1024 (2) 5.97 × 1023 (3) 6.97 × 1030 (4) 1.27 × 1023

Q.6 How many grams of sodium will have the samenumber of atoms as atoms present in 6 g of

magnesium? (Atomic masses : Na = 23u ; Mg =24u)

(1) 1.5 g (2) 3 g (3) 5.75 g (4) 2.5 g

Q.7 How many moles of Cr are there in 85g of Cr2S

3? (Atomic masses : Cr = 52 u , S =32 u)

(1) 2.5 (2) 1.5 (3) 0.75 (4) 0.85

Q.8 The volume in litres of 20 g of hydrogen gas at STP :

(1) 224 litres. (2) 100 litres (3) 520 litres. (4) 1024 litres.

Q.9 The mass of oxygen contained in 1 kg of potassium nitrate (KNO3) :

(1) 125.7 g (2) 475.2 g (3) 2.5 g (4) 275.2 g

Q.10 Air is a mixture of O2 and N2 in which O2 is present 20% by mole and N2 is present 80% bymole. Findout the average molecular mass of air ?

(1) 28 (2) 28.8 (3) 30 (4) 31.2

Q.11 Hydrogen and oxygen combine to form H2O2 and H2O containing 5.93% and 11.2% Hydrogenrespectively. The data illustrates-(1) Law of conservation of mass (2) Law of constant proportions(3) Law of reciprocal proportions (4) Law of multiple proportions

Q.12 1.2 gm of Mg (At. mass 24) will produce MgO equal to -(1) 0.05 mol (2) 40 gm (3) 40 mg (4) 4 gm


PAGE# 14


Q.13 The atomic mass of calcium is 40. How many atoms are present in 2g calcium ?

(1) 3.0115 × 1022 (2) 3.1115 × 1022 (3) 4.0115 × 1022 (4) 3.1125 × 1022

Q.14 The point given as full stop at the end of a sentence by a graphite pencil weighs 2 × 10–12 g. How many

carbon atoms are present in such a point ?

(1) 1.003 × 1011 (2) 1.004 × 1011 (3) 1.002 × 1011 (4) 1.001 × 1011

Q.15 One mole of CO2 contains :

(1) 6.023 × 1023 g-atom of CO2 (2) 3.02 × 1023 atom of oxygen

(3) 18.1 × 103 molecule of CO2 (4) 6.023 × 1023 atom of carbon

Q.16 Which has the maximum number of atoms :

(1) 24 g C (2) 56 g Fe (3) 27 gAl (4) 108 gAg

Q.17 Mass of 1 atom of Hydrogen is -

(1) 1.66×10–24g (2) 10–22 g (3) 10–23 g (4) 10–25 g

Q.18 Which of the following contains the largest number of atoms -

(1) 11g of CO2 (2) 4g of H2 (3) 5g of NH3 (4) 8g of SO2

Q.19 One mole of P4 molecules contains -

(1) 1 molecule (2) 4 molecules

(3) 1/4 × 6.022 × 1023 atoms (4) 24.088 × 1023 atoms

Q.20 The total number of protons , electrons and neutrons in 12gm of 6C12 is -

(1) 1.084 × 1025 (2) 6.022 × 1023 (3) 6.022 × 1022 (4) 18


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ANSWER KEY EXERCISE

Q.1 1 Q.2 3 Q.3 4 Q.4 1 Q.5 2 Q.6 3 Q.7 4Q.8 1 Q.9 2 Q.10 2 Q.11 4 Q.12 1 Q.13 1 Q.14 2Q.15 4 Q.16 1 Q.17 1 Q.18 2 Q.19 4 Q.20 1

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TOPIC : MOLE CONCEPT - Microsoft · NTSE_WORKSHOP Chemistry_Mole Concept (d) Gay Lussac’s Law of gaseous volume ... (By Avogadro's Law) 2 ... Mole in Terms of volume :