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3SOME BASIC CONCEPTS OF CHEMISTRY

1.0 INTRODUCTIONChemistry deals with the composition, structureand properties of matter. These aspects can bebest described and understood in terms of basicconstituents of matter: atoms and molecules. Thatis why chemistry is called the science of atoms andmolecules. Can we see, weight and perceive theseentities? Is it possible to count the number of at-

oms and molecules in a given mass of matter andhave a quantitative relationship between the massand number of these particles (atoms and mol-ecules)? We will like to answer some of these ques-tions in this Unit. We would further describe howphysical properties of matter can be quantitativelydescribed using numerical values with suitable units.

MATTER

PHYSICAL

CHEMICALSolid Liquid Gas

Pure substance Mixture

Elements Compounds

Homogeneous Heterogeneous

(definite shape& volume)

(definite volume butno definite shape)

(no definite shape & volume)

(fixed ratio of massesof constituents)

(Consists of only onetype of atoms

(Made of two or moreatoms of different elements)

(same composition throughout& component are indistinguishable.

eg. gases, liquid solution, alloys, atmospheric air nearly homogeneous)

(composition isnot uniform

eg. sand & water)

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4 SOME BASIC CONCEPTS OF CHEMISTRY

Classification of universeUniverse is classified into two types, matter andenergy.(A) MATTER : The thing which occupy spaceand having mass which can be felt by our five sensesis called matter.Matter is further classified into two categories :1) Physical classification2) Chemical classificationPHYSICAL CLASSIFICATIONIt is based on physical state under ordinary condi-tions of temperature and pressure, so on the basisof two nature of forces matter can be classifiedinto the following three ways :(a) Solid (b) Liquid (c) Gas(a)Solid: A substance is said to be solid if it pos-sesses a definite volume and a definite shape, e.g.sugar, iron, gold, wood etc.(b)Liquid : A substance is said to be liquid if itpossesses a definite volume but not definite shape.They take up the shape of the vessel in which theyare put.e.g. water, milk, oil, mercury, alcohol etc.(c) Gas : A substance is said to be gas if it neitherpossesses a definite volume nor a definite shape.This is because they fill up the whole vessel in whichthey are put.e.g. hydrogen(H

2), oxygen(O

2), carbon dioxide

(CO2) etc.

CHEMICAL CLASSIFICATIONIt may be classified Into two types :(A) Pure Substance (B) Mixture(A) Pure Substance : A material containing onlyone type of substance. Pure Substance can not beseparated into simpler substance by physicalmethod.e.g. : Element = Na, Mg, Ca ........... etc.Compound = HCl, H

2O, CO

2, HNO

3..........etc.

Pure substance is classified into two types :(a) Element (b) Compound1) Element: The pure substance containing onlyone kind of atoms.It is classified into 3 types (depend on physicaland chemical property)(i) Metal → Zn, Cu, Hg, Ac, Sn, Pb etc.(ii) Non-metal → N

2, O

2, Cl

2, Br

2, F

2, P

4, S

8 etc.

(iii) Metalloids → B, Si, As, Te etc. .

(ii) Compound: It is defined as pure substancecontaining more than one kind of elements or at-oms which are combined together in a fixed pro-portion by weight and which can be decomposedinto simpler substance by the suitable chemicalmethod. The properties of a compound are com-pletely different from those of its constituent ele-ment. e.g. HCl, H

2O, H

2SO

4, HClO

4, HNO

3 etc.

(B) Mixture : A material which contain more thanone type of substances and which are mixed inany ratio by weight is called as mixture. The prop-erty of the mixture is the property of its compo-nents. The mixture can be| separated by simplephysical method.Mixture is classified into two types :(i) Homogeneous mixture : The mixture, in whichall the components are present uniformly is calledas homogeneous mixture. Components of mixtureare present in single phase.e.g. Water + Salt, Water + Sugar, Water + alco-hol, .(ii) Heterogenous mixture : The rriixture in whichall the components are present non-uniformly e.g.Water + Sand, Water + Oil, blood, petrol etc.

EX. 1. Which is an example of matter according tophysical state at room temperature and pres-sure.1) solid 2) liquid 3) gas 4) all of these

Sol. Ans. 4) According to the physical state at roomtemperature and pressure, the matter is present in3 state solid, liquid & gas

EX. 2. What are the types of the compound.1) Organic compound 2) Inorganic compound3) Both 1) and 2) 4) None of these

Sol. Ans. 3) Compound is divided in 2 types. Inor-ganic compound & Organic compound

EX. 3. Which of the following example of a Homo-geneous mixture.1) Water + Alcohol 2) Water + Sand3) Water + Oil 4) None of these

Sol. Ans. 1) Water and alcohol are completely mixedand form uniform solution.

EX. 4. Which mixture is called as solution.1) Heterogeneous mixture2) Homogeneous mixture3) Both 1) and 2)4) None of these

Sol. Ans. 2) Homogeneous mixture is called as solution.

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EX. 5. Which of the following is a compound1) graphite 2) producer gas

3) cement 4) marble

Sol. Ans. 4) Marble = CaCO3 = compound.

EX. 6. Which of the following statements is/are true1) An element of a substance contains only onekind of atoms.

2) A compound can be decomposed into its com-ponents.

3) All homogeneous mixtures are called as solu-tions.

4) All of these

Sol. Ans. 4)

EX. 7. A pure substance can only be1) A compound

2) An element

3) An element or a compound

4) A heterogenous mixture

Sol. Ans. 3)

EX. 8. Which one of the following is not a mixture1) Tap water 2) Distilled water

3) Salt in water 4) Oil in water

Ans. 2)

1.1 S.I. UNITS (INTERNATIONAL SYSTEMOF UNITS)Different types of units of measurements have beenin use in different parts of the world e.g. kilograms,pounds etc. for mass ; miles, furlongs, yards etc.for distance.

To have a common system of units throughout theworld. French Academy of Science, in 1791, in-troduced a new system of measurements calledmetric system in which the different units of a physi-cal quantity are related to each other as multiplesof powers of 10, e.g. 1 km = 103 m, 1cm = 10-2 metc. This system of units was found to be so con-venient that scientists all over the world adoptedthis system for scientific data.

(A) Seven Basic Units : The seven basic physi-cal quantities in the International System of Units,their symbols, the names of their units (called thebase units) and the symbols of these units are givenin Table.

TABLE : SEVEN BASIC PHYSICAL QUAN-TITIES AND THEIR S.I. UNITSPhysical Symbol S.I. Unit SymbolQuantityLength l metre mMass m kilogram kgTime t second sElectriccurrent I ampere AThermodynamictemperature T kelvin KLuminousintensity I

Ucandela cd

Amount of thesubstance n mole mol(B) Prefixes Used With Units : The S.I. systemrecommends the multiples such as 103,106,109 etc.and fraction such as 10-3,10-6,10-9 etc., i.e. thepowers are the multiples of 3. These are indicatedby special prefixes. These along with some otherfractions or multiples in common use, along withtheir prefixes are given below in Table and illus-trated for length (m)TABLE : SOME COMMONLY USED PRE-FIXES WITH THE BASE UNITS

Prefix Symbol Multiplication Example Factorded d 10-1 1 decimetre (dm) = 10-1 mcenti c 10-2 1 centimetre (cm) = 10-2 mmilli m 10-3 1 millimetre (mm) = 10-3 mmicro M 10-6 1 micrometre (pm) = 10-6 mnano n 10-9 1 nanometre (nm) = 10-9 mpico P 10-12 1 picometre (pm) = 10-12 mfemto f 10-16 1 femtometre (fm) = 10-l5matto a 10-18 1 attometre (am) = 10-18 mdeka da 101 1 dekametre(dam) = 101 mhecto h 102 1 hectometre (hm) = 102 mkilo k 103 1 kilometre (km) = 103 mmega M 106 1 megamerte(Mm) = 106 mgiga G 109 1 gigametre (Gm) = 109 mtera T 1012 1 teramerte (Tm) = 1012 mpeta P 1015 1 petametre (Pm) = 1015 mexa E 1018 1 exametre (Em) = 1018 m

As volume is very often expressed in litres, it isimportant to note that the equivalence in S.I. unitsfor volume is as under: 1 litre (1L) = 1 dm3 = 1000 cm3

and 1 millilitre (lmL)= 1 cm3 = lcc

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(C) SOME IMPORTANT UNIT CONVERSIONS1. Length: 1 mile = 1760 yards

1 yard = 3 feet1 foot = 12 inches1 inch = 2.54 cmlA0 = 10-10m or 10-8cm

2. Mass1 Ton = 1000 kg1 Quintal = 100 kg1 kg = 2.205 Pounds (lb)1 kg = 1000 g1 gram = 1000 milli gram1 a.m.u. = 1.67x 10g

3. Volume1L = 1dm3 = 103m3=103 cm3

= 103 mL = 103 cc1 mL = 1 cm3 = 10-6 m3

= 1 cc4. Energy:

1 calorie = 4.184 joules� 4.2 joules1 joule = 107 ergs1 litre atmosphere(L-atm) = 101.3 joule1 electronvolt (eV) = 1.602 x 10-19 joule

5. Pressure1 atmosphere (atm) = 760 torr

= 760 mm of Hg= 76 cm of Hg

=1.01325x105pascal(Pa)= 1.1325x 10s N/m2

Some More Prefixes :Semi =1/2 Mono =1Sesqui= 3/2=1.5 Di or Bi =2Tri =3 Tetra =4Penta =5 Hexa =6Hepta =7 Octa=8Nona=9 Deca =10Undeca=11 Do deca=12Trideca=13 Tetra deca =14Pentadeca=15 Hexa deca =16Hepta deca=17 Octa deca = 18Nonadeca =19 Eicoso/Icoso = 20

GOLDEN KEY POINTSThe unit named after a scientist is started with asmall letter and not with a capital letter e.g. unit offorce is written as newton and not as Newton.Likewise unit of heat and work is written as jouleand not as Joule.Symbols of the units do not have a plural ending like's'. For example we have 10 cm and not 10 cms.Words and symbols should not be mixed e.g. we

should write either joules per mole or J mol-1 andnot joules mol-1

Prefixes are used with the basic units e.g. kilome-ter means 1000 m (because meter is the basic unit).Exception. Though kilogram is the basic unit ofmass, yet prefixes are used with gram because inkilogram, kilo is already a prefix.A unit written with a prefix and a power is a powerfor the complete unit e.g. cm3 means (centimeter)3

and not centi (meter)3.EX. 9. Which one of the following forms part of

seven basic SI units :1) Joule 2) Candela 3) Newton 4) Pascal

Sol. Ans. 2)EX. 10 Convert 2 litre atmosphere into erg.Sol. 2 litre atmosphere = 2 x 101.3 joule = 2x 101.3x

107 erg. = 202.6 X 107 erg.1 litre atmosphere=101.3J

EX. 11. Convert 2 atm into cm of Hg.Sol. 2 atm = 2 x 76 cm of Hg = 152 cm of Hg

1 atmosphere = 76 cm of HgEX. 12 Convert 20 dm3 into mL.Sol. 20 dm3 = 20 litre = 20 x 1000 mL = 2 x 104 mL

1 dm3 = 1 litre = 1000 mLEX. 13. Convert 59 F into 0C.Sol. °C = 5/9 (F - 32) = 5/9 (59 - 32) = 5/9 x 27 = 150C

1.2 MOLE CONCEPTIn SI Units we represent mole by the symbol 'mol'.It is defined as follows:(i) A mole is the amount of a substance that con-tains as many entities (atoms, molecules or otherparticles) as there are atoms in exactly 12g of thecarbon - 12 isotope.It may be emphasised that the mole of a substancealways contains the same number of entities, nomatter what the substance may be. In order to de-termine this number precisely, the mass of a car-bon 12 atom was determined by a mass spec-trometer and found to be equal to 1.992648 x 10-

23g. Knowing that 1 mole of carbon weighs 12g,the number of atoms in it is equal to :

/

. /

g molCg C atom−×

12

23 12

12

1 992648 10

= 6.0221367 x 1023 atoms/mol(ii) In a simple way, we can say that mole has6.0221367 x1023 entities (atom, molecules or ions etc.)The number of entities in 1 mol is so important that

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It is given a separate name and symbol, known as'Avogadro constant' denoted by N

A .

Here entities may represent atoms, ions, moleculesor other subatomic entities. Chemists count thenumber of atoms and molecules by weighing. In areaction we require these particles (atoms, mol-ecules and ions) in a definite ratio. We make useof this relationship between numbers and massesof the particles for determining the stoichiometryof reactions .Formula to get moles are following :

(i) weight(g)

Number of moles(n)=molar mass

Where molar mass = gram atomic mass or grammolecular mass or gram ionic mass

(ii) ( )V

Number of moles(n)=.L

22 4 (Where V - Vol-

ume of gas In L at NTP or STP)

(iii) N

Number of moles(n)=AN (Where N =

Number of particles)

A

number of atomsmole atoms=

N and

A

number of moleculesmole molecules =

N

SOME RELATED DEFINITIONS:Atomic Mass (Relative Atomic Mass)It is defined as the number which indicates howmany times the mass of one atom of an element isheavier in comparison to l/l2 th part of the mass ofone atom of C-12.Atomic mass unit (a.m.u.): The quantity 1/12th

mass of an atom of C12 is known as atomic massunit. Since mass of 1 atom of C -12 = 1.9924 x10-23 g1/12th part of the mass of 1 atom

-231.9924×10 g=

12

-2423

1=1.67×10 g = 1 a.m.u=

6.023×10It may be noted that the atomic masses as ob-tained above are the relative atomic masses andnot the actual masses of the atoms. These masses

on the atomic mass scale are expressed in termsof atomic mass units (abbreviated as amu). Today,'amu' has been replaced by 'u' which is known asunified mass.One atomic mass unit (amu) is equal to 1/12th ofthe mass of an atom of carbon -12 Isotope.Thus the atomic mass of hydrogen is 1.008 amuwhile that of oxygen is 15.9994 amu (or taken as16 amu).Gram Atomic Mass (or Mass of 1 Gram Atom)When numerical value of atomic mass of an ele-ment is expressed in grams then the value becomesgram atomic mass.gram atomic mass = mass of 1 gram atom = massof 1 mole atom= mass of Na atoms=mass of 6.023 x 1023 atoms.Ex. gram atomic mass of oxygen = mass of 1 gatom of oxygen = mass of 1 mol atom of oxygen.

= mass of NA atoms of oxygen =

16A

A

N gN

⎛ ⎞× =⎜ ⎟

⎝ ⎠16

Molecular Mass (Relative Molecular Mass)The number which indicates how many times themass of one molecule of a substance is heavier incomparison to 1/12th part of the mass of an atomof C-12.Gram Molecular Mass (Mass of 1 Gram Mol-ecule)When numerical value of molecular mass of thesubstance is expressed in grams then the valuebecomes gram molecular mass.gram molecular mass = mass of 1 gram molecule= mass of 1 mole molecule= mass of Na molecules = mass of 6.023 x 1023

moleculesEx. gram molecular mass of H

2SO

4= mass of 1

gram molecule of H2SO

4= mass of 1 mole mol-

ecule of H2SO

4 = mass of Na molecules of H

2SO

4

AA

g N gN

⎛ ⎞× =⎜ ⎟

⎝ ⎠

9898

Actual MassThe mass of one atom or one molecule of a sub-stance is called as actual mass.Ex. (i) Actual mass of O

2 = 32 amu = 32 x 1.67 x

10-24 g - Actual mass(ii) Actual mass of H

2O = (2 + 16) amu = 18 x

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1.67 x 10-24g = 2.99 x 10-23 gAtomicity - Total number of atoms in a moleculeof elementary substance is called as atomicity.Molecule AtomicityH

22

O2

2O

33

NH3

4EX. 14. Find out the volume and mole in 56 g nitro-

gen at STP SolutionSol. Molecular weight of N

2 is 28 g

(a) Calculation of volume∵ 28 g of N

2 occupies = 22.4 litre at STP

∴ 56 g of N2 occupies =

.22 4

28x561itre = 44.8

litre at STP(b) Calculation of mole∵ 28 g of N

2 = 1 mol of N

2

∴56 g of N2 = 1

28x56 = 2 mol of N

2

EX. 15. Calculate the volume and mass of 0.2 molof O3 at STP.

Sol. (a) Calculation of volume ∵ volume of 1 mole ofO

3 at STP = 22.4 litre

∴ volume of 0.2 mole of O3 at STP = 22.4 x 0.2 =

4.48 litre (b) Calculation of mass ∵ mass of 1 mol of O

3= 48 g

∴ mass of 0.2 mol of O3- 48 x 0.2 gm = 9.6 g

EX. 16. Find out the moles & mass in 1.12 litre O3at STP.

Sol. (a) Calculation of mole:∵ at STP 22.4 litre of O

3 contain = 1 mol of O

3

∴ at STP 1.12 litre of O3 contain =

.

1

22 4x1.12=

0.05 mol of O3

(b) Calculation of mass :Molecular weight of O

3 = 48 g

∵ weight of 22.4 litre of O3 at STP is = 48 g

∴ weight of 1.12 litre of O3 at STP is

= .

48

22 4x1.12 = 2.4 g

EX. 17. Find out the mass of 1021 molecules of Cu.Sol. For Cu (i.e. mono atomic substance) number of

atoms = number of molecules

21

23A

N 10Number of moles of Cu= =

N 6.023×10

weight weight= =

Atomic weight 63.5

weight of Cu = 21

23

10= 63.5=0.106g

6.023×10×

EX. 18. Calculate the number of molecules andnumber of atoms present in 1 g of nitrogen ?

Sol. Number of moles (n)

A

w

Nweight 1= = Number of molecules (N)

M 28 28⇒

1molecule of N2 gas contain = 2 atoms

A A A2

N N Nmolecules of N gas contain=2× = atoms

28 28 14EX. 19. Calculate the number of moles in 11.2 litre

at STP of oxygen.Sol. Number of moles of O

2 (n)

=11

= =0.5 mol. .

V22 4 22 4

EX. 20. 1/2 g molecule of oxygen. Find (i) mass, (ii)number of molecules, (iii) volume at STP. (iv)No. of oxygen atoms.

Sol.(i)

w

weight weightn=1/2mol= = weight of oxygen =16g

M 32⇒

(ii)

A

A

NNn=1/2mol= Number of moles of oxygen(N) =

N 2⇒

(iii) n=1/2mol= V=11.2 litre.

V⇒

22 4(iv) 1 molecule of O

2 contain = 2 oxygen atoms.

A2

AA

Nmolecules of O contain

2N

= ×2 = N oxygen atoms2

1. The modern atomic weight scale is based on.1) C12 2) O16 3) H1 4) C13

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2. Gram atomic weight of oxygen is1) 16 amu 2) 16 g 3) 32 amu 4) 32 g

3. Molecular weight of SO2 is1) 64 g 2) 64 amu 3) 32 g 4) 32 amu

4. 1 amu is equal to1) 1/12 of C-12 2) 1/14 of O-163) 1 g of H

24) 1.66 x 10-24 kg

5. The actual molecular mass of chlorine is1) 58.93 x 10-24 g 2) 117.86 x 10-24 g3) 58.93 x 10-24 kg 4) 117.86 x 10-24 kg

1) 1 2) 2 3) 2 4) 1 5) 2

RELATION BETWEEN MOLECULARWEIGHT AND VAPOUR DENSITY :Vapour density (V.D): Vapour density of a gas isthe ratio of densities of gas & hydrogen at the sametemperature & pressure.

2

gas

H

Density of gasVapour Density (V.D)=

Density of hydrogen

d m(mass)(g)= d=

d V(volume)(mL)

⎧⎨⎩

(m ) for certain V litre volumeV.D=

(m ) for certain V litre volumegas

H2

If N molecules are present in the given volume of agas and hydrogen under similar condition of tem-perature and pressure.

(m ) for certain V litre volumeV.D=

(m ) for certain V litre volume

(m ) of 1molecule molecular mass of gas

(m ) of 1 molecule 2= =

gas

H

gas

H

2

2

Molecular mass of gas (Mw) = 2 x V.D

RELATION BETWEEN MOLAR MASS(Mw) & VOLUME :

2

gas

H

dAt STP M =2 V.D=2

d

(m ) for certain V litre volume=2

(m ) for certain V litre volume

× ×

×

w

gas

H2

w2

mass of 1 litre gasor M =2×

mass of 1 litre H

w

mass of 1 litre gasor M =2×

. g0 089

Mw(g)= 22.4 x mass of 1 litre gas

2H

gd =0.000089

mLm m

= =V 1000ML

V=1litre=1000mL

then2Hm =0.089g

Mw(g) = Mass of 22.4 litre gas orMw(g) - 22.4 litre (at STP)

GRAM MOLECULAR VOLUME (GMV)At NTP, the volume of 1 mole of gaseous sub-stance is 22.4 litre is called as gram molecular vol-ume.

At NTP, 2Hd = 0.000089 g/mL = mass/volume =

mass/1000 mLIf volume - 1 litre = 1000 mL then mass = 0.089 g0.089g H

2 occupies = 1 litre at STP

2 g or 1 mol H2 occupies = 00gg at STP

1 mole of any gaseous substance occupy 22.4 li-tre of volume at NTP or STP1 mol s 22.4 litre (at STP)

EX. 21. Calculate the number of atoms of chlorinein 2.08 gm of BaCl2.(Atomic weight of Ba =137,Cl = 35.5)

Sol. Number of moles of BaCl2 (n) =

W A

weight 2.08 N= =0.01mol=

M 208 N

Number of molecules of BaCl2 (N) = 0.01 N

A

1 molecule of BaCl2 contain = 2 chlorine atoms.

0.01 NA molecules BaCl

2 contain = 2 x 0.01 N

A

Chlorine atoms. = 2 x 10-2 NA Chlorine atoms.

EX. 22. Calculate the number of molecules andnumber of atoms present in 1.2 g of ozone.

Sol. Number of moles of O3 (n) =

W

weight 1.2 1= = mol

M 48 40

⇒ number of molecules of O3 (N)= AN

mol40

∵ 1 molecule of O3 contain = 3 atoms, AN

40

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molecules O3 contain = A3N

40atoms.

EX. 23. Calculate the number of atoms present inone drop of water having mass 1.8 g.

Sol. Number of moles of H2O (n)

1.80.1

18w

weight molM

= =

Number of molecules of H2O (N) = 0.1 N

A

∵1 molecule of H2O contain = 3 atoms

∴0.1 NA molecules H2O contain = 3 x (0.1 N

A)

= 0.3 NA atoms.

EX. 24. Calculate the number of atoms present inone litre of water (density of water is 1 g/mL).

Sol.. 1 litre = 1000 mL = 1000 gMoles of H

2O

(n)=1000

55.518w A

weight NmolM N

= = = ⇒ number

of molecules of H2O (N)=55.5NA

∵1 molecule of H2O contain = 3 atoms

∴ 55.5 NA molecules H

2O contain = 3 x (55.5

NA) atoms = 166.5 N

A atoms

EX. 25. At NTP the density of a gas is 0.00445 g/mL then find out its V.D. and molecular mass.

Sol.2

0.004450. 50

0.000089

Density of gasV DDensity of H

= = =

Molecular mass = 2 x V.D.= 2 x 50 = 100EX. 26. Weight of 1 litre gas is 2 g then find out its

V.D. and molecular massSol.

2

0.002 /1000

MassDensity of gas g mLVolume

= = =

2

0.00200. 22.4

0.000089

Density of gasV DDensity of H

= = =

Molecular mass = 2 x V.D. = 44.8GOLDEN KEY POINTS

Term molar mass means mass of 1 mol particles.Vapour density is calculated with respect to H2gasunder similar conditions of temperature and pressure.

Relative density =

gas B

Density of gas ADensity of

=

Specific gravity : It is density of material with re-spect to water.

Vapour density, relative density and specific grav-ity are ratios so they are unitless.The term STP means 273.15 K (00C) and 1 barpressure NTP means 273.15 K (00C) and 1 atm.

1. Calculate the number of atoms in 11.2 litre ofSO2 gas at STP :(l) N

A/2 2) 3N

A / 2 3) 3N

A4) N

A

2. Which of the following has maximum mass1) 0.1 gram atom of carbon2) 0.1 mol of ammonia3) 6.02 x 1022 molecules of hydrogen4) 1120 cc of carbon dioxide at STP

3. The total number of electrons present in 18mL of water1) 6.02 x 1022 2) 6.02 x 1023

3) 6.02 x 1024 4) 6.02 x 1025

4. The volume of 1.0 g of hydrogen at NTP is1) 2.24 L 2) 22.4 L 3) 1.12 L 4) 11.2 L

5. 11 grams of a gas occupy 5.6 litres of volumeat STP. The gas is1) NO 2) N

2O

43) CO 4) COz

6. At NTP, 5.6 litre of a gas weight 8 gram. Thevapour density of gas is1)32 2)40 3)16 4)8

7. The vapour densities of two gases are in theratio of 1 : 3. Their molecular masses are inthe ratio of1) 1:3 2)1:2 3).2:3 4)3:1

1) 2 2) 4 3) 4 4) 45) 4 6) 3 7) 1

1.3 PERCENTAGE COMPOSITION, EMPIRI-CAL FORMULA & MOLECULAR FOR-MULAPercentage formula (% by mass)

(In a molecule or compound) Mass % of an element

Number of atoms (Atomicity) atomic mass= ×100

molecular mass

×

If number of atom =1 : Molecular mass = mini-mum molecular massEmpirical Formula

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The empirical formula of a compound express thesimplest whole number ratio of atoms of variouselementspresent in 1 molecule of the compound,Ex.

MolecularFormula H

2O

2 CH

4 C

2H

6 C

2H

4O

2

2:2 1:4 2:6 2:4:2 1:1 1:4 1:3 1:2:1

EmpiricalFormula HO CH

4 CH

3 CH

2O

Molecular FormulaThe molecular formula of a compound representsthe actual number of atoms present in 1 moleculeof the compound i.e. it shows the real formula ofits 1 molecule.Relationship between Empirical & Molecu-lar FormulaMolecular Formula = nx Empirical Formula[Where n = natural no. (1, 2, 3.....)]

or Molecular Formula

n=Empirical Formula or

Molecular Formula massn=

Empirical Formula mass

Determination of Empirical FormulaFollowing steps are involved to determine the em-pirical formula of the compounds -1) First of all find the % by weight of each elementpresent in 1 molecule of the compound2) The % by weight of each element is divided byits atomic weight. It gives atomic ratio of elementspresent in the compounds.3) Atomic ratio of each element is divided by theminimum value of atomic ratio as to get simplestratio of atoms.4) If the value of simplest atomic ratio is fractionalthen raise the value to the nearest whole number,or Multiply with suitable coefficient to convert itinto nearest whole number.(5) Write the Empirical formula as we get the sim-plest ratio of atoms.

EX. 28. Find out percentage composition of eachelement present in glucose ?

Sol.12 6

% of C = 100 40%180

×× =

12 1% of H = 100 6.66%

180

×× =

16 6% of O = 100 53.33%

180

×× =

EX. 29. In a compound x is 75.8% and y is 24.2%by weight present. If atomic weight of x and yare 24 and 16 respectively. Then calculate theempirical formula of the compound.

Sol. Elements % Atomic % Simplest Ratio

weight Atomic ratioweight

x 75.8% 24 75.8

3.124

= 3.1

=21.5

2

y 24.2% 16 24.2

=1.516

1.5

=11.5

1

Empirical formula = x2y

EX. 30. In a compound Carbon = 52.2%, Hydro-gen = 13%, Oxygen = 34.8% are present andmolecular mass of the compound is 92. Cal-culate molecular formula of the compound ?

Sol.Elements % Atomic % Simplest Ratio

weight Atomic ratio weight

C 52.2 2 52.2

=4.35=4.412

4.4

=22.2

2

H 13 1 13

=131

13

=5.922

6

O 24.8 16 34.8

=2.216

2.2

=12.2

1

Empirical formula = C2H

6O

Empirical formula mass = 12 x2 + 16 + 6 = 46

Molecular Formula massn=

Empirical Formula mass92

=246

=

molecular formula = 2 x(C2H

6O) = C

4H

12O

2

1. A hydrocarbon contain 80%C. The vapourdensity of compound is 30. Empirical formulaof compound is1) CH

32) C

2H

63) C

4H

124) C

4H

8

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2. Two elements X (Atomic weight=75) and Y(Atomic weight=16) combine to give a com-pound having 75.8% of X, The empirical for-mula of compound is(DXV 2) X

2Y 3) X

2Y

24)X

2Y

3

3. In a compound element A (Atomic weight -12.5) is 25% and element B (Atomic weight =37.5) is 75% by weight. The Empirical formulaof the compound is1) AB 2) A

2B 3) A

2A

24) A

2B

3

1) 1 2) 4 3) 1

1.4 STOICHIOMETRY BASED CONCEPT(PROBLEMS BASED ON CHEMICALREACTION)One of the most important aspects of a chemicalequation is that when it is written in the balancedform, it gives quantitative relationships between thevarious reactants and products in terms of moles,masses, molecules and volumes. This is called sto-ichiometry (Greek word, meaning to measure anelement). For example, a balanced chemical equa-tion along with the quantitative information con-veyed by it is given below:

CaCOs + 2HC1 → CaCl2 + H

2O + CO

2

1 Mole 2 Mole 1 Mole 1 Mole 1 Mole40+12 2(1+35.5) 40+2x35.5 2x1+16 12 + 2 x 16+3x16= 100 g = 73 g = 111 g =18g = 44g or 22.4L

at STPThus,(i) 1 mole of calcium carbonate reacts with 2 molesof hydrochloric acid to give 1 mole of calcium chlo-ride, 1 mole of water and 1 mole of carbon dioxide.(ii) 100 g of calcium carbonate react with 73 ghydrochloric add to give 111 g of calcium chlo-ride, 18 g of water and 44 g (or 22.4 litres at STP)of carbon dioxide.

1 3 2 Stoichiometric

coefficientN

2 + 3H

2 → 2NH3

1 mole + 3 mole → 2 mole22.4 litre + 3 x 22.4 litre→2 x 22.4 litre (at STP)1 litre + 3 litre → 2 litre

1000 mL + 3000 mL → 2000 mL1 mL + 3 mL → 2 mL28 gm + 6 gm → 34 g (According to the law of conservation of mass)Gram can not be represented according to sto-ichiometry.The quantitative information conveyed by a chemi-cal equation helps in a number of calculations. Theproblems involving these calculations may be clas-sified into the following two different types:(a) Single reactant based(b) More than one reactant based(A) SINGLE REACTANT BASED :1) Mass - Mass Relationships i.e. mass of one ofthe reactants or products is given and the mass ofsome other reactant or product is to be calculated.2) Mass - Volume Relationships i.e. mass/volumeof one of the reactants or products is given and thevolume/mass of the other is to be calculated.3) Volume - Volume Relationships i.e.volume ofone of the reactants or the products Is given andthe volume of the other is to be calculated.General method : Calculations for all the problemsof the above types consists of the following steps :(i) Write down the balanced chemical equation.(ii) Write the relative number of moles or the rela-tive masses (gram atomic or molecular masses) ofthe reactants and the products below their formula.(iii) In case of a gaseous substance, write down22.4 litres at STP below the formula in place of 1mole(iv) Apply unitary method to make the requiredcalculations.Quite often one of the reactants is present in largeramount than the other as required according to thebalanced equation. The amount of the productformed then depends upon the reactantwhich hasreacted completely. This reactant is called the lim-iting reactant. The excess of the otheris leftunreacted.Combustion reaction : (Problem based on com-bustion reactions):For balancing the combustion reaction: First of allbalance C atoms, Then balance H atom, Finallybalance Oxygen atom.For Example : Combustion reaction of C2H6 :C

2H

6 + O

2 → CO

2 + H

2O (skeleton equation)

balance C atoms

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C2H

6 + O

2 → 2CO

2 + H

2O

Now balance H atomsC

2H

6 + O

2 → 2CO

2 + 3H

2O

Now balance Oxygen atomsC

2H

6 + 7/2O

2 → CO

2 + H

2O

TYPE-I (INVOLVING MASS-MASS RELA-TIONSHIP)

EX. 31. How much iron can be theoretically obtainedin the reduction of 1 kg of Fe2O3

Sol.

2 3 2 2

1 3 2 3

3 2 3Fe O H Fe H O+ → +

5.850.1

58.5w

weightn molM

= = =

The equation shows that 2 mol of iron are ob-tained from 1 mol of ferric oxide.Hence, the obtained no. of moles of

2 100012.5

160

56

Fe mol

weight weightAtomic weight

×= =

= =

Weight of iron obtained = 12.5 x 56 g = 700 gEX. 32. What amount of silver chloride is formed

by the action of 5.850 g of sodium chloride onan excess of silver nitrate?

Sol.3 3

1 1 1 1

NaCl AgNO AgCl NaNO+ → +

5.850.1

58.5w

weightn molM

= = =

1 mol of AgCl is obtained with 1 mol of NaClHence, the number of moles of AgCl obtained with0.1 mol of NaCl = 0.1 mol

0.1143.5W W

weight weight weightn molM M

= ⇒ = =∵

0.1 143.5 14.35weight g g⇒ = × = TYPE-II (MASS - VOLUME RELATIONSHIP)EX. 33. At 1000C for complete combustion of 3g

ethane the required volume of O2 & producedvolume of CO2 at STP will be.

Sol.

2 6( ) 2( ) 2( ) 2 ( )

2 7 4 6

2 7 4 6g g g gC H O CO H O+ → +

3 10.1

30 10w

weightn moleM

= = = =

(a) Required moles of O2 =

70.1 0.35

2mol× =

volume of O2 at STP = 0.35 x 22.4 = 7.84 litre

(b) Produced moles of CO2 =

40.1 0.2

2mol× =

volume of CO2 at STP = 0.2 x22.4 = 4.48 litre

EX. 34. In the following reaction, if 10 g of H2, re-act with N2. What will be volume of NH3 atSTP.N2 + 3H2 → 2NH3

Sol. 1 3 2N

2 + 3H

2 → 2NH

3

10g

105

2w

weightn molM

= = =

Produced moles of NH3 =

2 105

3 3× = ,Volume of

NH3 at STP =

1022.4 74.67

3litre× =

TYPE-III (VOLUME-VOLUME RELA-TIONSHIP)

EX. 35. At 1000C for complete combustion of 1.12litre of butane (C4H10), the produced volumeof H2O(g)& CO2 at STP will be.

Sol. 1 13/2 3 4C

4H

10(g) + 13/2 O

2(g) → 4CO2(g)

+ 5H2O

(g)

1.12 litreVolume of H

2O

(g) at STP = 5 x 1.12 = 5.6 litre

Volume of CO2(g)

at STP = 4x 1.12 = 4.48 litreEX. 36. At 250C for complete combustion of 5 mol

propane (C3H8). The required volume of O2at STP will be.

Sol. For C3H

8, the combustion reaction is

1 5 3 4C

3H

8(g) + 5O

2(g) → 3CO2(g)

+ 4H2O

(g)

5 litreRequired moles of O

2 = 5 x 5 = 25 mol = v/22.4

Volume of O2 gas at STP(V)=25x22.4 = 560 litre

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(B) MORE THAN ONE REACTANTBASED :Limiting reagent (L.R.) conceptLimiting Reagent (L.R.) : The reactant which iscompletely consumed in a reaction is called as re-agent.

Ex. 1 2 1 2 A + 2B → C + 2D

given 3 mol 9 mol 3-3 =0 mol 9-6 = 3 mol 3 mol 6 molL.R. = A

given value (may moles, volume, or molecules)X=

Stoichiometry Co-efficient

Least value of reactant indicates the limitingreagents.

Ex. A + B → P3/1 = 3 9/2 = 4.53 < 4.5 So A is L.R.

EX. 37. A + 5B→C + 3D. In this reaction which isa L.R.

Sol. A + 5B → C + 3DGiven 10mol 10mol

For A For B10/1 = 10 10/5 = 22 < 10 So B is L.R.

EX. 38. H2(g) + 1/2 O2(g)→H2O(g); In the above re-action what is the volume of water vpour atSTP.Given 4 g of H2 and 32 g of O2

Sol. 1 1/2 1H

2(g) + 1/2 O

2(g) → H2O

(g)

4 g 32g For H2 For O

2

n=4/2 = 2mol n=32/32 =1 mol 2/1 = 2 1/ 1/2=2 molMoles of H

2O

(g)= 2 mol=V/22.4

2=2 So Both H2 & O

2 are L.R. volume of

H2O

(g)at STP = 22.4 x 2 = 44.8 litre

EX. 39. At NTP, In a container 100 mL N2 and 100mL of H2 are mixed together. Then find outthe produced volume of NH3.

Sol. Balanced equation will be N

2 + 3H

2 → 2NH3

Given 100mL 100mLFor determination of Limiting reagent. Now di-vide the given quantities by stoichiometry coeffi-cients100/1 =100 100/3 = 33.3 (Limiting reagent)In this reaction H

2 is limiting reagent so reaction

will proceed according to H2.

As per stoichiometry from 3 mL of H2 produces ;

volume of NH3 = 2 mL

That is from 100 mL of H2 produced volume of

NH3 =2/3 x 100 = 66.6 mL

1. 1.5 moles of O2 combine with Mg to form ox-ide MgO. The mass of Mg (At. mass 24) thathas combined is:1) 72 g 2) 36 g 3) 24g 4) 94g

2. What quantity of lime stone on heating willgive 56 kg of CaO1)1000 kg 2) 56 kg 3) 44 kg 4) 100 kg

3. For reaction A + 2 B → C. The amount ofproduct formed by starting the reaction with5 moles of A and 8 moles of B is1) 5 mol 2) 8 mol 3) 16 mol 4) 4 mol

1) 1 2) 4 3) 4

1.5 EQUIVALENT WEIGHTThe equivalent weight of a substance is the num-ber of parts by weight of the substance that com-bine displace directly or indirectly 1.008 parts byweight of hydrogen or 8 parts by weight of oxygenor 35.5 parts by weight of chlorine or 108 partsby weight of Ag.(a) Calculation of Equivalent Weight

(i) Equivalent weightAtomic weight

valency factor=

(ii) Equivalent weight of ions

=formula weight of ion

valency=

(iii) Equivalent weight of ionic compound =equivalent weight of cation + equivalent weightof anionEx. Equivalent weight of H

2SO

4 = Equivalent

weight of H+ + Equivalent weight of Anion(SO4-2)

= 1 + 48 = 49(iv) Equivalent weight of acid / base]

Molecular weight=

Basicity /Acidity

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(v) Equivalent weight of salt =

Molecular weight=

Total charge on cation or anion

Ex. Na2SO

4 (salt) 2Na+ + SO

4-2

Total charge on cation or anion is 2molecular weight of Na

2SO

4 is=(2x23 + 32+16x4)

= 142Equivalent weight of Na

2SO

4 = 142/2 = 71

(vi) Equivalent weight of an oxidizing or reducingagent

Molecular weight of the substance

=Number of electrons gain/lost by one molecule

(b) Concept of gram equivalent and law ofchemical equivalence :

Number of gram equivalent ( )W

= gram

E

(gram)W × valence factor

=M

=n×valence factor; where

number of gram equivalent of soluteNormality=

volume of solution in(L)

⎛ ⎞⎜ ⎟⎝ ⎠

According to it in a reaction equal gram equivalentof reactant are reacts to give same number of gramequivalent of products.For a reactionaA + bB → cC + dDNumber of gram equivalent of A = Number of gramequivalent of B = Number of gram equivalent of C= Number of gram equivalent of D

(c) METHODS FOR DETERMINATION OFTHE EQUIVALENT WEIGHTS(i) Hydrogen displacement method : Thismethod is used for those elements which can evolvehydrogen from acids, i.e., active metals.equivalent weight of metal =

2

weight of metal1.008

weight of H gas(displaced)×

(ii) Oxide formation method : A known mass ofthe element is changed into oxide directly or indi-rectly. The mass of oxide is noted.Mass of oxygen = (Mass of oxide - Mass of ele-ment)equivalent weight of element =

2

weight of element8

weight of O×

(iii) Chloride formation method : A known massof the element is changed into chloride directly orindirectly. The mass of the chloride is determined.Equivalent weight of element =

2

weight of element35.5

weight of Cl×

(iv) Metal to metal displacement method : Moreactive metal can displace less active metal from itssalts solution. The mass of the displaced metal bearthe same ratio as their equivalent weights.

1 1

2 2

m Em E

=

(v) Double decomposition method : this methodis based on the following points -(a) The mass of the compound reacted and themass of product formed are in the ratio of theirequivalent masses.(b) The equivalent mass of the compound (elec-trovalent) is the sum of equivalent masses of itsradicals.(c) The equivalent mass of a radical is equal to theformula mass of the radical divided by its charge.AB + CD→ AD (ppt.) + CB

Mass of AB Equivalent mass of AB=

Mass of AD Equivalent mass of AD

Equivalent mass of A+Equivalent mass of B

=Equivalent mass of A+Equivalent mass of D

(vi) Silver salt method: This method is used forfinding the equivalent weight of carbonic (organic)adds. A known mass of the RCOOAg is changedinto Ag through combustion. The mass of Ag isdetermined.

Equivalent mass of RCOOAg Weight of RCOOAg

Equivalent mass of Ag Mass of Ag=

equivalent weight of RCOOAg =

Weight of RCOOAg108

weight of Ag×

(vi) By electrolysis:1 1

2 2

w Ew E

=

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Where wl & w

2 are deposited weight of metals at

electrodes and E1 and E

2 are equivalent weight

respectively.1.6 METHODS FOR CALCULATION OF

ATOMIC WEIGHTS AND MOLECULARWEIGHTS(a) Methods for Determination of AtomicWeight(i) Atomic weight = equivalent weight x va-lency(ii) Dulong and Petit's law - This law is appli-cable only for solids (except Be, B, Si, C)Atomic mass x specific heat (in cal/ gram x0C)≈ 6.4

or atomic mass (approximate) = 6.4

specific heat

(iii) Law of isomorphism : Isomorphous sub-stances form crystals which have same shape andsize and can grow in the saturated solution of eachother.Examples of isomorphous compounds -1) H

2SO

4 and K

2,CrO

4

2) ZnSO4.7H

2O and FeSO

4.7H

2O and

MgSO4.7H

2O

3) KC1O4 and KMnO

4

4)K2SO

4.Al(SO

4)

3.24H

2Oand

K2SO

4.Cr

2(SO

4)

3.24H

2O

Conclusions -Masses of two elements that combine with samemass of other elements in their respective com-pounds are in the ratio of their atomic masses.

Mass of one element(A) that combines with a

certain mass of other element

Mass of one element(B) that combines with

same mass of other element

Atomic mass of A=

Atomic mass of BThe valencies of the elements forming isomorphouscompounds are the same.(iv) Volatile chloride methodRequired condition - chloride of element shouldbe vapour.Required data - (i) Vapour density of chloride,(ii) Equivalent weight of element.

Let the valency of the element be x. The formulaof its chloride will be MCl

x.

Molecular weight = Atomic weight of M + 35.5 xAtomic weight = Equivalent weight x valency or A = E x xMolecular weight = E x + 35.5 x or 2 xV.D.

= x(E + 35.*5) or x=2 V.D.

x=E+35.5

×

(v) Specific heat method : If CP

VCγ= is given,

thenCase I. If γ = 5/3 = 1.66 Atomicity will be one

Case II. If γ = 7/5 =1.4 Atomicity will be two

Case III. If γ = 4/3 = 1.33 Atomicity will be three

Atomic weight = Molecular weight

Atomicity

(b) Method for Determination of MolecularWeight:(i) Molecular weight = 2 x V.D.(ii) It is sum of atomic weights of elements in agiven compound.

EX. 40. Specific heat of a metal is 0.031C calo

g×

,

and its equivalent weight is 103.6. Calculatethe exact atomic weight of the metal.

Sol. According to Dulong and Petit's law - approxi-mate atomic weight = 6.4/0.031 = 206.45Valency of metal =

Approximate atomic weight 206.45

=1.99=2Equivalent weight 103.6

=

So, the exact atomic weight of the element =Equivalent weight x valency= 103.6 x2 = 207.2

EX. 41. A chloride of an element contains 49.5%chlorine. The specific heat of the element is

0.064C calo

g×

. Calculate the equivalent mass,

valency and atomic mass of the element.Sol. Mass of chlorine in the metal chloride = 49.5

Mass of metal = (100 - 49.5) = 50.5Equivalent weight of metal =

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weight of metal 50.5

35.5 35.5=36.21weight of chlorine 49.50

× = ×

Now according to Dulong and Petit's law,Approximate at. wt. of the metal =

6.4 6.4100

specific heat 0.064= =

Valency =

Approximate atomic weight 100

2.7 3Equivalent weight 36.21

= = =

Hence, exact atomic weight = 36.21 x 3 = 108.63EX. 42. The oxide of an element contains 67.67%

of oxygen and the vapour density of its vola-tile chloride is 79. Calculate the atomic weightof the element.

Sol. Calculation of equivalent weight: weight of oxygen= 67.67 gweight of element = 100 - 67.67 = 32.33 g∴ 67.67 g of oxygen combines with 32.33 g ofelement∴8 g of oxygen combines with

32.33 83.82

67.67

×= = g of element

∴Equivalent weight of the element = 3.82Suppose M represents one atom of the elementand x is its valency. The molecular formula of thevolatile chloride would be MCl

x.

Formula weight of chloride = 3.82 x x + 35.5 x =39.32 xBut molecular weight of Chloride = 2 x V.D.

2 7939.32 2 79 4

39.32x x ×

⇒ = × ⇒ = =

Now atomic weight = Equivalent weight x valencyof element = 3.82 x4 = 15.28

EX. 43. Vapour density of a gas is 16. If the ratio ofspecific heat at constant pressure and spe-cific heat at constant volume is 1.4. Then findout its atomic weight.

Sol. Given : 1.4P

V

CC

γ= = and vapour density = 16

We know that Molecular weight=2 x vapour densityMolecular weight = 2 x 16 = 32Here γ = 1.4 so atomicity will be 2.

Atomic weight = Molecular weight 32

16Atomicity 2

= =

GOLDEN KEY POINTSEquivalent weight of a species changes with reac-tion in which it gets involved.Amount of substance which loses or gains 1 moleelectrons or 96500 coulomb, electricity will alwaysbe its equivalent weight.Victor Mayer's method is used to determine mo-lecular weight of volatile compound.

1. Molecular weight of dibasic acid is W. Itsequivalent weight will be1) W/2 2) W/3 3) W 4) 3W

2. 0.126 g of an acid requires 20 ml of 0.1 NNaOH for complete neutralization. Eq. wt. ofthe acid is 1) 45 2) 53 3) 40 4) 63

3. In a metal oxide 32% oxygen is present whatwill be equivalent mass of metal ?1) 17 2) 34 3) 32 4) 52

4. 1 mol O2 will be equal to1) 4 g equivalent oxygen2) 2 g equivalent oxygen3) 32 g equivalent oxygen4) 8 g equivalent oxygen

5. Volume of one gram equivalent of H2 at NTP is1) 5.6 L 2) 11.2 L 3) 22.4 L 4) 44.8 L

1) 1 2) 4 3) 1 4) 1 5) 2

1.7 LAWS OF CHEMICAL COMBINATION(a) Law of Mass Conservation (Law of Indestructi-

bility of Matter)"It was given by Lavoisier and tested by Landolt"According to this law, the mass can neither be cre-ated nor be destroyed in a balanced chemical re-action or physical reaction. But one form is changedinto another form is called as law of mass conser-vation.If the reactant is completely converted in prod-ucts, then the sum of the mass of reactants is equalto the sum of the mass of products.

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Total mass of reactants= Total mass of products.If reactants are not completely consumed then therelationship will be :Total mass of reactants =Total mass of products + Mass of unreacted

reactants H

2 + Cl

2 → 2HClEx. Mass in (g) 2 71 2(1 + 35.5)

2+ 71 = 73 g 73 gEX. 44. What weight of BaCl2 would react with 24.4

g of sodium sulphate to produce 46.6 g ofbarium sulphate and 23.4 g of sodium chlo-ride ?

Sol. Barium chloride and sodium sulphate react to pro-duce barium sulphate and sodium chlorideaccording to the equation:BaCl

2 + Na

2SO

4 → BaSO4 + 2NaCl

x g 24.4 g 46.6 g 23.4 gLet the weight of BaCl

2 be x g. According to law

of conservation of mass :Total mass of reactants = Total mass of productsTotal mass of reactants = (x + 24.4) gTotal mass of products = (46.6 + 23.4) gEquating the two masses x + 24.4 = 46.6 + 23.4

x = 46.6 + 23.4 - 24.4 or x = 45.6 gHence, the weight of BaCl

2 is 45.6 g

EX. 45. 10g of CaCO3 on heating gives 4.4 g ofCO2 then determine weight of produced CaOin quintal

Sol. CaCO3→CaO + CO

2

10 g x g 4.4 gAccording to few of conservation of mass10 = 4.4 + x l quintal = 100kg 10 - 4.4 = x 1 kg = 1000g x = 5.6 g

weight of CaO(x) =35.6 5.6 10

1000

kg kg−× = ×

-3 -51=5.6×10 × quintal=5.6×10 quintal

100(b) Law of Definite Proportion / Law of Constant

Composition"It was given by Proust."According to this law, a compound can be ob-tained from different sources. But the ratio of eachcomponent (by weight) remain same.i.e. it doesnot depend on the method of its preparation or thesource from which it has been obtained.

For example : molecule of ammonia always hasthe formula NH

3. That is one molecule of ammo-

nia always contains, one atom of nitrogen and threeatoms of hydrogen or 17.0 g of NH

3 always con-

tains 14.0 g of nitrogen and 3 g of hydrogen.Ex. Water can be obtained from different sourcesbut the ratio of weight of H and O remains same.

2 : 16or 1 : 8

H2 OSea waterTap waterRiver waterGanga JalRain water

2H2O 2H2 O2+

EX. 46. Weight of copper oxide obtained by treating2.16 g of metallic copper with nitric add andsubsequent ignition was 2.70 g. In another ex-periment, 1.15 g of copper oxide on reductionyielded 0.92 g of copper. Show that the resultsillustrate the law of constant composition.In I experiment In II experiment

weight of Cu = 2.16 g weight of CuO = 1.15 gweight of CuO = 2.7 g weight of Cu = 0.92 gweight of O

2 = 2.7 - 2.16 weight of O

2= 1.15 - 0.92

= 0.54 g = 0.23 g Cu : O Cu : O2.16 : 0.54 0.92 : 0.23

2.16

0.54 :

0.54

0.54

0.92

0.23 :

0.23

0.23Thus the ratio of the masses of copper and oxygenin the two experiment is the same. Hence the givendata illustrate the law of constant proportion.

EX. 47. In an experiment 2.4 g of FeO on reductionwith hydrogen gives 1.68 g of Fe. In anotherexperiment 2.9 g of FeO gives 2.03 g of Fe onreduction with hydrogen. Show that the abovedata illustrate the law of constant proportion.

Sol. In I experiment In II experimentWeight of FeO = 2.4 g Weight of FeO = 2.9 gWeight of Fe = 1.68 g Weight of Fe = 2.03 gWeight of Oxygen = Weight of Oxygen =2.4 - 1.68 = 0.72 g 2.9 - 2.03 = 0.87 g Fe : O Fe : O

1.68 : 0.72 2.03 : 0.87

1.68

0.72 :

0.72

0.72

2.03

0.87 :

0.87

0.87

2.33 : 1 2.33 : 1

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Thus the ratio of the masses of iron and oxygen inthe two experiment is the same. Hence the givendata illustrate the law of constant proportion.

(c) Law of Multiple Proportion"It was given by John Dalton"According to law of Multiple proportion if two el-ements combine to form more than one compoundthan the different mass of one element which com-bine with a fixed mass of other element bear asimple ratio to one another.The following examples illustrate this law.(i) Nitrogen and oxygen combine to form five ox-ides, which are : Nitrous oxide (N

2O), nitric oxide

(NO), nitrogen trioxide (N2O

3), nitrogen tetraoxide

(N2O

4) and nitrogen pentoxide (N

2O

5).

Weights of oxygen which combine with the fixedweight of nitrogen in these oxides are calculatedas under:Oxide Ratio of weights pf nitrogen and oxygen ineach compoundN

2O 28 : 16 NO 14 : 16 N

2O

3 28 : 48

N2O

4 28 : 64 N

2O

5 28 : 80

Number of parts by weight of oxygen which com-bine with 14 parts by weight of nitrogen from theabove are 8,16,24,32 and 40 respectively. Theirratio is 1 : 2 : 3 : 4 : 5, which is a simple ratio.Hence, the law is illustrated.(ii) Sulphur combines with oxygen to from twooxides SO

2 and SO

3, the weights of oxygen which

combine with a fixed weight of sulphur, i.e. 32 partsby weight of sulphur in two oxides are in the ratioof 32 : 48 or 2 :, 3 which is a simple ratio. Hencethe law of multiple proportions is

EX. 48. Hydrogen peroxide and water contain5.93% and 11.2% of hydrogen respectively.Show that the data illustrate the law of mul-tiple proportions.

Sol. Compound H2O

2Compound H

2O

H : O H : O 5.93 : 94.07 11.2 : 88.8

5.93

5.93 :

94.07

5.93

11.2

11.2 :

88.8

11.2

1 : 15.86 1 : 7.92Thus the ratio of weighs of oxygen which combinewith the fixed weight (1.0 gram) of hydrogen in H

2O

2

and H2O is 15.86 : 7.92 = 2:1 (Which is simple

ratio). So the law of multiple proportion is illustrated.

EX. 49. Carbon combines with hydrogen in P, Q, R.The % of hydrogen in P, Q, R are 25, 14.3,7.7 respectively. Which law of chemical com-bination is illustrated ?

Sol. P Q R H : C H : C H : C 25 : 75 14.3 : 85.7 7.7 : 92.3

1 : 75

25 1 :

85.7

14.3 1 :

92.3

7.7

1 : 3 1 : 6 1 : 12Ratio of C in compounds P, Q and Ris = 3:6: 12 =1:2:4Which is a simple ratio so the data illustrate thelaw of multiple proportion.(d) Law of Gaseous Volume"It was given by Gay Lussac"According to this law, in the gaseous reaction, thereactants are always combined in a simple ratio byvolume and form products, which is simple ratioby volume at same temperature and pressure.

Ex.1 One volume of hydrogen combines with chloride.one volume of chlorine to produce 2 volumes ofhydrogen chloride.Simple ratio =1:1:2.

1 1 2←Stoichiometry H

2(g) + Cl

2(g) → 2 HCl

(g)

1 Volume + 1 Volume 2 VolumeEx.2 One volume of nitrogen combines with 3 volumes

of hydrogen to from 2 volumes of ammonia.Simple ratio 1:3:2 1 3 2 N

2(g) + 3H

2(g) → 2NH3(g)

1 Volume + 3 Volume 2 VolumeSpecial Note : This law is used only for gaseousreaction. It relate volume to mole or molecules.But not relate with mass.

EX. 50. For the gaseous reaction : H2(g) + Cl2(g)→2HCl(g). If 40 mL of hydrogen completely re-acts with chlorine then find out the requiredvolume of Chlorine & volume of producedHCl(g) ?

Sol. According to Gay Lussac's Law: 1 1 2H

2(g)+Cl

2(g)→2HCl(g)

∵1 mL of H2(g)

will react will 1 mL of Cl2(g)

and 2mL of HC1

(g) will produce

∴ 40 mL of will react with 40 mL of Cl2(g)

and 80

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mL of HC1(g)

will producerequired volume of Cl

2(g) = 40 mL

produced volume of HCl(g)

= 80 mLEX. 51. For the gaseous reaction :

H2(g)+Cl2(g)→2HCl(g) . If initially 20 mL of H2(g)and 30 mL of C2(g) are present then find out thevolume of HCl(g) and unreacted part of Cl2(g).

Sol. According to Gay-Lussac's Law1 1 2H

2(g) + Cl

2(g) → 2HCl

(g)

∵1 mL of H2(g)

will react will 1 mL of Cl2(g)

and 2mL of HC1

(g) will produce

∴ 20 mL of will react with 40 mL of Cl2(g)

and 80mL of HC1

(g) will produce

Given volume of Cl2(g)

is 30 mL but its 20 mL re-acts with H

2(g). So 10 mL of Cl

2(g) remains

unreacted.(e) Avogadro's law"Equal volume of all gases contain equal numberof molecules at same temperature and pressure".

Ex. 1 1 2 H

2(g) + Cl

2(g) → 2HCl(g)

1 Volume 1 Volume 2 Volume N molecules N molecules 2N molecules 1/2 molecule 1/2 molecule 1 molecule

(1atom) (1atom)It is correct due to molecule is divisible.

QUESTIONS BASED ON MOLES1. The number of atoms present in 16 g of oxygen

isA) 11.56.02 10× B) 233.01 10×C) 11.53.01 10× D) 236.02 10×

2. No. of atoms in 4.25 g. of NH3 is approx.A) 1 × 1023 B) 1.5 × 1023

C) 2 × 1023 D) 6 × 1023

3. Which of the following contains maximumnumber of oxygen atoms ?A) lg of O B) lg of O

2

C) lg of O3

D) all have the same number of atoms4. The number of atoms present in 0.5g atom of

nitrogen is same as the atoms inA) 12g of C B) 32g of SC) 8g of oxygen D) 24g of Mg

5. Which of the following contains maximumnumber of atoms ?A) 4g of H

2B)16g of O

2

C) 28g of N2

D)18g of H2O

6. Number of neutrons present in 1.7 g ofammonia isA) N

AB) N

A/10 ×4

C) (NA/10)×7 D)N

A× l0×7

7. 5.6 L of oxygen at STP containsA) 6.02×1023 atoms B) 3.01×1023 atomsC) 1.505×1023 atoms D) 0.7525× l023atoms

8. Number of oxygen atoms in 8 g of ozone is

A) 236.02 10× B) 236.02 10

2

×

C) 236.02 10

3

×D)

236.02 10

6

×

9. The number of atoms in “n” mole of gas canbe given by

A) n ×Av. No.×atomicity B) n Av.No.

Atomicity

×

C)

Av.No. Atomicity

n

×

D) None10. Sum of number of protons, electrons and

neutrons in 12g of 126 C is

A) 1.8 B) 12.044× l023

C) 1.084× l025 D)10.84× l023

11. The weight of one atom of Uranium is 238 amu.Its actual weight is .... g.A) 1.43× l026 B) 3.94× l0–22

C) 6.99× 10–23 D) 1.53× l0–22

12. The actual weight of a molecule of water isA) 18 g B) 2.99 × l0–23 gC) both 1) & 2) are correctD) 1.66× l0–24g

13. What is the mass of a molecule of CH4

A) 16g B) 2226.6 10 g×

C) 232.66 10 g−× D) 16 NAg

14. Which of the following has the highest mass ?A) 1 g atom of C

B) 1

2 mole of CH

4

C) 10 mL of waterD) 3.011×1023 atoms of oxygen

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15. Which of the following contains the leastnumber of molecules ?A) 4.4 g CO

2B) 3.4 g NH

3

C)1.6gCH4

D)3.2gSO2

16. The number of molecule in 4.25 g of NH3 isA) 1.505 × l023 B)3.01× l023

C) 6.02× 1023 D) None of these17. Elements A and B form two compounds B2A3

and B2A. 0.05 moles of B2A3 weight 9.0 g and0.10 mole of B2A weight 10 g atomic weight ofA and B areA) 20 and 30 B) 30 and 40C) 40 and 30 D) 30 and 20

18. 5.6 L of oxygen at NTP is equivalent to

A) 1 mole B) 12 mole

C) 1

4 mole D) 1/8 mole

19. 4.4 g of an unknown gas occupies 2.24 L ofvolume at STP. The gas may beA)N

2O B) CO

C)CO2

D) l&3 both20. Which contains least no. of molecules

A) 1 g CO2

B) 1 g N2

C) 1 g O2

D) 1 g H2

21. If V mL of the vapours of substance at NTPweight W g. Then molecular weight ofsubstance is

A) (W / V) 22400× B) V

22.4W×

C) (W V) 22400− × D) W 1

V 22400

××

22. If 3.01× 1020 molecules are removed from 98mg of H2SO4, then the number of moles ofH2SO4 left areA) 30.1 10−× B) 30.5 10−×C) 31.66 10−× D) 29.95 10−×

23. A gas is found to have the formula (CO)x. It’sVD is 70 the value of x must beA) 7 B) 4 C) 5 D) 6

24. Vapour density of gas is 11.2. Volume occupiedby 2.4 g of this at STP will beA) 11.2L B) 2.24 LC) 22.4 L D) 2.4 L

25. The volume of a gas in discharge tube is1.12× l07 mL at STP. Then the number ofmolecule of gas in the tube isA) 3.01×104 B) 3.01× l015

C) 3.01× 1012 D) 3.01× 1016

26. A person adds 1.71 gram of sugar (C12H22O11) inorder to sweeten his tea. The number of carbonatoms added are (mol. mass of sugar = 342)A) 3.6 ×1022 B) 7.2×1021

C) 0.05 D) 6.6× l022

27. The total number of ions persent in 1 mL of0.1 M barium nitrate Ba(NO3)2 solution isA) 6.02× l018 B) 6.02×1019

C) 3.0× 6.02× l019 D) 3.0×6.02× l018

28. The weight of 1 mole of a gas of density 0.1784gL–1 at NTP isA) 0.1784 g B) l gC) 4g D) 4 amu

29. Given that one mole of N2 at NTP occupies22.4 litre the density of N2 isA) 1.25 g L–1 B) 0.80 g L–1

C) 2.5gL– 4)1.60g L–1

30. The number of gram molecules of oxygen in6.02× 1024 CO molecules isA) 10 g molecules B) 5 g moleculesC) 1 g molecules D) 0.5 g molecules

QUESTIONS BASED ON PERCENTAGE,EMPERICAL FORMULA &MOLECUALR FORMULA

31. A compound of X and Y has equal mass of them.If their atomic weights are 30 and 20respectively.A) X

2Y

2B) X

3Y

3

C) X2Y

3D) X

3Y

2

32. An oxide of sulphur contains 50% of sulphurin it. Its emperical formula isA) SO

2B) SO

3C) SO D) S

2O

33. A hydrocarbon contains 80% of carbon, thenthe hydrocarbon isA) CH

4B) 2 4C H C) 2 6C H D) 2 2C H

34. Emperical formula of glucose isA) C

6H

12O

6B) C

3H

6O

3

C) 2 4 2C H O D) CH2O

35. An oxide of metal M has 40% by mass ofoxygen. Metal M has atomic mass of 24. Theemperical formula of the oxide

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A)M2O B)M

2O

3 C) MO D) M

3O

4

36. A compound contains 38.8% C, 16.0% H and45.2% N. The formula of the compound wouldbeA) CH

3NH

2B) CH

3CN

C) C2H

5CN D) CH

2(NH)

2

37. The simplest formula of a compound containing50% of element X(at wt. = 10) and 50% ofelement Y(atwt. = 20) isA)XY B)X

2Y C)XY

2 D)X

3Y

38. Which of the following compounds has sameempirical formula as that of glucoseA) CH

3CHO B) CH

3COOH

C) CH3OH D) C

2H

6

39. A gas is found to contain 2.34 g of Nitrogenand 5.34 g of oxygen. Simplest formula of thecompound isA)N

2O B) NO C) N

2O

3 D) NO

2

40. 2.2 g of a compound of phosphorous andsulphur has 1.24 g of ‘P’ in it. Its empericalformula isA) P

2S

3B) P

3S

2C) P

3S

4D) P

4S

3

41. On analysis, a certain compound was found tocontain iodine and oxygen in the ratio of254:80. The formula of the compound is(At mass I = 127, O = 16)A) IO B)I

2O C) I

5O

2 D)I

2O

5

42. The number of atoms of Cr and O are4.8× 1010 and 9.6× 1010 respectively. Itsempirical formula isA) Cr

2O

3B)CrO

2C) Cr

2O

4D) CrO

s

43. Insulin contains 3.4% sulphur ; the minimummolecular weight of insulin isA)941.176 B)944C) 945.27 D) None

44. A giant molecule contains 0.25% of a metalwhose atomic weight is 59. Its moleculecontains one atom of that metal. Its minimummolecular weightA) 5900 B) 23600

C) 11800 D) 100 59

0.4

×

45. Caffine has a molecular weight of 194. Itcontains 28.9% by mass of nitrogen, Numberof atoms of nitrogen in one molecule of it.A) 2 B) 3 C) 4 D) 5

QUESTIONS BASED ONSTOICHIOMETRY

46. In a gaseous reaction of the typeaA bB cC dD+ ⎯⎯→ + ,which statement is wrong ?A) a litre of A combines with b litre of B to giveCand DB) a mole of A combines with b moles of B to giveCand DC) a g of A combines with b g of B to give Cand DD) a molecules of A combines with b molecules ofB to give C and D

47. Assuming that petrol is octane (C8H18) and hasdensity 0.8 g mL–1, 1.425 litre of petrol onA) 50 mole of O

2B) 100 mole of O

2

C) 125 mole of O2

D) 200 mole of O2

48. 9g of Al will react, with 2 2 3

32Al O Al O

2+ →

A) 6g O2

B) 8g O2

C) 9g O2

D) 4g O2

49. The equation ;

(s) 2 2 3(s)

32Al O (g) Al O

2+ → shows that

A) 2 mole of Al reacts with 3

2 mole of O

2 o

3

2to

produce 7

2 mole of 2 3Al O

B) 2g of Al reacts with 3

2g of O

2 to produce one

mole of 2 3Al O

C) 2 g of AI reacts with 3

2litre of O

2 to produce 1

mole of A12O

3

D) 2 mole of Al reacts with 3

2mole of O

2 to

produce 1 mole of Al2O

3

50. 1L of CO2 is passed over hot coke. when thevolume of reaction mixture becomes 1.4L, thecomposition of reaction mixture isA) 0.6 L CO B) 0.8 L CO

2

C) 0.6LCO2 and 0.8 L CO D) None

51. 26 cc of CO2 are passed over red hot volumeof CO evolved is

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(l) 15cc B)10cc C) 32 cc D) 52 cc52. If 1/2 moles of oxygen combine with A to form

A12O3 then weight of Aluminium in the reactionis (Al= 27).A) 27g B)18g C) 54g D) 40.5 g

53. The number of litres of air required to burn 8litres of C2H2 is approximately.A) 40 B) 60 C) 80 D) 100

54. If 0.5 mole of BaCl2 is mixed with 0.2 mole ofNa3PO4 the maximum number of moles ofBa3(PO4)2 that can be formed is3BaCl2 + 2Na3 PO4 → Ba3 (PO4)2 + 6NaClA) 0.7 B) 0.5 C) 0.3 D) 0

55. If 8 mL of uncombined O2 remain afterexplodinc O2 with 4 mL of hydrogen, thenumber of mL of O2 originally wereA) 12 B) 2 C) 10 D) 4

56. 4 g of hydrogen are ignited with 4 g of oxygen.The weight of water formed isA) 0.5g B)3.5g C) 4.5 g D) 2.5 g

57. For the reaction A + 2B ⎯⎯→C, 5 mole of AAand 8 mole of B will produceA) 5 mole of C B) 4 mole of CC) 8 mole of C D) 13 mole of C

58. If 1.6 g of SO2 and 221.5 10× molecules of H2Sare mixed and allowed to closed vessel untilthe reaction2H2S + SO2 ⎯⎯→3S + 2H2O,proceeds to completion. Which of the followingstatement is true ?A) Only ‘S’ and ‘H

2O’ remain in the reaction

vesses.B) ‘H

2S’ will remain in excess

C) ‘SO2’ will remain in excess

4) none59. 12L of H2and 11.2Lof Cl2 are mixed and

exploded. The composition by volume ofmixture isA) 24L of HCl(g)B) 0.8 L Cl

2 and 20.8 L HC1 (g)

C) 0.8 L H2 and 22.4 L HCl (g)

D) 22.4 L HCl (g)60. 10mL of gaseous hydrocarbon on combustion

given 40mL of CO2(g) and 50mL of H2O (vap).The hydrocarbon isA) C

4H

5B) C

8H

10C) C

4H

8D) C

4H

10

61. 500 mL of a gaseous hydrocarbon when burnexcess of O2 gave 2.5 L of CO2 and 3.0 L ofwater vapours under same conditions.Molecular form of the hydrocarbon isA) C

4H

8B) C

4H

10C) C

5H

10D) C

5H

12

QUESTIONS BASED ON EQUIVALENTWEIGHTS

62. Molecular weight of tribasic acid is W. Itsequivalent weight will be

A) W

2B)

W

3C) W D) 3W

63. A, E, M and n are the atomic weight,equivalent weight, molecular weight andvalency of an element. The correct relation is

A) A E n= × B) M

AE

=

C) M

An

= D) M A n= ×

64. Sulphur forms two chlorides S2Cl2 and SCl2.The equivalent mass of sulphur in SCl2 is 16.The equivalent weight of sulphur in S2Cl2 isA) 8 B) 16 C) 32 D) 64

65. If equivalent weight of S in SO2 is 8 thenequivalent weight of S in SO3 is

A) 8 2

3

×B)

8 3

2

×C) 8 2 3× × D)

2 3

8

×

66. Which property of an element is not variableA) Valency B) Atomic weightC) Equivalent weight D) None

67. One g equivalent of a substance is present inA) 0.25 mole of O

2B) 0.5 mole of O

2

C) 1.00 mole of O2

D) 8.00 mole of O2

68. In a compound AxBy,A) Mole of A = mole of B = mole of Ax ByB) eq. of A = eq of B = eq. of AxByC) yx mole of A = yx mole of B = (x + y) x mole ofAxByD) y xmole of A = y xmole of B

69. 0.45g of acid (molecular wt. = 90) was exactlyneutralised by 20mL. of 0.5N NaOH. Basicityof the acid isA) 1 B) 2 C) 3 D) 4

70. 0.5 g of a base was completely neutralised by100 mL. of 0.2 N acid. Equivalent weight ofthe base is

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A) 50 B) 100 C) 25 D) 12571. 0.126 g of an acid requires 20 mL of 0.1 N

NaOH for complete neutralisation. Equivalentweight of the acid isA) 45 B) 53 C) 40 D) 63

72. 2g of a base whose equivalent weight is 40reacts with 3g of an acid. The equivalent weightof the acid isA) 40 B) 60 C) 10 D) 80

73. Equivalent weight of a divalent metal is 24.The volume of hydrogen liberated at STP by12 g the same metal when added to excess ofan acid solution isA) 2.8 litres B) 5.6 litresC) 11.2 litres D) 22.4 litres

74. 0.84g of a metal carbonate react’s exactly with40mL of N/2 H2SO4. The equivalent weight ofthe metal carbonate isA) 84 B) 64 C) 42 D) 32

75. 1.0g of a metal combines with 8.89g ofBromine. Equivalent weight of metal is nearly(at.wt. of Br = 80)A) 8 B) 9 C) 10 D) 7

76. H3PO4 is a tribasic acid and one of its salt isNaH2PO4. What volume of 1M NaOH solutionshould be added to 12g NaH2PO4 to onvert itinto Na3PO4? (at.wt of P = 31)A) 100 mL B) 200 mL C) 80 mL D) 300 mL

77. 0.84 g. of metal hydride contains 0.04 g ofhydrogen. The equivalent wt. of metal isA) 80 B) 40 C) 20 D) 60

78. A1 g of an element give A2 g of its oxide. Thefor complete neutralisation. Equivalent weightof the acid is

A) 2 1

1

A A8

A

−× B)

2 1

2

A A8

A

−×

C) 1

2 1

A8

A A×

− D) 2 1(A A ) 8− ×

79. When an element forms an oxide in whichoxygen is 20% of the oxide by mass, theequivalent mass of the element will be.A) 32 B) 40 C) 60 D) 128

80. If 1.2g of a metal displaces 1.12 litre ofhydrogen at NTP, equivalent mass of the metalwould be.A) 1.2 11.2× B) 12

C) 24 D) 1.2 + 11.281. One g of hydrogen is found to combine with

80g of bromine. One g of calcium (valency =2) combines with 4g of bromine. Theequivalent weight of calcium isA) 10 B) 20 C) 40 D) 80

82. 2.8g of iron displaces 3.2g of copper from asolution of copper suphate solution. If theequivalent mass of iron is 28, then equivalentmass of copper will be.A) 16 B) 32 C) 48 D) 64

83. A metal oxide is reduced by heating it in astream of hydrogen. It is found that aftercomplete reduction 3.15g of the oxide haveyielded 1.05g of the metal. We may concludethatA) Atomic weight of the metal is 4B) Equivalent weight of the metal is 8C) Equivalent weight of the metal is 4D) Atomic weight of the metal is 8

84. If m1g of a metal A displaces m2g of anothermetal B from its salt solution and if theirequivalent weights are E2 and E1 respectivelythen the equivalent weight of A can beexpressed by

A) 1

22

mE

m× B)

22

1

mE

m×

C) 1

12

mE

m× D)

21

1

mE

m×

85. 14 g of element X combines with 16 g ofoxygen. On the basis of this information, whichof the following is a correct statementA) The element X could have an atomic wt. of 7and its oxide is XOB) The element X could have an atomic wt. of 14and its oxide is X

2O

C) The element X could have an atomic wt. of 7and its oxide is X

2O

D) The element X could have an atomic wt. of 14and its oxide is XO

2

86. If 2.4 g of a metal displace 1.12 litre hydrogenat normal temperature and pressureequivalent weight of metal would beA) 12 B) 24C)1.2 11.2× D)1.2 11.2÷

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87. 45g of acid of mol. wt . 90 neutralised by 200mLof 5N caustic potash. The basicity of acid isA) 1 B) 2 C) 3 D) None

88. The weights of two elements which combinewith one another are in the ratio of theirA) Atomic weight B) Molecular weightC) Equivalent weight D) None

89. The oxide of a metal has 32% oxygen. It’sequivalent weight would beA) 34 B)32 C) 17 D) 16

90. 1.6 g of Ca and 2.60 g of Zn when treated withan acid in excess separately, produced thesame amount of hydrogen. If the equivalentweight of Zn is 32.6, what is the equivalentweight of Ca.A) 10 B)20 C)40 D)5

91. 74.5g of a metallic chloride contains 35.5g ofchlorine. The equivlaent mass of the metal isA) 19.5 B) 35.5 C) 39.0 D) 78.0

QUESTIONS BASED ON CALCULATIONOF ATOMIC WEIGHTS ANDMOLECULAR WEIGHTS

92. The equivalent weight of an element is 4. It’schloride has a V.D. 59.25. Then the valency ofthe elements isA) 4 B) 3 C) 2 D) 1

93. Vapour density of metal chloride is 77.Equivalent weight of metal is 3, then its atomicweight will beA) 3 B) 6 C) 9 D) 12

94. Specific heat of a solid element is 0.1 cal g–1

oC and its equivalent weight is 31.8. Its exactatomic weight isA) 31.8 B) 63.6 C) 3.18 D) 95.4

95. The specific heat of an element is 0.214 calg–1 oC . The approximate atomic weight isA)0.6 B)12 C)30 D) 65

96. A metal M forms a sulphate which isisomorphous with MgSO47H2O. If 0.6538g ofmetals M displaced 2.16g of silver from silvernitrate solution, then the atomic weight of themetal M isA) 32.61 B) 56.82 C) 65.38 D) 74.58

97. The carbonate of a metal is isomorphous withMgCO3 and contains 6.091% of carbon.Atomic weight of the metal is nearlyA) 48 B) 68.5 C) 137 D) 120

98. 71 g of chlorine combines with a metal giving111g of its chloride. The chloride isisomorphous with MgCl2.6H2O. The atomicmass of the metal isA)20 B)30 C)40 D)69

99. The atomic weight of a metal (M) is 27 and itsequivalent weight is 9, the formula of itschloride will beA) MCl B) MCl

2C) M

3Cl D) MCl

3

100. The chloride of a metal contains 71% chlorineby weight and the vapour density of its is 50,the atomic weight of the metal will beA)29 B)58 C)35.5 D)71

101. The specific heat of a metal M is 0.25. Its eq.wt. is 12. What is it’s correct at wt.A) 25.6 B) 36 C) 24 D) 12

102. The density of air is 0.001293 g ml–1. It’s vapourdensity isA)143 B)14.3 C)1.43 D)0.143

103. Relative density of a volatile substance withrespect to CH4 is 4. Its molecular weight wouldbeA) 8 B) 32 C) 64 D) 128

104. The ratio of specific heat at constant pressureto specific heat at constant volume is 1.4, thenits atomic weight isA)8 B)16 C)24 D)32

105. The weight of substance that displace 22.4 litreair at NTP isA) Mol. wt. B) At.wtC) Eq. wt. D) All

106. 0.39 g of a liquid on vapourisation gave 112mL of vapour at STP. Its molecular weight isA)39 B)18.5 C)78 D)112

107. In victor Mayer’s method 0.2 g of a volatilecompound on volatilisation gave 56 mL. ofvapour at STP. Its molecular weight isA) 40 B) 60 C) 80 D) 120

108. 510 mg of a liquid on vapourisation in VictorMayer’s apparatus displaces 67.2 cc of dry air(at NTP). The molecular weight of liquid isA) 130 B) 17 C) 1700 D) 70

109. 5 litre of gas at STP weighs 6.25 g. What is itsgram molecular weight ?A) 1.25 B) 14 C) 28 D) 56

110. 0.44 g of a colourless oxide of nitrogenoccupies 224 mL. at STP. The compound isA) N

2O B) NO C)N

2O

4D) NO

2

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111. One litre of a certain gas weighs 1.16 g at STP.The gas may posibly beA) C

2H

2B) CO C) O

2D) NH

3

112. Equivalent weight of bivalent metal is 32.7.Molecular weight of its chloride isA)68.2 B)103.7 C)136.4 D)166.3

113. The oxide of an element possess the molecularformula M2O3. If the equivalent mass of themetal is 9, the molecular mass of the oxidewill beA)27 B)75 C)102 D)18

QUESTIONS BASED ON LAWS OFCHEMICAL COMBINATION

114. The law of multiple proportion was proposedbyA) Lavoisier B) DaltonC) Proust D) Gaylusac

115. Which one of the following pairs of compoundsillustrate the law of multiple proportions ?A) H

2O, Na

2O B) MgO, Na

2O

C) Na2O, BaO D) SnCl

2, SnCl

4

116. In the reaction N2 + 3H2 ⎯⎯→2 NH3, ratio byvolume of N2, H2 and NH3 is 1: 3: 2. ThisillustratesA) Difinite proportionB) Multiple proportionC) Law of conservation of massD) Gaseous volumes

117. Different proportions of oxygen in the variousA) Equivalent proportionB) Multiple proportionC) Constant proportionD) Conservation of matter

118. Oxygen combines with two isotopes of carbon12C and 14C to form two sample of carbon di-oxide. The data illustrates.A) Law of conservation of massB) Law of multiple proportionsC) Law of gaseous volumeD) None of these

119. The law of conservation of mass holds goodfor all of the following exceptA) All chemical reactionsB) Nuclear reactionsC) Endothermic reactionsD) Exothermic reactions

120. Number of molecules in 100 mL of each of O2,NH3and CO2 at STP areA) in the order CO

2 < O

2 < NH

3

B) in the order NH3 < O

2 < CO

2

C) the sameD) NH

3 = CO

2 < O

2

121. The emperical formula of an organic compoundcontaining carbon and hydrogen is CH2. Themass of one litre of this organic gas is exactlyequal to that of one litre of N2 at sametemperature and pressure. Therefore, themolecular formula of the organic gas isA) C

2H

4B) C

3H

6

C) C6H

12D) C

4H

8

122. Four one litre flasks are seperately filled withthe gases hydrogen, helium oxygen and ozoneat same room temp, and pressure. The ratioof total number of atoms of these gases presentin the different flasks would beA) 1 : 1 : 1 : 1 B) 1 : 2 : 2 : 3C) 2 : 1 : 2 : 3 D) 2 : 1 : 3 : 2

123. A container of volume V, contains 0.28 g of N2gas. If same volume of an unknown gas undersimilar condition of temperature and pressureweighs, 0.44g, the molecular mass of the gas isA) 22 B) 44 C) 66 D) 88

124. A and B are two identical vessels. A contains15 g ethane at 1 atm and 298 K. The vessel Bcontains 75 g of a gas X2 at same temperatureand pressure. The vapour density of X2 isA) 75 B) 150 C) 37.5 D) 45

125. When 100 g of ethylene polymerizes topolythylene according to equationnCH2 = CH2 → -(- CH2 - CH2 -)n-. The weightof polyethylene produced will be

A)n

g2

B)100g C)100

gn

D)100ng

126. If law of conservation of mass was to hold true,then 20.8 g of BaCl2 on reaction with 9.8 g ofH2SO4 will produce 7.3 g of HCl and BaSO4equal toA)11.65 g B) 23.3 gC) 25.5 g D) 30.6 g

127. A chemical equation is balanced according tothe law ofA) Multiple proportions B) Constant proportionsC) Gaseous volume D) Conservation of mass

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128. Two flasks A & B of equal capacity containNH3 and SO2 gas respectively under similarconditions which flask has more no.of moles.A) A B) BC) Both have same moles D) None

1) D 2) D 3) D 4) C 5) A6) C 7) B 8) B 9) A 10) C11) B 12) B 13) C 14) A 15) D16) A 17) C 18) C 19) D 20) A21) A 22) B 23) C 24) D 25) C26) A 27) C 28) C 29) A 30) B31) C 32) A 33) C 34) D 35) C36) A 37) B 38) B 39) D 40) D41) D 42) B 43) A 44) B 45) C46) C 47) C 48) B 49) D 50) C51) D 52) B 53) D 54) D 55) C56) C 57) B 58) C 59) C 60) D61) D 62) B 63) A 64) C 65) A66) B 67) A 68) B 69) B 70) C71) D 72) B 73) B 74) C 75) B76) B 77) C 78) C 79) A 80) B81) B 82) B 83) C 84) C 85) C86) B 87) B 88) C 89) C 90) C91) C 92) B 93) D 94) B 95) C96) C 97) C 98) C 99) D 100) A101) C 102) B 103) C 104) B 105) A106) C 107) C 108) D 109) C 110) A111) A 112) C 113) C 114) B 115) D116) D 117) B 118) D 119) B 120) C121) A 122) C 123) B 124) A 125) B126) B 127) D 128) C

3. 1) 1g of 1

16 AO N= × atoms

2) 1 gm of 2

12

32 AO N= × atoms

3) 1 gm of 3

13

48 AO N= × atoms

Hence all have the same no. of atoms

4. No. of atoms in 0.5 gm atoms of nitrogen = 0.5NA

8 gm of oxygen 8

16=

= 0.5 NA atoms of oxygen

6. 1 mol. of NH3 contains 7 AN× neutrons

1.7 gms of 3

1.70.1

17NH = = mol.

∴ 0.1 mol contains 0.1 7 AN⇒ × × neutrons

7. Moles of 2

5.6 1

22.4 4O = =

∵ 1 mol of 2O contains 2 AN atoms

1

4∴ mol of 2O contains

2

4 2A AN N= atoms

8. ∵ 1 mol of 3O contains 3 AN atoms of oxygen

8 gm 3

8

48O = moles

1

6= mol

1

6∴ mol of 3O contains

13

6 2A

ANN⇒ × =

17. Let the atomic weight of A and B are a & b

wt. of 0.05 moles of 2 3 9B A = and

wt. of 0.1 mole of 2 10B A = gms

then wt. of 1 moles of 2 3 180B A = and

wt. of 1 mole of 2 100B A =

∵ wt. of 1 mole of compound = mol wt.

mol. wt. of 2 3 : 2 3 180B A b a+ = .....(i)

mol. wt. of 2 : 2 100B A b a+ = .....(ii)

solve eq. (i) & (ii) and get value of a & b

23. Mol. wt. 2 . .V D= ×

2 70 140= × ×

mol. wt. of ( ) 28xCO x=

Hence 28 140x =24. mol. wt. 2 . .V D= ×

2 11.2 22.4⇒ × =

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moles of gas 2.4

[22.4

= ∵ 1 mole occupies = 22.4 L]

∴ vol. of gas = 2.4 L

25. Moles of gas ( )

( )71.12 10

22400

mLmL

−×⇒

No. of molcules of gas 71.12 10

22400 AN−×

= ×

26. moles of sugar ( )12 22 11

1.71

342C H O =

∵ In one mole 12 22 11C H O no. of C atom = 12NA

1.71

342∴ mole 12 22 11C H O no. of C atom

1.7112

342 AN= ×

27. Moles of ( ) ( )3 2Ba NO M V Lit= ×

3 40.1 10 10 mol− −= × =

∵ 1 mol ( )3 2Ba NO contains 3 mol ions

410−∴ mol ( )3 2Ba NO contains 43 10−× mol ions

43. % of 1 . . 100at wt of SS

Minimum molecular mass× ×

=

1 32 1003.4

m× ×

=

Molecular mass ( ) 32 100

3.4m ×

=

44. % of metal

1 . .100

at wt of metalMinimum molecular weight

×= ×

1 59 1000.25

m× ×

=

59 100 10033600

0.25m × ×= =

45. wt. of nitrogen in caffine 194 28.9

100

×=

No. of atom in one molecule 194 28.9

2100 28

×= ×

×

64.

( )

2 2ln

2 32 2 35.5

32 35.5

32

S ClRatio of wt

E s

= × ×∴

∴ =

67. 0.25∵ mole of 2 20.25 32 8O gm of O= × =

8. 1

8

Wgm eqE

= = =

70. gm eq. of acid = gm eq. of Base

W N VE= ×

73. Here gm eq. of metal = gm eq. of H2

1 2

2 2

W WE E

= 22

120.5

24 1

W W gm= =

22 22.4gm H L=

20.5 5.6gm H L=

74. gm eq. of metal carbonate = gm eq. of H2SO

4

( )W N V LE= ×

0.84 1 40

2 1000E= ×

75. Eq. wt. of Br = 80

∵ 8.89 gm Br is combines with 1 gm of a metal

∴ 80 gm Br is combines with 1

808.99

×

of a metal (Eq. wt. of metal)

(∵ elements combines in the ratio of their volumes)

76. gm eq. of NaH2PO

4 ≡ gm eq. of NaOH

W N VE= ×

[NaH2PO

4 is acidic salt & its V.F = 2,

12060

2E = = ]

77. wt. of metal hydride = 0.84 gm

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wt. of hydrogen = 0.04 gm

∴ wt. of metal = 0.84 – 0.04 = 0.80 gm.

Eq. wt. of metal 2

.

.

wt of metalwt of H

=

79. If wt. of oxide = 100

then wt. of oxygen = 20

wt. of element = 100 – 20 = 80

Eq. wt. of element .

8.

wt of elementwt of oxygen

= ×

80. Vol. of hydrogen 1.12

1.12 2 0.122.4

L g g= ⇒ × =

∵ 0.1 gm H2 is displaced by

→ 1.2 gm of metal

∴ 1 gm H2 is displaced by

→ 12 gm = eq. wt. of metal

81. ∵ One gm of hydrogen is combined with 80 gmof Br

∴ Eq. wt. of Br = 80

and we known that element combines in the ratioof their eq. wt

∵ 4 gm of Br combines with 1 gm of Ca

∴ 80 (eq. wt) of Br combines with 20 gm ofCa(eq. wt.)

83. Eq. wt .

8.

wt of metalwt of oxygen

= ×

94. According to Dulong & Petti’s rule

Atomic weight × sp. heat � 6.4

(Approx.) at wt. 6.4

640.1

=�

Eq. wt = 31.8

valency of element ( ) 642

31.8n = =

Atomic weight n E= ×96. Formula of metal sulphate will be MSO

4.7H

2O

Valency of metal (M) n = 2

( ). . ( ) .

. . 108 .

Eq wt of metal E wt of metalEq wt of Ag wt of Ag

==

At. wt of the metal (M) n E= ×

97. Formula of metal carbonate = MCO3

at. wt. of metal = A

mol wt. of MCO3 = A + 60

% of carbon 12

1000 6.09160A

= × =+

Calcualte A

98. Wt. of Chlorine = 71

Wt. of Chloride = 111 gm.

Wt. of metal = 111 - 71 = 40 gm.

Eq. wt. of metal 40

35.5 2071

= × =

∵ The Chloride is isomorphous with MgCl2.6H

2O

Hence its formula will be MCl2.6H

2O

∴ Valency of metal (M) = 2

∴ At. Wt = E × Valency

100. Valency of metal ( ) ( )2 .

35.5

V DxE×

=+

Hence At. wt. E x= ×

102. ( )2

.0.000089 /

density of gasV Ddensity of H gm ml

=

103.4

1.

1

mass of mol gasV Dmass of mol CH

=

416

m= 64m =

So, 64

. . 322 2

mV D = = =

V.D. of substance = 4

but V.D. of CH4 = 8,

∴ V.D. of substance = 32 ∴ mol. wt. 2 32= ×104. mol. wt of a gas 2 . . 2 16 32V D= × = × =

1.4,γ = atomaticity = 2

∴ at. wt. = 16

106. ∵ wt. of 112 mL of vapour at STP = 0.39 gm

∴ wt. of 22400 mL (1 mol) at STP

0.3922400

112= × gm (mol. wt.)

108. vol. of vapour = 67.2 cc

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wt. of liquid = 510 mg

22400 cc vapour mass will be equal to molar mass.

109. ∵ at STP wt. of 5L of gas = 6.25 gm

∴ at STP wt. of 22.4 L of gas

6.2522.4

5= × gm (gm molecular wt.)

110. At STP 224 mL occupies → 0.44 gms.

∴ At STP 22400 mL occupies ⇒ 44 gm.

(1 mol.)

and wt. of N2O is 44 gm.

121. ∵ mass of 1 L of this organic gas

= mass of 1 L of N2

∴ mass of 22.4 L (1 mol) organic gas

= mass of 22.4 L of N2 (mol wt) ⇒ 28 gm

∴ mol. wt of organic gas = 28 gm

empirical formula 2,CH= It’s wt. = 14

282

14n = =

mol. formula 2 4C H=

122. Equal volume of gases contains equal no. ofmolecules

Molecules

Atoms

2 2 3

1 : 1 : 1 : 1

2 : 1 : 2 : 3

H He O O

123. ∵ equal volumes contains equal molecules (ormoles)

∴ moles of 2N = moles of gas

(Let mol. mass of the gas = m)

0.28 0.44

28 m=

124. Equal volume contains equal molecules moles ofethane = moles of gas X

2

15 75

30 m=

mol. wt. (m) = 150 . . 75V D∴ =

AIPMT 20081. What volume of oxygen gas (O2) measured at

0oC and 1 atm, is needed to burn completely1L, of propane gas (C3H8) measured under thesame conditionsA) 5L B) 10L C) 7L D) 6L

2. Volume occupied by one molecule of water(density =1 gcm–3) isA) 23 33.0 10 cm−× B) 23 35.5 10 cm−×C) 23 39.0 10 cm−× D) 23 36.023 10 cm−×

3. How many moles of lead (II) chloride will beformed from a reaction between 6.5g of PbOand 3.2g of HCl ? (Atomic wt. of Pb = 207)A) 0.011 B) 0.029 C) 0.044 D) 0.333

4. An organic compound contains carbon,hydrogen and oxygen. Its elemental analysisgives C, 38.71% and H, 9.67%. The empiricalformula of the compound would beA) CHO B) CH

4O C) CH

3O D) CH

2O

AIPMT 20095. 10g of hydrogen and 64g of oxygen were filled

in a steel vessel and exploded. Amount ofwater produced in this reaction will beA) 1 mol B) 2mol C) 3 mol D) 4mol

AIPMT 20106. The number of atoms in 0.1 mol of a triatomic

gas is (NA = 23 16.02 10 mol−× )A) 221.800 10× B) 226.026 10×C) 231.806 10× D) 233.600 10×

AIPMT Mains 20117. Which has the maximum number of molecules

among the following ?A) 64g SO

2B) 44g CO

2

C) 48g O3

D) 8g H2

NEET UG 20138. An excess of AgNO3 is added to 100mL of a

0.01M solution of dichlorotetraaquachromium(III) chloride. The number ofmoles of AgCl precipitated would beA) 0.01 B) 0.001 C) 0.002 D) 0.003

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9. Equal masses of H2,O2 and methane havebeen taken in a container of volume V attemperature 27°C in identical conditions. Theratio of the volumes of gases H2: O2 : methanewould beA) 8 : 16 : 1 B) 16 : 8 : 1C) 16 : 1 : 2 D) 8 : 1 : 2

10. When 22.4 litres of H2(g) is mixed with 11.2 litof Cl2(g), each at S.T.P., the moles of HCl (g)fermed is equal toA) 1 mol of HCl (g) B) 2 mol of HCl (g)C) 0.5 mol of HCl (g) D) 1.5 mol of HCl (g)

11. 1.0g of magnesium is burnt with 0.56g O2 in aclosed vessel. Which reactant is left in excessand how much ?(At. wt. Mg = 24 ; O = 16)A) Mg, 0.16g B) O

2, 0.16g

C) Mg, 0.44g D) O2, 0.28g

AIPMT 201512. A mixture of gases contains H2 and O2 gases

in the ratio of 1: 4 (w/w). What is the molarratio of the two gases in the mixture ?A) 4: 1 B) 16 :1 C) 2 :1 D) 1 : 4

Re-AIPMT 201513. The number of water molecules is maximum in

A) 18 gram of water B) 18 moles of waterC) 18 molecules of water D) 1.8 gram of water

14. If avogadro number NA, is changed from6.022x1023 mol-1 to 6.022x1020mol-1, this wouldchangeA) the ratio of chemical species to each other in abalanced equationB) the ratio of elements to each other in a compoundC) the definition of mass in units of gramsD) the mass of one mole of carbon

15. 20.0 g of a magnesium carbonate sampledecomposes on heating to give carbon dioxideand 8.0g magnesium oxide. What will be thepercentage purity of magnesium carbonate inthe sample ? (At. Wt. : Mg = 24)A) 60 B)84 C) 75 D) 96

16. What is the mass of the precipitate formedwhen 50 mL of 16.9% solution of AgNO3 ismixed with 50 mL of 5.8% NaCl solution ? (Ag= 107.8, N = 14,0 = 16, Na = 23, Cl =35.5)A) 7g B) 14 g C) 28 g D) 3.5 g

NEET-II 201617. Suppose the elements X and Y combine to form

two compounds XY2 and X3Y2. When 0.1 moleof XY2 weighs 10 g and 0.05 mole of X3Y2weighs 9g, the atomic weights of X and Y areA) 40, 30 B) 60, 40 C) 20, 30 D) 30, 20

NEET 201818. In which case is the number of molecules of

water maximum?1) 18 mL of water2) 0.18 g of water3) 0.00224 L of water vapours at 1 atm and 273 K4) 10-3 mol of water

19. Magnesium reacts with an element (X) to forman ionic compound. If the ground stateelectronic configuration of (X) is 1s2 2s2 2p3,the simplest formula for this compound is1)Mg

2X

32) MgX

23) Mg

2X 4) Mg

3X

2

1) A 2) A 3) B 4) C 5) D 6) C 7) D8) B 9) C 10) A 11) A 12) A 13) B 14) D15) B 16) A 17) A 18) A 19) D

EXERCISE-II

3.2 22

1 2 1 1

PbO HCl PbCl H Omol mol mol mol

+ ⎯⎯→ +

Given moles of 6.5

0.029223

PbO mol= =

moles of 3.2

0.08736.5

HCl mol= =

here PbO is limiting reagent

∵ 1 mol PbO gives 1 mol of PbCl2

∴ 0.029 mol PbO gives 0.029 of PbCl2

18. 18 mL waterAs d

H2O = 1 g/mL So WH

2O = 18g

nH2O =18/18 = 1

molecules = 1 x Na2) 0.18 g of water

nH2O = 0.18/18 = 0.01

(molecules)H2O

= 0.01 x Na

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3) (VH2O(g)

)STP

=0.00224 LnH

2O =V/22.4 = 0.00224/22.4 = 0.0001

molecules = 0.0001 x Na

4) nH2o = 10-3

(molecules)H2O

= 10-3 x Na19. Magnesium ion = Mg+2

X = Nitrogen Nitrogen ion = N-3

2Mg + 3N −

( )3 2 3 2/Mg N Mg X

1. Number of HCl molecules present in 10mL of0.1M solution isA) 236.022 10× B) 226.023 10×C) 216.022 10× D) 206.022 10×

2. The volume of a gas at 0oC and 700mmpressure is 760cc. The no.of molecules presentin this volume isA) 221.88 10× B) 236.022 10×C) 2318.8 10× D) 2218.8 10×

3. Rearrange the following (I to IV) in the order ofincreasing masses and choose the correct answer.(Atomic masses: N = 14, O = 16, Cu =63)I. 1 molecule of oxygenII. 1 atom of NitrogenIII. 1 x 10–10 ×molecular weight of oxygen)IV. 1 x 10–10 × (g atomic weight of copper)A) II<I<III<IV B) IV<III<II<IC) II < III < I < IV D) III<IV<I<II

4. The number of moles of carbon dioxide whichcontain 8 g of oxygen isA) 0.5 mole B) 0.20 moleC) 0.40 mole D) 0.25 mole

5. If 224 mL of a triatomic gas has a mass of 1gat 273K and 1 atm pressure, then the mass ofone atom isA) 238.30 10 g−× B) 232.08 10 g−×

C) 235.53 10 g−× D) 236.24 10 g−×6. The maximum number of molecules are

present inA) 0.5g of H

2 gas B) 5L of N

2 gas at STP

C) 10g of O2 gas D) 15L of H

2 gas at STP

7. How many moles of magnesium phosphate,3 4 2Mg (PO ) will contain 0.25 mole of oxygen

atoms ?A) 22.5 10−× B) 0.02

C) 23.125 10−× D) 21.25 10−×8. The mass of carbon anode consumed (giving

only carbondioxide) in the production of 270Kgof aluminium metal from bauxite by the Hall’sprocess is

2 3 22Al O 3C 4Al 3CO+ → +A) 180 Kg B) 270 Kg C) 240 Kg D) 90 Kg

9. 22.4 litre of water vapour at NTP, Whencondensed to water occupies an approximatevolume ofA) 18 litre B) 1 litre C) 1 mL D) 18 mL

10. 0.01 mole of iodoform (CHI3) reacts with Agto produce a gas whose volume at NTP is

3 2 22 6 6 ( )CHI Ag C H AgI s+ → +A) 224mL B) 112mLC) 336 mL D) None of these

11. The minimum quantity in grams of H2S neededto precipitate 63.5 g of Cu2+ will be nearly

22 2Cu H S CuS H+ + → +

A) 63.5g B) 31.75 gC) 34 g D) 20 g

12. 2.76g of silver carbonate on being stronglyheated yields a residue weighing

2 3 2 21Ag CO 2Ag CO O2→ + +

A) 2.16g B) 2.48 gC) 2.32 g D) 2.64 g

13. The volume of gas at NTP produced by 100 gof CaC2with water

2 2 2 2 2CaC 2H O Ca(OH) C H+ → +A) 70 litre B) 35 litreC) 17.5 litre D) 22.4 litre

14. 90 mL of pure dry O2 is subjected to silentelectric discharge. If only 10% of it is convertedto O3, volume of the mixture of gases (O2 andO3) after the reaction will be .......... and afterpassing through turpentine oil will be ......A) 84 mL and 78 mL B) 81 mL and 87 mLC) 78 mL and 84 mL D) 87 mL and 81 mL

15. Element ‘A’ reacts with oxygenic A2O3. If0.359 gram of ‘A’ react to give 0.559 gram ofthe compound, atomic weight of ‘A’ will be

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A) 51 B) 43.08 C) 49.7 D) 47.916. 1.12 mL of a gas is produced at STP by the

acti of 4.12 mg of alcohol ROH with methylmagnesium iodide. The molecular mass ofalcohol is

3 4R OH CH MgI CH Mg(OR)I− + → +A) 16 B) 41.2 C) 82.4 D) 156.0

17. CaCO3 is 90% pure. Volume of CO2 collectedSTP when 10 g of CaCO3 is decomposed isA) 2.016 litres B) 1.008 litresC) 10.08 litres D) 20.16 litres

18. 50 g CaCO3 will react with ..........g of 20% HClby weightA) 36.5g B)73g C) 109.5 g D) 182.5 g

19. Two oxides of a metal contains 50% and 40%of the metal respectively. The formula of thefirst oxide is MO.Then the formula of thesecond oxide isA) MO

2B) M

2O

3C) M

2O D) M

2O

5

20. A gas mixture of 3 litres of propane and butaneon complete combustion at 25oC produced 10litres of CO2 Initial composition of the propane& butane in the gas mixture is.A) 66.67%, 33.33% B) 33.33%, 66.67%C) 50%, 50% D) 60%, 40%

21. The atomic mass of an element is 27. If valencyis 3, the vapour density of the volatile chloridewill beA) 66.75 B) 6.675 C) 667.5 D) 81

22. Two elements X (at-mass 16) and Y (at-mass14) combine to form compounds A, B and C.The ratio of different masses of Y whichcombines with a fixed mass of X in A, B and Cis 1 : 3 : 5. If 32 parts by mass of X combineswith 84 parts by mass of Y in B, then in C 16parts by mass of X will combine withA) 14 parts by mass of YB) 42 parts by mass of YC) 70 parts by mass of YD) 84 parts by mass of Y

23. 1 L of a hydrocarbon weighs as much as onelitre of CO2 Under similar conditions. Then themolecular formula of the hydrocarbon is.A) C

3H

8B) C

2H

6C) C

2H

4D) C

3H

6

24. There are two oxides of sulphur. They contain50% and 60% of oxygen respectively by weights.The weights of sulphur which combine with 1 g ofA)1 : 1 B) 2:1 C) 2 : 3 D) 3 : 2

25. A litre of air containing 1% Ar is repeatedlypassed over hot Cu and hot Mg till noreduction of volume takes place. The finalvolume of Ar shall beA) 0 mL B) 230 mL C) 770 mL D) 10 mL

26. An organic compound having molecular mass60 is found to contain C=20%, H=6.67% andN=46.67% while rest is oxygen. On heating itgives NH3 along with a solid residue. The solidresidue give violet colour with alkaline coppersulphate solution. The compound isA) (NH

2)

2CO B) CH

3CH

2CONH

2

C) CH3NCO D) CH

3CONH

2

27. Perecentage composition of an organiccompound is as followsC=10.06, H=0.84 , Cl=89.10Which of the following corresponds to itsmolecular formula if the vapour density is 60.0A) CH

2 Cl

2B) CHCl

3

C) CH3C1 D) None

28. The ratio of masses of oxygen and nitrogen in aparticular gaseous mixture is 1 : 4. The ratio ofA) 1:8 B) 3 : 16 C) 1 : 4 D) 7 : 32

29. A gaseous hydrocarbon gives upon combustion0.72 g of water and 3.08 g of CO2. Theempirical formula of the hydrocarbon isA) C

2H

4B) C

3H

4C) C

6H

6D) C

7H

8

1) D 2) A 3) A 4) D 5) C 6) D 7) C8) D 9) D 10) B 11) C 12) A 13) B 14) D15) B 16) C 17) A 18) D 19) B 20) A 21) A22) C 23) A 24) D 25) D 26) A 27) B 28) D29) D

1. 0.1HCl M→310 10 10V mL L−= = ×

moles of HCl M V= ×3 30.1 10 10 10− −= × × =

No. of molecules 310 AN−=

2. Given at NTP, 1 1 2 2

1 2

PV PVT T

=

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2700 760 760

273 273

mm cc mm Vk k× ×

=

2 700V cc= at NTP

moles of gas 700 1

22400 32mol= =

No. of molecules 1

32 AN= ×

I. 1 molecules of oxygen

⇒ 32 amu 2432 1.67 10 .gm−= × ×II. 1 atom of Nitrogen

⇒ 14 amu 2414 1.677 10 .gm−= × ×

III. 101 10−× (gm molecular wt. of oxygen)101 10 32 .gm−= × ×

IV. 101 10−× (gm atomic wt. of Copper)101 10 63.5 .gm−= × ×

Hence increasing order of massesII < I < III < IV

4. 1 mole of CO2 contains 32 oxygen

∵ 32 gm Oxygen is present in 1 mole of CO2

∴ 8 gm Oxygen is present in 14 mol of CO

2

5. ∵ wt. of 224 mL = 1 gm∴ wt of 22400 mL = 100 gm (mol. wt)

wt. of one atom 100

3amu=

7. 1 mol Mg3(PO

4)

2 contains ⇒ 8 moles of oxygen

atoms∵ 8 mo of oxygen atoms are presnt in 1 mol ofMg

3(PO

4)

2

∴ 0.25 mol of oxygen atoms are present in

10.25

8× mol of Mg

3(PO

4)

2

8. Reaction is

( ) ( )

2 3 22 3 4 3

3 4

3 12 4 27

Al O C Al COmol mol

g g

+ ⎯⎯→ +

× ×

∵ In the production of ( )4 27 g× of Al mass of

C consumed 3 12g= ×

∴ In the production of 270 kf of Al mass of C

consumed ( )( )3 12

270 904 27

kg×

= × =×

9. 22.4 L of water vapour = 1mol water = 18 gmwater density of water =1 gm/mLvolume of water = 18 mL.

10.32 6 6

2 1

CHI Ag CH CH AgImol mol

+ → ≡ +

∵ 2 mol iodoform produces 1 mol of gas= 22400 mL.∴ 0.01 mol iodoform produces 1 mol of a gas= 112 mL.

11.

22 2

1 1

63.5 34

Cu H S CuS Hmol mol

gm gm

+ + ⎯⎯→ +

12.

2 3 2 2

12

21 . 2

2.76 2.16 .

Ag CO Ag CO O

mol mol

or gm gm

→ + +

↓ ↓

[Oxidation of Ag is unstable & converted in Agon heating]

13.

2 2

1 64 . 1 . 22.4

CaC H O HC CH CaO

mol gm mol L

+ ⎯⎯→ ≡ +

↓ ↓⇒ ⇒

64 .gm∵ of 2CaC produces

= 22.4 L gas at NTP∴ 100 gm of CaC

2 will produce

22.4100

64L= × gas at NTP

14. 2 33 2O O→given vol. of O

2 = 90 mL

vol. of O2 which is converting into O

3 is 10% of it

90 109 .

100mL×

= =

After reaction ( )90 9 81 .mL− =

3

29 6

3mLO× =

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vol. of the mixture of gases ( )2 3&O O

81 6 87 mL= + =turpentine oil absorb O

3, so after passing through

turpentine oil vol. will be → 81 mL15. Compound is A

2O

3. Hence velency of A is 3

wt. of A = 0.359 gm.wt. of its oxide = 0.559 gm∴ wt. of oxygen = 0.559-0.359 = 0.20 gm.

∴ Eq. wt. of 0.359

8 14.360.20

A = × =

∴ At wt. of 3 14.36 43.08A = × =

16.( )

3 4

1 1 22.4

ROH CH MgI CH ROMgI

mol mol L

+ → +

↓ ↓

According to stoichawmetery moles of

( ) ( )22.4W

W gm V LROH

M= = moles of gas ( )4CH

17.

3 2

100 22.4

10 2.24

1 0.224

CaCO CaO COgm L

gm Lgm L

→ +

3[ CaCO∵ is 90% pure so 10 gms of 3CaCO

contains 9 gm pure 3CaCO ]

∴ vol. of CO2 evolved from (10 - 1 = 9 gm)

3CaCO

( )2.24 0.224 2.016L L= − =18. gm eq. of CaCO

3 = gm eq. of HCl

1 2

1 2

W WE E

= or 250

50 36.5

W=

2 36.5W gm HCl=

HCl∵ is 20% by wtso 20 gm pure HCl is present in 100 gm

∴ 36.5 gm pure HCl is present in 100

36.520

gm×

19. Let at wt of metal = xoxide :- I

M O

50

x :50

16formula = MOHence x= 16In oxide :II M Oformula of II oxide 2 3M O

40 60

16 16

2 : 3

20. 3 8 4 10 3C H C H L+ =

If vol. of 3 8C H in mixture = aL

than vol. of ( )4 10 3C H a L= −

3 8 2 2 25 4

1 . 3 .

3

C H O CO H Ovol vol

aL a L

+ ⎯⎯→ +

( ) ( )

3 10 2 2 2

134 5

21 . 4 .

3 4 3

C H O CO H O

vol vola L a L

+ ⎯⎯→ +

− −

Total vol. CO2 produce ( )3 4 3 10a a= + − =

2a L=

Hence 3 8 2C H L= 4 10 1C H L=22. Ratio of different mass of Y in A B C is

1 : 3 : 5In B: ∵ 32 parts by mass of X combines with 84parts by mass of Y∴ 16 parts by mas of X combines with 42 partsby mass of YIn C : mass of X = 16and ratio of Y in B & C is 3 : 5

42 : 7023. wt. of 1 L of hydrocarbon = wt. of 1 L of CO

2

∴ wt. of 22.4 L of hydrocarbon = wt. of 22.4 Lof CO

2 (mol. wt of CO

2) = 44 gm.

C3H

8 have mol. wt = 44

24. Oxides

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50 : 50 60 : 40

21: 1 1 :

3

I IIO S O S

ratio of wt. of sulphur 1 : 2/33 : 2

25. 1 L of air contains 1% of Ar

∴ Vol. of Ar in 1 L air 1000 1

100

×= ml = 10 mL

Since Ar is inert gas there fore it’s volume remainsconstant in air (10 mL)

Directions for Assertion & ReasonquestionsThese questions consist of two statements each,printed as Assertion and reason. While answeringthese Questions you are required to choose anyone of the following four responses.A) If both Assertion & Reason are True & theReason is a correct explanation of the Assertion.B) If both Assertion & Reason are True but Reasonis not a correct explanation of AssertionC) If Assertion is True but the Reason is False.D) If both Assertion & Reason are false.

1. Assertion : 16g each O2 and O

3 contains AN

2

and AN

3 atoms respectively..

Reason : 16g O2 and O

3 contains same no. of

molecules.A) A B) B C) C D) D

2. Assertion : Volume occupied by 1mol H2O

(I) is

equals to 22400cc at NTP.Reason : 1mol of any substance occupies 22.4Lvolume at N.T.P.A) A B) B C) C D) D

3. Assertion : 44g of CO2 28g of CO have same

volume at STP.Reason : Both CO

2 and CO are formed by C

and oxygen.4. Assertion : Equivalent wt. of Cu in both CuO and

Cu2O is 31.75

Reason : Equivalent wt. of an element is constant.A) A B) B C) C D) D

5. Assertion : On compressing a gas to half thevolume, the number of moles is halved.Reason : Number of moles present decreases withdecrease is volume.A) A B) B C) C D) D

6. Assertion : The mass of the products fomed in areaction depends upon the limiting reactant.Reason: Limiting reactant reacts completely in thereactionA) A B) B C) C D) D

7. Assertion : Carbon and oxygen combinedtogether only in one fixed ratio.Reason : In a chemical compound the elementsare combined together in a fixed ratioA) A B) B C) C D) D

8. Asser tion : At same temp & pressure 1lit O2 and

1lit SO2 contains equal no.of molecules.

Reason : According to avogadros hypothesis equalvolume of all gases under similar condition oftemperature and pressure contains equal numberof molecules.A) A B) B C) C D) D

9. Assertion : Law of conservation of mass holdgood for nuclear reaction.Reason: Law states that mass can be neithercreated nor destroyed in a chemical reaction.A) A B) B C) C D) D

10. Assertion : The balancing of chemical equationsis based on law of conservation of massReason : Total mass of reactants is equal to totalmass of products in a chemical reaction.A) A B) B C) C D) D

11. Assertion : Pure water obtained from differentsources such as, river, well, spring, sea etc. alwayscontains hydrogen and oxygen combined in the ratio1 : 8 by massReason : A chemical compound always containselements combined together in same proportion bymass.A) A B) B C) C D) D

1) D 2) D 3) B 4) D 5) D 6) A 7) D8) A 9) D 10) A 11) A

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