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MOLE CONCEPT

I N D E X

Topic Page No.

PHYSICAL CHEMISTRY

MOLE CONCEPT

01. Chemical Classification of Matter 02

02. Laws of Chemical Combinations 02

03. Methods to calculate moles 06

04. Basic Reaction Stoichiometry 08

05. Concentration terms 14

06. Methods of expressing concentration of solution 15

07. Solved Example 21

08. Problems Related with Mixing& Dilution 27

09. Summary 29

09. Exercise - 1 30

Exercise - 2 41

Exercise - 3 44

Exercise - 4 45

10. Answer Key 46

11. Hints/Solution 48

MEDICAL ENTRANCE

MOLE CONCEPT

BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road No.-1, I.P.I.A., Kota-5

ACC_CHEMISTRY_Mole Concept 1

Syllabus :

• General Introduction: Importance and scope of chemistry.• Laws of chemical combination, Dalton’s atomic theory: concept of elements, atoms and molecules.• Atomic and molecular masses. Mole concept and molar mass; percentage composition and empirical

and molecular formula; concentration terms, stoichiometry and calculations based on stoichiometry.

Contents :

• Chemical Classification of Matter

• Laws of Chemical Combinations

• Mole concept

• Basic Reaction Stoichiometry

• Concentration terms

• Methods of expressing concentration of solution

• Solved Example

• Exercise

MOLE CONCEPT



MOLE CONCEPT

CHEMICALCLASSIFICATION OFMATTER

LAWS OFCHEMICALCOMBINATIONS Law of Conservation of Mass (Lavoisier)

Mass cannot be created or destroyed. In any physical or chemical process, the total mass of the systemremain conserved. This law is not applicable to the nuclear processes, where mass and energy are inter-

conversable by the Einstein’s equation, E =m.c2, (c is the speed of light)

Ex. 1.7 gram of silver nitrate dissolved in 100 gram of water is taken. 0.585 gram of sodium chloridedissolved in 100 gram of water is added to it and chemical reaction occurs. 1.435 gm ofAgCl and 0.85

gm NaNO3are formed. Show that these results illustrate the law of conservation of mass.

Sol. Total mass before chemical change = mass ofAgNO3+ Mass of NaCl + Mass of water

= 1.70 + 0.585 + 200 = 202.285 gram

Total mass after the chemical reaction = mass ofAgCl + Mass of NaNO3+ Mass of water

= 1.435 + 0.85 + 200 = 202.285 gram

Thus in the given reaction

Total mass of reactants = Total mass of the products.

Law of Constant Composition or Definite Proportion (Proust)

The composition of a compound always remains fixed and it is independent to the source from which the

compound is obtained. This law cannot be applied to the compound obtained by using different isotopesof the elements, selectively.

Special Highlights

BRAIN TEASERS: GENERAL MISTAKE:

TEACHER’S ADVICE:



Ex. Thefollowingareresultsof analysisof twosamplesof thesameor twodifferentcompoundsofphosphorus

andchlorine.Fromtheseresults,decidewhether the twosamplesare fromthesameordifferentcompounds.

Also state the law, which will be obeyed by the given samples.

Compound AmountP Amount Cl

CompoundA 1.156 g 3.971 g

Compound B 1.542 g 5.297 g

Sol. The mass ratio of phosphorus and chlorine in compoundA,

mP

: mCl

= 1.156 : 3.971 = 0.2911 : 1.000

the mass ratio of phosphorus and chlorine in compound B,

mP

: mCl

= 1.542 : 5.297 = 0.2911 : 1.000

As the mass ratio is same, both the compounds are same and the samples obey the law of definite

proportion.

Law of Multiple Proportion (Dalton)

If two elements combine to form more than one compound, then for the fixed mass of one element, the

mass of other element combined will be in simple ratio.

e.g. CO and CO2

12:16 12:32

ratio = 16 : 32

1 : 2

H2S and H2S2

2:32 2:64

ratio = 32 : 64

1 : 2

Law of Reciprocal Proportion (Richter)

If the two elements combine separately with a third element, the mass ratio of the first two elements

combined with a fixed mass of the third element will be equal to or in simple ratio to the mass ratio of first

two elements in a compounds formed by their direct combination.

e.g. C combines with O to form CO2 and with H to form CH4. In CO2 12g of C reacts with 32g of O,

whereas in CH4 12g of C reacts with 4g of H. Therefore when O combines with H, they should combine

in the ratio of 32 : 4 (i.e. 8:1) or in simple multiple of it. The same is found to be true in H2O molecules.

The ratio of weight of H and O in H2O is 1:8.

Gay Lussac’s Law of Volume Combination

This law is applicable to only those chemical reactions in which at least two reaction components aregaseous. According to this law the volumes of all gaseous reactants reacted and the volumes of all

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

e.g. H2 (g) + Cl2 (g) 2HCl (g)

1 unit vol. 1 unit vol. 2 unit vol.

ratio = 1 : 1 : 2

N2 + 3H2 2NH3

1 unit vol. 3 unit vol. 2 unit vol.

ratio = 1 : 3 : 2



"WONDER THESE LAWS ARE USEFUL?"

"These are no longer useful in chemical calculations now but gives an idea of earlier

methods of analysing and relating compounds by mass."

Can you think of the cases where the above laws will not be applicable.

MOLE CONCEPT

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

species106.0221 mole 23

Ex 2 mole of Cl– ion = 2 × 6.022 × 1023 ions of Cl–

3 moles of H2O= 3 × 6.022 × 1023 Molecules of H

2O

1 mole of atoms is also termed as 1 gm-atom, 1 mole of ions is termed as 1 gm-ion

and 1 mole of molecule termed as 1 gm-molecule.

Atomic mass –Atomic mass of an element can be defined as the number which indicates how many

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

1th part of the mass of one atom

of Carbon-12.

Atomic mass =]12-carbonofatomanofMass[

12

1

]elementtheofatomanofMass[

=amu1

amuinatomanofMass

Atomic mass unit (amu) – The quantity [12

1× mass of an atom of C–12] is known as atomic mass

unit. The actual mass of one atom of C-12 = 1.9924 × 10–26 kg

So, 1amu = kg12

109924.1 26= 1.66 × 10–27 k.g. = 1.66 × 10–24 gm =

AN

1gm.

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

Gram atomic mass – The gram atomic mass can be defined as the mass of 1 mole atoms of an element.It is also called molar mass of an element.



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

16gm.

Mass of NA

oxygen atom = A

A

N.N

16= 16 gram

Ex (a) What is the mass of one atom of Cl?

(b) What is the atomic mass of Cl?

(c) What is the gram atomic mass of Cl?

Sol. (a) Mass of one atom of Cl = 17 amu.

(b) Atomic mass of Cl =1amu

amuinatomanofMass=

amu1

amu17= 17.

(c) Gram atomic mass of Cl = [Mass of 1 Cl atom × NA]

= 17 amu × NA

=AN

17× N

Agram = 17 gram

Ex (a) What is the mass of one atom of He ?

(b) What is the atomic mass of He ?

(c) What is the gram atomic mass of He ?

Sol. (a) Mass of one atom of He = 4 amu.

(b) Atomic mass of He =1amu

amuinatomanofMass=

amu1

amu4= 4.

(c) Gram atomic mass of He = [Mass of 1 He atom × NA]

= 4 amu × NA

=AN

4× N

Agram = 4 gram

Molecular mass : Molecular mass is the number which indicates how many times the mass of one

molecule of a substance is heavier in comparison to th12

1part of the mass of one atom of C-12.

Molecular mass =]12-CofatomanofMass[

12

1

substancetheofmoleculeoneofMass

=amu1

substancetheofmoleculeoneofMass

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

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

N

32

A

.

Mass of NA

molecules of O2= gmN

N

32A

A

= 32 gm



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

(b) What is the molecular mass of HNO3?

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

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

(b) Molecular mass of HNO3= 63

amu1

amu63

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

3× N

A

= 63 amu × NA

=AN

63gm × N

A= 63 gram

Methods to calculate moles :

Volum

eW

eight

Mole

No. ofparticles

(i) From number of particles

AN

molecules]es [atoms/of ParitclGiven no.eNo. of mol

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

Sol. No. of mole =A

24

N

10022.6 = 23

24

10022.6

10022.6

= 10 mole

(ii) From given Mass –(a) For atom :

No. of mole =massatomicGram

gm)substance(theofmassGiven

(b) For molecules :

No. of mole =massmolecularGram

gm)substance(theofmassGiven

Ex. How many molecules of water are present in 9 gram of water?

Sol.2310023.6

18

9 = 3.011 × 1023



(ii) From the given volume of a gasOne mole of Gas occupies 22.4 L at NTP and 22.7 L at STP.

22.7

e)K (in litr273&bar1gas atvolume ofn at S.T.P..

OR

22.4

(in litre)m and 273Kgas at 1atvolume ofn at N.T.P..

If volume is given under any other condition of temperature and pressure, then use the ideal gas equationto find the no. of mole.

RT

PV)n(moleof.No (P in atm, V in L, R = 0.0821 atmLK–1mol–1, T in K)

1 mole of any gas at STP occupies 22.7 litre.

Units of Pressure1 atm = 76 cm Hg = 760 torr = 1.01325 bar = 1.01325 × 105 pa. (N/m2)

= 760 mm of Hg

Units of temp :

273CK

Units of R :R = 0.0821 litre-atm/mole.K

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

Units of volume :

ml10m10litre1cm10dm1 33333 3

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

Sol. No. of mole = 25.04

1

4.22

6.5

Gases do not have volume. What is meant by "Volume of gas"?

Do not use this expression (PV = nRT) for solids/liquids.

How would I calculate moles if volume of a solid is given?



Significance of Chemical Equations &Avogadro’s Law :

Achemical equationdescribes thechemicalprocessbothqualitativelyandquantitatively.Thestoichiometric

coefficients in the chemical equation give the quantitative information of the chemical process. These

coefficients represent the relative number of molecules or moles of the reactants and products, e.g.,

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

2 molecules 2 molecules 3 molecules

or 2 N molecules 2 N molecules 3 N molecules

or 2 moles 2 moles 3 moles

Avogadro’s principle states that under the same conditions of temperature and pressure, equal volumes

of gases contain the same number of molecules. Thus, for homogeneous gaseous reactions, the

stoichiometric coefficients of the chemical equation also signify the relative volumes of each reactant and

product under the same conditions of temperature and pressure, e.g.,

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

1 molecule 1 molecule 2 molecule

or 1 mole 1 mole 2 mole

or 1 volume 1 volume 2 volume

(T & P constant)

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

BASIC REACTION STOICHIOMETRY :

Suppose a Reaction :

2NH3(g) N2(g) + 3H2(g)

2 mole 1 mole 3 mole

34 gm 28 gm 6 gm

On the basis of above reaction we can say :

2 moles of NH3 will react to form one mole of N2 and 3 mole of H2

34 gm of NH3 on decomposition will give 28 gm of N2 and 6 gm of H2

Like if we have to answer following question :

Ex How many moles of H2 will be produced on decomposition of 0.8 moles of NH3?

Sol. As with 2 moles of NH3 , 3 moles of H2 are produced

with 0.8 moles of NH3 ,2

3× 0.8 = 1.2 moles of H2 will be produced.

Ex. During decomposition of NH3 , 14 gm of N2 is produced, what weight of H2 gas will be produced?

Sol.. With 28 gm of N2 , 6 gm of H2 are produced

with 14 gm of N2, 1428

6 = 3 gm of H2 will be produced



Limiting Reagent :

The reactant which gives least amount of product on being completely consumed is known as limitingreagent.

OR

Limiting reagent is the reactant that is completely consumed when a reaction goes to completion.

It comes into the picture when reaction involves two or more reactants. For solvingsuch reactions, first step is to calculate Limiting Reagent.

Amount of product formed or amount of other reactant left is determined with respectto limiting reagent only.

Calculation of Limiting Reagent:

(i) By calculating the required amount by the equation and comparing it with given amount.

[Useful when only two reactant are there]

(ii) By calculating amount of any one product obtained taking each reactant one by one irrespective of other

reactants. The one giving least product is limiting reagent.

(iii) If initially weight or volume of reactant is given then convert them in moles.

(iv) Divide given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting

reagent. [Useful when number of reactants are more than two.]Ex. If 20gm of CaCO3 is treated with 20gm of HCl, how many grams of CO2 can be generated according

to followingreaction?CaCO3(g) + 2HCl(aq) CaCl2(aq) + H2O() + CO2(g)

Sol. Mole of CaCO3 = 2.0100

20

Mole of HCl = 274.073

20

tcoefficientricStoichiome

Molefor CaCO3 = 2.0

1

2.0

tcoefficientricStoichiome

Molefor HCl = 137.0

2

274.0

So HCl is limiting reagent

According to reaction : 73 gm of HCl gives 44gm of CO2

20 gm HCl will give 2073

44 =12.054 gm CO2


ACC_CHEMISTRY_Mole Concept 1 0

Ex. 2 moles ofA, 3 moles of B and 4 moles of C are taken in vessel where they reacts according to reaction.

A + 2B + 3C 5D + 6E

Find the maximum number of moles of ‘E’. Which can produced and moles of reactant left?

Sol. A + 2 B + 3C 5D + 6E

moles 2 3 4

now divide with respective stoichiometric coefficient.

Ratio : 2/1 3/2 4/3

least ratio

C is limitingreagent

Now amount of product formed or amount of reactant left will be determined corresponding to C as C

is limitingreagent

From reaction stoichiometry

3 moles of C give, 6 moles of E

With 4 moles of C , 43

6 = 8 moles of E will be formed.

Also, 3 moles of C react with 1 mole ofA

4 moles of C will react with 43

1 =

3

4moles ofAA

Moles of A left = 2 –3

4=

3

2moles

Problems related with mixture

Ex. 4 grams of a mixture of CaCO3 and Sand (SiO2) is treated with an excess of HCl and 0.88 gm of CO2is produced. What is the percentage of CaCO3 in the original mixture?

Sol. CaCO3 + 2HCl CaCl2 + H2O + CO2SiO2 + HClNo reactionCaCO3 = x gm100 gm CaCO3 gives 44 gm CO2x gm CaCO3 gives0.88 gm CO2

88.0

44

x

100 x = 2 gram

% CaCO3 = 1004

2 = 50%

Percentage yield :In general when a reaction is carried out in the laboratory we do not obtain the theoretical amount ofproduct. The amount of product that is actually obtained is called the actual yield. Knowing the actualyield and theoretical yield, the % yield can be calculated by the following formula–

Percent yield = 100yieldlTheoritica

yieldActual



Ex. For the reaction

CaO + 2HCl CaCl2 + H2O

1.23 gram of CaO is reacted with excess of hydrochloric acid and 1.85 gm CaCl2 is formed. What is the

% yield of the reaction?

Sol. CaO + 2HCl CaCl2 + H2O

56 gm CaO will produce 111 gm CaCl2

1.23 gram of CaO will produce 23.156

111 = 2.43 gm CaCl2

Thus Theoretical yield = 2.3 gm

Actual yield = 1.85 gm

% yield = 10043.2

85.1 = 76%

Percent purity :

Depending upon the mass of the product, the equivalent amount of reactant present can be determinedwith the help of given chemical equation. Knowing the actual amount of the reactant taken and theamount of reactant calculate with the help of a chemical equation the purity can be determined.

% purity = 100akenreactant tofamountActual

equationchemicalthefromcalculatedreactantofAmount

The actual amount of any limiting reagent consumed in such incomplete

reactions is given by [% yield × given moles of limiting reagent][For reversible reactions]

For irreversible reaction with % yield less than 100, the reactants is

converted to product (desired) and waste.

Ex. Calculate the amount of (CaO) in kg that can be produced by heating 200 kg lime stone that is 90% pureCaCO3.

Sol. Pure CaCO3 = kg180100

90200

CaCO3 CaO + CO2

100 k.g. 56 k.g.

180 x

x

56

180

100

n x = 100.8 kg



Principle of atom conservation

The principle of conservation of mass, expressed in the concepts of atomic theorymeans the conservationof atoms. And if atoms are conserved, moles of atoms shall also be conserved. This is known as theprinciple of atom conservation. This principle is in fact the basis of the mole concept.

Ex. All carbon atoms present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How manygram of CO2 were obtained?

Sol. KH3(C2O4)2.2H2O CO2

Apply P.O.A.C. on carbon atom

4 × mole of KH3(C2O4)2.2H2O = 1 × mole of CO2

254

2544 = 1 × mole of CO2

mole of CO2 = 4

wt. of CO2 = 4 × 44 = 176 gram

Average atomic mass

Average atomic mass = moletotal

masstotal

Let a sample contains n1 mole of atoms with atomic wt M1 and n2 mole of atoms with atomic wt M2.then

21

2211av nn

MnMnM

Ex. Find the average atomic mass of a mixture containing 25% by mole Cl37 and 75% by mole Cl35 ?

Sol. n1 = 25 M1 = 37

n2 = 75 M2 = 35

5.357525

35753725Mav

Average molecular mass :(Weight of one mole of mixture)

Average molecular mass = moletotal

masstotal

Letasamplecontainsn1 moleofmoleculeswithmolecularwtM1 andn2 moleofmoleculeswithmolecular

wt. M2 , then :-

21

2211av nn

MnMnM

Ex. Air is a mixture of O2 and N2 in which O2 is present 20% by mole and N2 is present 80% by mole. Findout the average M. wt. of air ?

Sol. n1 = 20 M1 = 32

n2 = 80 M2 = 28

)nn(

MnMnM

21

2211av

gram8.28)8020(

28803220



Chemical Formula : It is of two types -

(i) Molecular formula : Chemical formula that indicate the actual number and type of atoms in a

molecule are called molecular formula eg. - Molecular formula of benzene is C6H

6

NaCl i not a molecular formula but actually is a formula unit.

(ii) Empirical formulae : The chemical formulae that give only the relative number of atoms of each

type in a molecule are called empirical formulae eg. - empirical formula of benzene is CH.

Check out the importance of each step involved in calculations of empirical formula.

Steps for writing the empirical formula

The percentage of the elements in the compound is determined by suitable methods and from the data

collected, the empirical formula is determined by the following steps:

(i) Divide the percentage of each element by its atomic mass. This will give the relative number

of moles of various elements present in the compound.

(ii) Divide the quotients obtained in the above step by the smallest of them so as to get a simple

ratio of moles of various elements.

(iii) Multiply the figures, so obtained by a suitable integer if necessary in order to obtain a whole

number ratio.

(iv) Finally write down the symbols of the various elements side byside and put the above number as

the subscripts to the lower right hand corner of each symbol. This will represent the empirical

formula of the compound.

If percentage is not given than from whatever given information try to calculate number

of moles of each element and then take their ratio to determine empirical formula.

Steps for writing the molecular formula

(i) Calculate the empirical formula as described above.

(ii) Find out the empirical formula mass by adding the atomic masses of all the atoms present in the

empirical formula of the compound.

(iii) Divide themolecularmass (determinedexperimentallybysomesuitablemethod)by theempirical

formula mass and find out the value of n.

Molecular formula = n × Emperical formula

where n = weightformulaEmperical

weightMolecular



Vapour density :Some times in numericals molecular mass of volatile substance is not given, instead vapour density isgiven. Vapour density can be defined as

PandTsameatHofDensity

PandTgivenaatgasofDensity.D.V

2

For gases : PV = nRT = RT.M

w

RTV

wPM

= dRTT d for a gas =

.RT

M.P gas

RT

M.PRT

M.PD.V

2H

gas =

2H

gas

M

M

2

MV.D

gas

Ex. A compound of nitrogen and oxygen was found to contain 7 : 16 by mass N and O respectively.Calculate molecular formula of the compound if V.D. is 46 ?

Sol. Let mass of N = 7 K gramMass of O = 16 K gram

Element wt. Atomic wt.Atomic

wtSimple ratio

N 7 K 14 K5.014

K7 1

K5.0

K5.0

O 16 K 16 K16

K16 2

K5.0

K

N : O = 1 : 2Empirical formula= NO2(Empirical formula)n = Molecularweight

46

D.V2

.wtformulaempirical

wt.Mn

=

46

462= 2

Molecular formula = (NO2)2 = N2O4

CONCENTRATION TERMS

A solution is a homogeneous mixture of two or more substances, the composition of which may vary

within limits. “Asolution is aspecial kindof mixture inwhichsubstances are intermixed so intimately that

they can not be observed as separate components”. The substance which is to be dissolved is called

solute while the medium in which the solute is dissolved to get a homogeneous mixture is called the

solvent.Asolution is termed as binary and ternary if it consists of two and three components respectively.



METHODS OF EXPRESSING CONCENTRATION OF SOLUTION :

Concentration of solution is the amount of solute dissolved in a known amount of the solvent or solution.

The concentration of solution can be expressed in various ways as discussed below.

What is solute and solvent in a solution?

If the mixture is not homogeneous, then none of the concentration

term is applicable.

General concentration terms :

(a) Density =Volume

Mass, Unit : g/cm3

(b) Relative density = substanceereferencDensity of

bstancef any suDensity o

(c) Specific gravity =C4water atDensity of

bstancef any suDensity o

(d) Vapour density = essurere and prtemperatusamegas atf HDensity o

end pressurerature afied tempat specif vapourDensity o

2

Which of these are temperature dependent?

Classify each of them as w/w, w/v, v/v ratio.

Percentage : It refers to the amount of the solute per 100 parts of the solution. It can also be called as

parts per hundred (pph). It can be expressed by any of following four methods.

(i) Weight to weight percentage (%w/w) =

100gutionWt. of sol

guteWt. of sol

e.g., 10% Na2CO

3solution w/w means 10g of Na

2CO

3is dissolved in 100g of the solution. (It means

10g Na2CO

3is dissolved in 90 of solvent)



(ii) Weight to volume percent (% w/v) =

100)(cmsolutionVolume of

guteWt. of sol3

e.g., 10% Na2CO

3(w/v) means 10g Na

2CO

3is dissolved in 100 cm3 of solution.

(iii) Volume to volume percent (% v/v) = 100cmsolutionVolume of

cmsoluteVolume of3

3

e.g., 10% ethanol (v/v) means 10cm3 of ethanol dissolved in 100cc of solution.

(iv) Volume to weight percent (%v/w) = 100solutionofWt.

soluteof.Vol

e.g., 10% ethanol (v/w) means 10cm3 of ethanol dissolved in 100g of solution.

Classify each of the given ratio as w/w, w/v, v/v and comment on their

temperature dependence.

Ex.: Concentrated nitric acid used as laboratory reagent is usually 69% by mass of nitric acid. Calculate thevolume of the solution which contains 23 g nitric acid. The density of concentrated acid is 1.41 g cm–3.

Sol. Given HNO3 is 69% by mass;density of HNO3 = 1.41 g cm–3.

Thus (i) 69 g HNO3 is present in conc. HNO3 = 100 g

23 g HNO3 is present in conc. HNO3 =69

100× 23 = 33.33 g

(ii) Volume of solution required =Density

Mass=

41.1

33.33= 23.64 mL

Parts per million (ppm) and parts per billion (ppb) : When a solute is present in very small quantity,

it is convenient to express the concentration in parts per million and parts per billion. It is the number of

parts of solute per million (106) or per billion (109) parts of solution. It is independent of the temperature.

ppm = 610solutionofmassTotal

componentsoluteofMass

ppm = 610solventofmassTotal

componentsoluteofMass (For dilute solution)

ppb =910

solutionofmassTotal

componentsoluteofMass

ppb =910

solventofmassTotal

componentsoluteofMass (For dilute solution)



Ex. Calculate the parts per million of SO2 gas in 250 ml water (density 1 g cm–3) containing5 × 10–4 g of SO2 gas.

Sol. Mass of SO2 gas = 5 × 10–4 g;Mass of H2O = Volume × Density = 250 cm3 × 1 g cm–3 = 250 g

Parts per million of SO2 gas =6

4

10g250

105

= 2

Strength :The strength of a solution is defined as the amount of solute in grams present in one litre (or

dm3) of the solution. It is expressed in g/litre or (g/dm3)

litresinsolutionofVolume

gramsinsoluteofMassStrength

Mole fraction ( ) :

Mole fraction may be defined as the ratio of number of moles of one component to the total number of

moles of all the components (solvent and solute) present in the solution. It is denoted by the letter . It

may be noted that the mole fraction is independent of the temperature. Mole fraction is dimensionless. If

a solution contains the components A and B and suppose that WA

gram of A and WB

gram of B are

present in it.

Number of moles ofAis given by,A

AA

M

Wn and the number of moles of B is given by,

B

BB

M

Wn

where MA

and MB

are molecular mass ofAand B respectively.

Total number of moles ofAand B = nA

+ nB

Mole fraction ofA,BA

AA

nn

n

Mole fraction of B,BA

BB

nn

n

The sum of mole fractions of all the components in the solution is always one.

1nn

n

nn

n

BA

B

BA

ABA

BA = 1

Thus, if we know the mole fraction of one component of a binary solution, the mole fraction of the other

can be calculated.



Ex. Find out the weights of acid and water required to prepare CH3COOH solution of 0.3 mole fraction.

Sol. 3.0COOHCH 3

7.03.01OH2

Wt. of CH3COOH = COOHCH3

× mol. wt.(CH3COOH) = 0.3 × 60 = 18

Wt. of water = OH2 × mol. wt. (H

2O) = 0.7 × 18 = 12.6 g

Mass fraction : Mass fraction of a component in a solution is the mass of the component divided by the

total mass of the solution. For a solution containing wA

gm ofAand wB

gm of B

Mass fraction ofBA

A

ww

wA

;

Mass fraction ofBA

B

ww

wB

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

number of millimoles per ml. of solution). Unit of molarity is mol/litre or mol/dm3. For example, a molar

(1M) solution of sugar means a solution containing 1 mole of sugar per litre of the solution. Solutions in

terms of molarity is generally expressed as,

1M = One molar solution, 2M = Molarity is two,2

Mor 0.5M = Semimolar solution,

Mathematically, molarity can be calculated byfollowing formulas :

(i)litresinsolutionofVol.

solute(n)ofmolesof.NoM

(ii)mlinsolutionofVol.

1000

soluteofwt.Mol.

gm)solute(inof.WtM

(iii)mlinsolutionofVol

soluteofmillimolesof.NoM

(iv) If molarity and volume of the solution are changed from M1, V

1to M

2,V

2. Then,

M1V

1= M

2V

2

(v) In balanced chemical equation, if n1moles of reactant-1 react with n

2moles of reactant-2 . Then,

n1A + n

2B Product

2

22

1

11

n

VM

n

VM

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

)VV(

VMVMM

21

2211



Molality (m) :

It is the number of moles of the solute per 1000g of the solvent. Unit of molality is mol/kg. For example,

a 0.2 molal (0.2m) solution of glucose means a solution obtained by dissolving 0.2 mole of glucose in

1000 gm of water. Molality (m) does not depend on temperature since it involves measurement of

weightof liquids.

Mathematically molality can be calculated byfollowingformulas,

(i)t in kgthe solvenWeight of

he solutemoles of tNumber ofm

(ii) 1000t in gmthe solvenWeight of

he solutemoles of tNumber ofm

(iii)gmsolvent inWeight of

1000

tef the soluMol. wt. o

uteWt. of solm

(iv)vent in gmWt. of sol

solutelimoles ofNo. of milm

Note : Molarity is temperature dependent but molality is temperature independent.

Ex. Calculate the molarity and molality of a solution of H2SO

4(sp. gr. = 1.98) containing 27% H

2SO

4by

mass.

Sol. Vol of 100 g of 27% H2SO

4=

d

.wt=

198.1

100ml

)litre(solutionof.vol

.wt.mol/.wtM

42SOH

=10098

1000198.127

= 3.3 mol L–1

kgsolventof.wt

.wt.mol/.wtm

42SOH

=98)27100(

100027

= 3.77 mol L–1

Formality (F) :Formalityof solution may be defined as the number of gram formula masses of the ionic

solute dissolved per litre of the solution. It is represented by F. Commonly, the term formality is used to

express the concentration of the ionic solids which do not exist as molecules but exist as network of ions.

A solution containingone gram formula mass of solute per litre of the solution has formality equal to one

and is called Formal solution. It may be mentioned here that the formality of a solution changes with

change in temperature.



litresinsoluionofVolume

soluteofmassesformulagramofmberNu)F(Formality

)(solutionVolume ofsolutela mass ofgram formu

(g)nic soluteMass of io

l

Ex. What will be the formality of KNO3 solution having strength equal to 2.02 g per litre ?Sol. Strength of KNO3 = 2.02 gL–1

g formula weight of KNO3 = 101 gWe know that :

Formality of KNO3 =3

-1

KNOof.wtformula.g

ginstrength l=

101

02.2= 0.02 F

Note : Although formality is different from molarity term but we generally use molarity termat place of formality as well



SOLVED EXAMPLE

Ex.1 How many carbon atoms are present in 0.35 mol of C6H12O6

Sol. 1 mol of C6H12O6 has = 6 NA atoms of C

0.35 mol of C6H12O6 has

= 6 × 0.35 NA atoms of C

= 2.1 NA atoms

= 2.1 × 6.023 × 1023 = 1.26 × 1024 carbon atoms

Ex.2 8 litre of H2 and 6 litre of Cl2 are allowed to react to maximum possible extent. Find out the finalvolume of reaction mixture. Suppose P and T remains constant throughout the course of reaction-

Sol. H2 + Cl2 2 HCl

Volume before reaction 8 lit 6 lit 0

Volume after reaction 2 0 12

Volume after reaction = Volume of H2 left + Volume of HCl formed

= 2 + 12 = 14 lit

Ex.3 How many years it would take to spend Avogadro's number of rupees at the rate of 1 millionrupees in one second -

Sol. 106 rupees are spent in 1sec.

6.023 × 1023 rupees are spent in

= 6

23

10

10023.61 sec

=36524606010

10023.616

23

years = 19.098 × 109 year

Ex.4 Naturally occurring chlorine is 75.53% Cl35 which has an atomic mass of 34.969 amu and 24.47%Cl37 which has a mass of 36.966 amu. Calculate the average atomic mass of chlorine-

Sol. Average atomic mass

=100

massatomicitsisotopeIIof%massatomicitsisotopeIof%

=75 53 34 969 24 47 36 96

100

. . . .x x

= 35.5 amu.

Ex.5 The density of O2 at NTP is 1.429g / litre. Calculate the standard molar volume of gas-

Sol. 1.429 gm of O2 gas occupies volume = 1 litre.

32 gm of O2 gas occupies =32

1429.= 22.4 litre/mol.

Ex.6 How many molecules are in 5.23 gm of glucose (C6H12O6)

Sol. 180 gm glucose has = N molecules

5.23 gm glucose has =180

10023.623.5 23= 1.75 × 1022 molecules



Ex.7 How many g of S are required to produce 10 moles and 10g of H2SO4 respectively?

Sol. 1 mole of H2SO4 has = 32g S

10 mole of H2SO4 has = 32 × 10 = 320 g S

Also, 98g of H2SO4 has = 32 g S

10 g of H2SO4 has = (32 × 10)/98

= 3.265 g S

Ex.8 P and Q are two elements which form P2Q3, PQ2 molecules. If 0.15 mole of P2Q3 and PQ2

weighs 15.9 g and 9.3g, respectively, what are atomic weighs of P and Q?

Sol. Let at. wt. of P and Q be a and b respectively,

Mol. wt. of P2Q3 = 2a + 3b

and Mol. wt. of PQ2 = a + 2b

(2a + 3b) × 0.15 = 15.9

and (a + 2b) × 0.15 = 9.3

Ex.9 The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 27ºC. Calculate the moleof NO2 in 100 mole mixture.

Sol. Mol. wt. of mixture of NO2 and N2O4 = 38.3 × 2 = 76.6

Let a mole of NO2 be present in 100 mole mixture

= wt. of NO2 + wt. of N2O4 = wt. of mixture,

a × 46 + (100 – a) × 92 = 100 × 76.6

a = 33.48 mole

Ex.10 How many moles of potassium chlorate to be heated to produce 5.6 litre oxygen at STP?

Sol. 2KClO3 2KCl + 3O2

Mole ratio for reaction 2 mole 2 mole 3 mole

3 × 22.4 litre O2 is formed by 2 mole KClO3

5.6 litre O2 is formed by4.223

6.52

= 1/6 mole KClO3

Ex.11 Zinc and hydrochloric acid react according to the reaction :

Zn(s) + 2HCl (aq.) ZnCl2(aq.) + H2(g)

If 0.30 mole of Zn are added to hydrochloric acid containing 0.52 mole HCl, how many moles

of H2 are produced?

Sol. Zn(s) + 2HCl(aq.) ZnCl2(aq.) + H2(g)

Initial moles 0.30 0.52 0 0

HCl is limiting reagent

Final moles )4.02

52.30.0( 0 0 0.26



Ex.12 How many molecules are present in 1m of water vapours at STP -

Sol. 22.4 litre water vapour at STP has

= 6.023 × 1023 molecules

1 × 10–3 litre water vapours at STP has

=4.22

10023.6 23× 10–3 = 2.69 × 10+19

Ex.13 A mixture of FeO and Fe3O4 when heated in air to constant weight gains 5% in its weight. Findout composition of mixture.

Sol. Let weight of FeO and Fe3O4 be a and b g, respectively.

2FeO +2

1O2 Fe2O3

2Fe3O4 +2

1O2 3Fe2O3

144 g FeO gives 160 g Fe2O3

a gm of FeO gives 160 × a /144 g Fe2O3

Similarly, weight of Fe2O3 formed by b gm of Fe3O4 =464

b3160

Now if a + b = 100 ; then 105464

b3160

144

a160

Solving these two equations : a = 21.06 and b = 78.94

Percentage of FeO = 21.06% and percentage of Fe3O4 = 78.94%

Ex.14 A plant virus is found to consist to uniform cylindrical particles of 150Å in diameter and 5000Ålong. The specific volume of the virus is 0.75 cm3/g. What is the molecular weight of virus.

Sol. Volume of virus = r2l =316816 cm10994.010500010

2

150

2

150

7

22

Weight of one virus = g10178.1g75.0

10884.0 1616

Mol. wt of virus = 1.178 × 10–16 × 6.023 × 1023 = 7.095 × 107

Ex.15 8g of sulphur are burnt to from SO2, which is oxidised by Cl2 water. The solution is treated withBaCl2 solution. The amount of BaSO4 precipitated is :

Sol.4

BaCl–24

Cl2

O BaSOSOSOS 222

One mole of S will give 1 mole of BaSO4

Mole of BaSO4 formed = moles of sulphur

=32

8=

4

1



Ex.16 An unknown compound contains 8% sulphur by mass. Calculate

(a) Least molecular weight of the compound and

(b) Molecular weight if one molecule contains 4 atoms of “S”

Sol. 100 gm molecule 8 gm sulphur

forminimum molecular weight at least 32 gm of sulphur

M.M compound =8

100× 32 = 400

If one molecule contains 4 atomsM.M. = 4 × 400 = 1600 Ans.]

Ex.17 If molecular weight of glucose-1-phosphate is 260 and its density is 1.5g/mL. What is the averagevolume occupied by each molecule of this compound ?

Sol. Volume occupied by one molecule =density

moleculeoneofmass=

5.1

1066.1260 –24= 2.88 × 10–22 mL

Ex.18 Amixture of 1.65 × 1021 molecules of X and 1.85 × 1021 molecules ofYweighs 0.638 g. If the molecular

weight of X is 42, what is the molecular weight ofY

Sol. mmix

= mx+ m

y

=yAxA

wt.mol.N

moleculesof.nowt.mol.

N

moleculesof.no

or, 0.638 = y23

21

23

21

M10023.6

101.8542

10023.6

1065.1

or, My

= 170.27

Ex.19 Calculate the molarityof the following solutions :(a) 4g of caustic soda is dissolved in 200 mL of the solution.(b) 5.3 g of anhydrous sodium carbonate is dissolved in 100 mL of solution.(c) 0.365 g of pure HCl gas is dissolved in 50 mL of solution.

Sol. (a) M = 5.0510

1

200

1000

40

4

(b) M = 5.0100

1000

106

3.5

(c) M =50

1000

5.36

365.0 = 2.0

5

10

55.36

365

Ex.20 Density of a solution containing 13% by mass of sulphuric acid is 1.09 g/mL. Then molarity of solutionwillbe

Sol. M = 1000100

09.1

98

13 = 1.445



Ex.21 Thedensityofasolutioncontaining40%bymassofHCl is1.2g/mL.Calculate themolarityof thesolution.

Sol. M = 1000100

2.1

5.36

40 = 15.13

365

12040

Ex.22 15 gof methyl alcohol is present in 100 mLof solution. If densityof solution is 0.90 g mL–1. Calculate themass percentage of methyl alcohol in solution

Sol. g15W OHCH3

lnsoW = 100 × 0.9 = 90

Wsolvent = 90–15 = 75

% of solute =90

15× 100 =

3

50

6

100 = 16.67

Ex.23 A 6.90 M solution of KOH in water contains 30% by mass of KOH. What is density of solution ingm/ml.

Sol. 30 g of solute in 100 g of solution6.9 mole of KOH 1000 mL of solution6.9 × 56 g 1000 mL of solution

30 g =569.6

1000

× 6.9 × 56

= 23 × 56 × 10–3 = 1.288 g ml–1

Ex.24 The average concentration of Na+ ion in human body is 3 to 4 gm per litre. The molarity of Na+ ion isabout.

Sol. M1 =21

3=

7

1= 0.12

M2 = 19.021

4

Avg. =

2

MM 21= 0.15

Ex.25 What is the concentration of chloride ion, in molarity, in a solution containing 10.56 gm BaCl2.8H2O perlitre of solution ? (Ba = 137)

Sol. BaCl2 Ba+2 + 2Cl¯1M 1 M 2M

¯ClM = 100014471137

100056.102

=

352

12.21= 0.06

Ex.26 The mole fraction of solute in aqueous urea solution is 0.2. Calculate the mass percent of solute ?Sol. 0.2 mole solute = 0.2 × 60 = 12 g

0.8 mole solvent = 0.8 × 18 = 14.4

Mass percentage =4.1412

12

× 100

= 45.45 %



Ex.27 The concentration of Ca(HCO3)2 in a sample of hard water is 500 ppm. The density of water sample is1.0 gm/ml. Calculate the molarity of solution ?

Sol. M = 6

3

10

101

162

500 = 3 × 10–3

Ex.28 0.115 gm of sodium metal was dissolved in 500 ml of the solution in distilled water. Calculate themolarity of the solution ?

Sol. M =500

1000

23

115.0

=23

10230 3= 0.01

Ex.29 How much BaCl2 would be needed to make 250 ml of a solution having the same concentration of Cl–

as one containing 3.78 gm NaCl per 100 ml ? (Ba = 137)Sol. M of NaCl = 0.646

M of BaCl2 =2

646.0

Wt of BaCl2 = 16.8 g

Ex.30 Calculate molality (m) of each ion present in the aqueous solution of 2M NH4Cl assuming 100%dissociation according to reaction.

NH4Cl (aq) 4NH (aq) + Cl– (aq)

Given : Density of solution = 3.107 gm / ml.Sol. 2M = 2

M = 5.532107.31000

10002

=1070.3107

2000

=3000

2000=

3

2= 0.667



PROBLEMS RELATED WITH MIXING& DILUTION

Ex.31 Find out the volume of 98% w/w H2SO4 (density = 1.8 gm/ ml), must be diluted to prepare 12.5 litresof 2.5 M sulphuric acid solution.

Sol. M1V1 = M2V2

100010098

8.198

V = 2.5 × 12.5

1 × 18 × V = 2.5 × 12.5V = 1.736

Ex.32 500 ml of 2 M NaCl solution was mixed with 200 ml of 2 M NaCl solution. Calculate the final volumeand molarityof NaCl in final solution if final solution has density 1.5 gm/ml.

Sol. M1V1 + M2V2 = MV500 × 2 + 2 × 200 = M (700)1000 + 400 = M (700)

700

1400= M

M= 2

Ex.33 Calculate the molarity and molality of a solution of H2SO

4(sp. gr. = 1.98) containing 27% H

2SO

4by

mass.

Sol. Vol of 100 g of 27% H2SO

4=

d

.wt=

198.1

100ml

)litre(solutionof.vol

.wt.mol/.wtM

42SOH

=10098

1000198.127

= 3.3 mol L–1

kgsolventof.wt

.wt.mol/.wtm

42SOH

=98)27100(

100027

= 3.77 mol L–1

Ex.34 Derive the relation between molality (m) and mole fraction of solute, 2

Sol. Molality, m means, m mole of solute in 1000 g of solvent which is equal to 1000/M1 molwhere M1 = molar mass of the solvent.

Mole fraction, 2 = solventofMolessoluteofMoles

soluteofmoles

=1000mM

mM

M

1000m

m

1

1

1

Or11

1

2 mM

10001

mM

1000mM1

12

2

12 mM

10001

mM

10001

1



Or12

1

mM

1000

1 = mole fraction of solvent [

1 + 2 =1

1 = 1 – 2]

Hence m =11

2

M

1000

Ex.35 The molality and molarity of a solution of H2SO

4are 94.13 and 11.12 respectively. Calculate the density

of the solution.

Sol. d = M

1000

.wt.mol

m

1= 11.12

1000

98

13.94

1= 1.2079 g/ml



SUMMARY

The study of chemistry is very important as its domain encompasses every sphere of life. Chem-ists study the properties and structure of substances and the changes undergone by them.All substancescontain matter which can exist in three states –solid, liquid or gas. The constituent particles are held indifferent ways in these states of matter and they exhibit their characteristic properties. Matter can also beclassified into elements, compounds or mixtures.An element contains particles of only one type whichmay be atoms or molecules. The compounds are formed where atoms of two or more elementscombine in a fixed ratio to each other. Mixtures occur widely and many of the substances present aroundus are mixtures.

When the properties of a substance are studied, measurement is inherent. The quantification ofproperties requires a system of measurement and units in which the quantities are to be expressed. Manysystems of measurement exist out of which the English and the Metric Systems are widely used. Thescientific community, however, has agreed to have a uniform and common system throughout the worldwhich is abbreviated as SI units (International System of Units).

Since measurements involve recordingof data which are always associated with a certain amountof uncertainty, the proper handling of data obtained by measuring the quantities is very important. Themeasurements of quantities in chemistry are spread over a wide range of 10–31 to 10+23. Hence, aconvenient system of expressing the numbers in scientific notation is used. The uncertainty is takencare of by specifying the number of significant figures in which the observations are reported. Thedimensional analysis helps to express the measured quantities in different systems of units. Hence, it ispossible to interconvert the results from one system of units to another.

The combination of different atoms is governed by basic laws of chemical combination – thesebeing the Law of Conservation of Mass, Law of Definite Proportions, Law of Multiple Propor-tions, Gay Lussac’s Law of Gaseous Volumes and Avogadro Law.All these laws led to the Dalton’satomic theory which states that atoms are building blocks of matter. The atomic mass of an element isexpressed relative to 12C isotope of carbon which has an exact value of 12u. Usually, the atomic massused for an element is the average atomic mass obtained by taking into account the natural abundanceof different isotopes of that element. The molecular mass of a molecule is obtained by taking sum of theatomic masses of different atoms present in a molecule. The molecular formula can be calculated bydetermining the mass per cent of different elements present in a compound and its molecular mass.

The number of atoms, molecules or any other particles present in a given system are expressedin the terms of Avogadro constant (6.022 1023). This is known as 1 mol of the respective particles orentities.

Chemical reactions represent the chemical changes undergone by different elements and compounds.Abalanced chemical equation provides a lot of information. The coefficients indicate the molar ratios andthe respective number of particles taking part in a particular reaction. The quantitative study of thereactants required or the products formed is called stoichiometry. Using stoichiometric calculations,the amounts of one or more reactant(s) required to produce a particular amount of product can bedetermined and vice-versa. The amount of substance present in a given volume of a solution is expressedin number of ways, e.g., mass per cent, mole fraction, molarity and molality.



EXERCISE-1

[SINGLE CORRECT CHOICE TYPE]

Q.1 Hydrogen and oxygen combine to form H2O2 and H2O containing 5.93% and 11.2% Hydrogenrespectively. The data illustrates –

(1) Law of conservation of mass (2) Law of constant proportions

(3) Law of reciprocal proportions (4) Law of multiple proportions

Q.2 If water samples are taken from sea, rivers, clouds, lake or snow, they will be found to contain H2 andO2 in the fixed ratio of 1 : 8. This indicates the law of –

(1) Multiple proportion (2) Definite proportion

(3) Reciprocal proportion (4) None of these

Q.3 The law of conservation of mass holds good for all of the following except:

(1)All chemical reactions (2) Nuclear reactions

(3) Endothermic reactions (4) Exothermic reactions.

Q.4 If law of conservation of mass was to hold true, then 20.8 gm of BaCl2 on reaction with 9.8 gm ofH2SO4 will produce 7.3 gm of HCl and BaSO4 equal to -

(1) 11.65 gm (2) 23.3 gm (3) 25.5 gm (4) 30.6 gm

Q.5 Percentage of copper and oxygen in sample of CuO obtained by different methods were found to besame. This proves the law of -

(1) Constant proportion (2) Multiple proportion

(3) Reciprocal proportion (4) None of these

Q.6 The haemoglobin of most mammals contains approximately0.33%of iron bymass. The molecularmassof haemoglobin is 67200. The number of iron atoms in each molecule of haemoglobin is-

(1) 3 (2) 4 (3) 2 (4) 6

Q.7 The atomic mass of calcium is 40. How many atoms are present in 2g calcium ?

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

Q.8 The point given as full stop at the end of a sentence by a graphite pencil weighs 2 × 10–12 g. How manycarbon atoms are present in such a point ?

(1) 1.003 × 1010 (2) 1.004 × 1011 (3) 1.002 × 1012 (4) 1.001 × 10–11

Q.9 Assume the human body contains 80% water. Calculate the number of molecules of water that are in thebody of a person who has a mass of 65 kg.

(1) 17.39 × 1026 (2) 17.39 × 1027 (3) 17.39 × 1226 (4) 18.39 × 1026

Q.10 Morphine contains 67.3% carbon, 4.6 % nitrogen (by mass) and remaining are the other constituents.Calculate the relative number of carbon and nitrogen atoms in morphine.

(1)1

17(2)

17

1(3)

1

16(4)

1

15



Q.11 Total number of atoms of all elements present in 1 mole of ammonium dichromate [(NH4)2Cr2O7] is

(1) 14 (2) 19 (3) 6 × 1023 (4) 114 × 1023

Q.12 11.2 litre of a gas at NTP weighs 14 g. The gases would be :

(1) N2 (2) CO (3) N2O (4) B2H6

Q.13 The number of molecules of water in 333g ofAl2(SO4)3.18H2O is

(1) 18.0 × 6.02 × 1023 (2) 9.0 × 6.02 × 1023

(3) 18.0 (4) 36.0

Q.14 The atomic weights of two elementsAand B are 40 and 80 respectively. If x g ofAcontains y atoms,

how many atoms are present in 2x g of B

(1)2

y(2)

4

y(3) y (4) 2y

Q.15 One mole of CO2 contains :

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

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

Q.16 Which has the maximum number of atoms :

(1) 24 g C (12) (2) 56 g Fe (56) (3) 27 gAl (27) (4) 108 g Ag (108)

Q.17 Mass of 1 atom of Hydrogen is -

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

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

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

Q.19 The total number of electrons present in 18 mL water (density 1 g/mL) is -

(1) 6.023 × 1023 (2) 6.023 × 1024 (3) 6.023 × 1025 (4) 6.023 × 1021

Q.20 One mole of P4 molecules contains -

(1) 1 molecule (2) 4 molecules

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

Q.21 The total number of protons , electrons and neutrons in 12gm of 6C12 is -

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

Q.22 The number of sodium atoms in 2 moles of sodium ferrocyanide Na4[Fe(CN)6], is-

(1) 2 (2) 6.023 × 1023

(3) 8 × 6.02 × 1023 (4) 4 × 6.02 × 1023



Q.23 Number of Ca+2 and Cl– ion in 111 g of anhydrous CaCl2 are -

(1) NA, 2NA (2) 2NA, NA (3) NA, NA (4) None

Q.24 16 g of SOx occupies 5.6 litre at NTP.Assuming ideal gas nature, The value of x is :

(1) 1 (2) 2 (3) 3 (4) None of these

Q.25 How many molesof magnesium phosphateMg3(PO4)2will contain 0.25 mole of oxygen atoms:

(1) 0.02 (2) 3.125 ×10–2 (3) 1.25 ×10–2 (4) 2.5 × 10–2

Q.26 How many moles of electron weigh one kilogram :

(1) 6.023 × 1023 (2)108.9

1× 1023 (3)

108.9

10023.6 54(4)

023.6108.9

1

×108

Q.27 2 moles of H2 at NTP occupy a volume of

(1) 11.2 litre (2) 44.8 litre (3) 2 litre (4) 22.4 litre

Q.28 4.48 litres of methane at N.T.P. correspond to-

(1) 1.2 x 1022 molecules of methane (2) 0.5 mole of methane

(3) 3.2 gm of methane (4) 0.1 mole of methane

Q.29 Mol. wt. = vapour density × 2, is valid for -

(1) metals (2) non metals (3) solids (4) gases

Q.30 5.6 litre of a gas at N.T.P. weighs equal to 8 gm, the vapour density of gas is -

(1) 32 (2) 16 (3) 8 (4) 40

Q.31 Equal masses of O2 , H2 and CH4 are taken in a container. The respective mole ratio of these gases incontainer is -

(1) 1 : 16 : 2 (2) 16 : 1 : 2 (3) 1 : 2 : 16 (4) 16 : 2 : 1

Q.32 The vapour density of gasAis four times that of B. If molecular mass of B is M, then molecular mass ofA is -

(1) M (2) 4M (3)M

4(4) 2M

Q.33 The density of air is 0.001293 gm/ml at S.T.P. It’s vapour density is -

(1) 147 (2) 14.7 (3) 1.47 (4) 0.147

Q.34 Nitric acid is manufactured by the Ostwald process, in which nitrogen dioxide reacts with water.

3 NO2 (g) + H2O (l) 2 HNO3 (aq) + NO (g)

How many grams of nitrogen dioxide are required in this reaction to produce 25.2 gm HNO3 ?

(1) 27.6 gm (2) 26.7 gm (3) 72.6 gm (4) None



Q.35 Flourine reacts with uranium to produce uranium hexafluoride, UF6, as represented by this equation

U(s) + 3F2(g)UF6(g)

How many fluorine molecules are required to produce 2.0 mg of uranium hexafluoride, UF6, from anexcess of uranium ? The molar mass of UF6 is 352 gm/mol.

(1) 2.0 ×1019 (2) 1.0 ×1019 (3) 1.0 ×1219 (4) 1.2 ×1019

Q.36 X gm ofAg was dissolved in HNO3 and the solution was treated with excess of NaCl. When 2.87 gm

ofAgCl was precipitated the value of x is -

(1) 1.08 gm (2) 2.16 gm (3) 2.70 gm (4) 1.62 gm

Q.37 What mass of calcium chloride in grams would be enough to produce 14.35 gm ofAgCl.

(Atomic mass : Ca = 40, Ag = 108) -

(1) 5.55 gm (2) 8.295 gm (3) 16.59 gm (4) 11.19 gm

Q.38 What is the maximum amount of nitrogen dioxide that can be produced by mixing 4.2 gm of NO(g) and3.2 gm of O2(g) ?

(1) 4.60 gm (2) 2.30 gm (3) 3.22 gm (4) 6.44 gm

Q.39 The percentage by volume of C3H8 in a gaseous mixture of C3H8, CH4 and CO is 20. When 100 mLof the mixture is burnt in excess of O2 , the volume of CO2 produced is :

(1) 90 mL (2) 160 mL (3) 140 mL (4) none of these

Q.40 A gaseous mixture contains CO2(g) and N2O(g) in a 2 : 5 ratio by mass. The ratio of the number of

molecules of CO2(g) and N2O(g) is :

(1) 5 : 2 (2) 2 : 5 (3) 1 : 2 (4) 5 : 4

Q.41 The percentage of nitrogen in urea is about –

(1) 38.4 (2) 46.7 (3) 59.1 (4) 61.3

Q.42 The mass of carbon present in 0.5 mole of K4[Fe(CN)6] is –

(1) 1.8 gm (2) 18 gm (3) 3.6 gm (4) 36 gm

Q.43 1.2 gm of Mg (At. mass 24) will produce MgO equal to –

(1) 0.05 mol (2) 40 gm (3) 40 mg (4) 4 gm

Q.44 Insulin contains 3.4% sulphur by mass. What will be the minimum molecularweight of insulin-

(1) 94.117 (2) 1884 (3) 941 (4) 976

Q.45 The percent of N in 66% pure (NH4)2SO4 sample is -

(1) 32 (2) 28 (3) 14 (4) None of these

Q.46 For the reaction, 2Fe(NO3)

3+ 3Na

2CO

3 Fe

2(CO

3)

3+ 6NaNO

3

Initially if 2.5 mole of Fe(NO3)

2and 3.6 mole of Na

2CO

3is taken. If 6.3 mole of NaNO

3is obtained

then % yield of given reaction is :

(1) 50% (2) 84% (3) 87.5% (4) 100%



Q.47 The conversion of oxygen to ozone occurs to the extent of 15% only. The mass of ozone that can be

prepared from 67.2 L of oxygen at S.T.P. will be -

(1) 14.4 gm (2) 96 gm (3) 640 gm (4) 64 gm

Q.48 An element (1) (at wt = 75) and another element (2) (at. wt. = 25) combine to form a compound. Thecompound contains 75% (1) by weight. The formula of the compound will be -

(1)A2B (2)A3B (3)AB3 (4)AB

Q.49 The empirical formula of a compound is CH. Its molecular weight is 78. The molecular formula of thecompound will be -

(1) C2H2 (2) C3H3 (3) C4H4 (4) C6H6

Q.50 Carbon reacts with chlorine to form CCl4. 36 gm of carbon was mixed with 142 g of Cl2. Calculatemass of CCl4 produced and the remaining mass of reactant.

(1) wc = 24 gm ;4CClw = 154 gm (2) wc = 21 gm ;

4CClw = 145 gm

(3) wc = 14 gm ;4CClw = 164 gm (4) None

Q.51 Sulphuric acid is produced when sulphur dioxide reacts with oxygen and water in the presence of acatalyst : 2SO2(g) + O2 (g) + 2 H2O(l)2 H2SO4 . If 5.6 mol of SO2 reacts with 4.8 mol of O2 anda large excess of water, what is the maximum number of moles of H2SO4 that can be obtained?

(1) 6.5 (2) 5.6 (3) 4.6 (4) 3.6

Q.52 A mixture containing 100 gm H2 and 100 gm O2 is ignited so that water is formed according to thereaction, 2H2 + O2 2H2O; How much water will be formed -

(1) 112.5 gm (2) 50 gm (3) 25 gm (4) 200 gm

Q.53 0.5 mole of H2SO4 is mixed with 0.2 mole of Ca(OH)2. The maximum number of moles of CaSO4formed is -

(1) 0.2 (2) 0.5 (3) 0.4 (4) 1.5

Q.54 For the reaction : A + 2BC

5 mole ofA and 8 mole of B will produce -

(1) 5 mole of C (2) 4 mole of C (3) 8 mole of C (4) 13 mole of C

Q.55 In the reaction 4A+ 2B + 3CA4B2C3 , what will be the number of moles of product formed, startingfrom one mole of A, 0.6 mole of B and 0.72 mole of C ?

(1) 0.25 (2) 0.3 (3) 0.24 (4) 2.32

Q.56 The amount of zinc required to produce 224mL of H2at NTP on treatment with dilute H

2SO

4will be –

(1) 0.65g (2) 6.5g (3) 65g (4) 0.065g

Q.57 When 0.5 mol of BaCl2 is added to 0.2 mol of Na3PO4, the number of moles of Ba3(PO4)2 formed is:

(1) 0.10 (2) 0.20 (3) 0.40 (4) 0.15



Q.58 Density of water is 1 g/m L. The concentration of water in mol/litre is

(1) 1000 (2)18 (3) 0.018 (4) 55.5

Q.59 How many grams of NaOH will be needed to prepare 250 mL of 0.1 M solution

(1) 1g (2) 10g (3) 4g (4) 6g

Q.60 The molarity of a glucose solution containing 36 g of glucose per 400 mL of the solution is

(1) 1.0 (2) 0.5 (3) 2.0 (4) 0.05

Q.61 The molality of 15% (wt./vol.) solution of H2SO4 of density 1.1 g/cm3 is approximately

(1) 1.2 (2) 1.4 (3) 1.8 (4) 1.6

Q.62 1000 gram aqueous solution of CaCO3 contains 10 gram of carbonate. Concentration of solution is

(1) 10ppm (2) 100ppm (3) 1000ppm (4) 10,000ppm

Q.63 200 ml of a solution contains 5.85 g dissolved sodium chloride. The concentration of the solution will be

(Na = 23; Cl = 35.5)

(1) 1 molar (2) 2 molar (3) 0.5 molar (4) 0.25 molar

Q.64 How many grams of NaOH will be required to neutralize 12.2 grams of benzoic acid

(1) 40 gms (2) 4 gms (3) 16 gms (4) 12.2 gms

Q.65 If one mole of a substance is present in 1 kg of solvent. Then

(1) It shows molar concentration (2) It shows molal concentration

(3) It shows normality (4) It shows strength gm / gm

Q.66 A mixture has 18g water and 414g ethanol. The mole fraction of water in mixture is (assume ideal

behaviourof the mixture)

(1) 0.1 (2) 0.4 (3) 0.7 (4) 0.9

Q.67 What is the molarity of H2SO4 solution, that has a density 1.84 gm/cc at 35°C and contains solute 98%

byweight

(1) 4.18 M (2) 8.14 M (3) 18.4 M (4) 18 M

Q.68 The maximum amount of BaSO4 precipitated on mixing 20 mL of 0.5 M BaCI2 with 20 mL of 1 M

H2SO4 is

(1) 0.25 mole (2) 0.5 mole (3) 1 mole (4) 0.01 mole

Q.69 The number of moles of a solute in its solution is 20 and total number of moles are 80. The mole fraction

of solute is :

(1) 2.5 (2) 0.25 (3) 1 (4) 0.75

Q.70 2.5 L of 1 M NaOH solution mixed with another 3Lof 0.5 M NaOH solution. Then find out molarity ofresultant solution.

(1) 0.80 M (2) 1.0 M (3) 0.73 (4) 0.50 M



Q.71 A sample of 98% acid has a density of 1.9 g cm–3. The amount of pure acid present in 2 L of the acid is:

(1) 3.724 kg (2) 372.4 g (3) 3.724 g (4) 100 g

Q.72 What is the [OH–] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of0.10 M Ba(OH)2?

(1) 0.12 M (2) 0.10 M (3) 0.40 M (4) 0.0050 M

Q.73 25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. Ifsodium carbonate dissociates completely, molar concentration of sodium ion, Na+ and carbonate ions,CO3

2– are respectively (Molar mass of Na2CO3 = 106 g mol–1)

(1) 0.477 M and 0.477 M (2) 0.955 M and 1.910 M

(3) 1.910 M and 0.955 M (4) 1.90 M and 1.910 M

Q.74 How much NaNO3 must be weighed out to make 50 ml of an aqueous solution containing 70 mg Na+

per ml ?

(1) 12.934 g (2) 1.29 g (3) 0.129 g (4) 1.59 g

Q.75 How many ions are present in 1 L of 0.6 M K4[Fe(CN)6]? (Avogadro's Number = 6 × 1023)

(1) 1.8 × 1024 (2) 1.8 × 1023 (3) 3.6 × 1023 (4) 3.6 × 1024

Q.76 360g of water is present in a one Lmixture of ethanol and water. Molarity of water in the mixture is –

(1) 20.0 (2) 36.0 (3) 18.0 (4) None of these

Q.77 Mole fraction of solvent in aqueous solution of NaOH having molality of 3 is :

(1) 0.3 (2) 0.05 (3) 0.7 (4) 0.95

Q.78 What volume of 0.1M H2SO

4is needed to completely neutralize 40mL of 0.2M NaOH solution -

(1) 10mL (2) 40mL (3) 20mL (4) 80 mL

Q.79 171 g of cane sugar (C12H22O11) is dissolved in 1 litre of water. The molarity of the solution is

(1) 2.0 M (2) 1.0 M (3) 0.5 M (4) 0.25 M

Q.80 Equal volumes of 0.1 M AgNO3 and 0.2 M NaCl are mixed. The concentration of NO3– ions in the

mixturewillbe

(1) 0.1 M (2) 0.05 M (3) 0.2 M (4) 0.15 M

Q.81 20 ml of HCl solution requires 19.85 ml of 0.01 M NaOH solution for complete neutralization. The

molarity of HCl solution is

(1) 0.0099 (2) 0.099 (3) 0.99 (4) 9.9

Q.82 At a particular place, 1000 mL of air weighs 1.248 g. The specific gravity of air in terms of hydrogen if

1g H2 occupies 11200 mLvolume will be :

(1) 13.98 (2) 10 (3) 16 (4) 2.47



Q.83 Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g/mL. Volume

of acid required to make 1L of 0.1 M H2SO4 solution is :

(1) 16.65 mL (2) 22.20 mL (3) 5.55 mL (4) 11.10 mL

Q.84 The charge on 1 gram ion ofAl3+ is :

(1)27

1NAe coulomb (2)

3

1NAe coulomb

(3)9

1NAe coulomb (4) 3 × NAe coulomb

Q.85 The density in grams per litre of a mixture containing an equal number of moles of methane and ethane at

NTP is :

(1) 1.03 (2) 1.10 (3) 0.94 (4) 1.20

Q.86 Equal weights of ethane and hydrogen are mixed in an empty vessel at 25°C. The fraction of the total

pressure exerted by hydrogen is :

(1)2

1(2)

1

1(3)

16

1(4)

16

15

Q.87 6 gm of carbon combines with 32 gm of sulphur to form CS2. 12 gm of C also combine with 32 gm ofoxygen to form carbondioxide. 10 gm of sulphur combines with 10 gm of oxygen to form sulphurdioxide. Which law is illustrated by them -

(1) Law of multiple proportions (2) Law of constant composition

(3) Law of Reciprocal proportions (4) Gay Lussac's law.

Q.88 The volume of 1.0 g of Hydrogen in liters at N.T.P. is -

(1) 2.24 (2) 22.4 (3) 1.12 (4) 11.2

Q.89 The specific gravity of the stainless steel spherical balls used in ball-bearings are 10.2. How many ironatoms are present in each ball of diameter 1 cm if the balls contain 84% iron, by mass ? The atomic massof iron is 56.

(1) 4.12 × 1021 (2) 4.82 × 1022 (3) 3.82 × 1022 (4) None

Q.90 The molar mass of N2O as well as CO2 is 44g mol–1.At 25°C and 1 atm, N2O contains n molecules of

gas. The number of molecules in 2.0 L of CO2 under the same conditions is :

(1)4

n(2)

8

n(3) 4n (4) n

Q.91 Two flasks P and Q of the same capacity contain helium and hydrogen respectively at 27°C and 1

atmospheric pressure. Flask P contains

(1) the same number of atoms as Q does (2) half the number of atoms as Q does

(3) twice the number of atoms as Q does (4) gas of the same weight as Q does



Q.92 Samples of 1.0 g ofAl are treated separately with an excess of sulphuric acid and an excess of sodium

hydroxide. The ratio of the number of moles of the hydrogen gas evolved is :

(1) 1 : 1 (2) 3 : 2 (3) 2 : 1 (4) 9 : 4

Q.93 A mixture of gas ''X'' (mol. wt. 16) and gas "Y" (mol. wt. 28) in the mole ratio a : b has a mean molecularweight 20. What would be mean molecular weight if the gases are mixed in the ratiob : a under identical conditions (gases are non reacting).

(1) 24 (2) 20 (3) 26 (4) 40

Q.94 Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and56 lit. O2 at 1 atm & 273 K according to given reaction, is

12C(s) + 11H2(g) + 11/2O2 (g) C12H22O11(s)

(1) 138.5 (2) 155.5 (3) 172.5 (4) 199.5

Q.95 An elementAis tetravalent and anotherelement B is divalent.The formula of the compound formed fromthese elements will be-

(1)A2B (2)AB (3)AB2 (4)A2B3

Q.96 Two elements X (At. mass 16) andY(At. mass 14) combine to form compoundsA, B and C.The ratio ofdifferent masses ofYwhich combine with a fixed mass of X inA, Band C is 1 :3 : 5. If 32parts bymass ofX combineswith 84 partsbymass ofYin B, then in C, 16partsbymass ofX will combine with-

(1) 14 parts by mass of Y (2) 42 parts by mass of Y

(3) 70 parts by mass of Y (4) 84 parts by mass of Y

Q.97 If one mole of ethanol (C2H5OH) completely burns to form carbon dioxide and water, the weight ofcarbon dioxide formed is about -

(1) 22 gm (2) 45 gm (3) 66 gm (4) 88 gm

Q.98 The moles of O2 required for reacting with 6.8 gm of ammonia.

(..... NH3 +..... O2 ..... NO + ..... H2O) is

(1) 5 (2) 2.5 (3) 1 (4) 0.5

Q.99 If isotopic distribution of C–12 and C–14 is 98% and 2% respectively, then the number of C–14 atomsin 12 gm of carbon is -

(1) 1.035 × 1022 (2) 3.01 × 1022 (3) 5.88 × 1023 (4) 6.02 × 1023

Q.100 If 3.01 × 1020 molecules are removed from 98 mg of H2SO4, then the number of moles of H2SO4 leftare –

(1) 0.1 × 10–3 (2) 0.5 × 10–3 (3) 1.66 × 10–3 (4) 9.95 × 10–2

Q.101 The density of mercury is 13.6 g/mL. Calculate the diameter of an atom of mercury assuming that eachatom of mercury is occupying a cube of edge length equal to the diameter of the mercury atom(Hg = 200)

(1) 9.2 × 10–8 cm (2) 2.9 × 10–6 cm (3) 2.7 × 10–8 cm (4) 2.9 × 10–8 cm



Q.102 If the density of water is 1 gm/cm3, then the volume occupied by one molecule of water is approximately-

(1) 18 cm3 (2) 22400 cm3 (3) 6.02×10–23 cm3 (4) 3.0×10–23 cm3

Q.103 How many gram glucose burn completely to produce sufficient CO2 needed for complete conversion of12 g NaOH into Na2CO3 :

(1) 5.5 g (2) 4.5 g (3) 6.5 g (4) 3.5 g

Q.104 1.35 gm of pureAmetal was quantitatively converted into 1.88 gm of pureAO. What is atomic weightof A –

(1) 40.75 (2) 50 (3) 60 (4) 70

Q.105 Mass of H2O in 1000 kg CuSO4.5H2O is - (Cu = 63.5)

(1) 360.7 kg (2) 36.05 kg (3) 3605 kg (4) 3.605 kg

Q.106 0.8 mole of a mixture of CO and CO2 requires exactly 40 gram of NaOH in solution for completeconversion of all the CO2 into Na2CO3. How manymoles more of NaOH would it require for conversioninto Na2CO3, if the mixture (0.8 mole) is completely oxidised to CO2

(1) 0.2 (2) 0.6 (3) 1 (4) 1.5

Q.107 4.4 gm of CO2 and 2.24 litre of H2 at STP are mixed in a container. The total number of moleculespresent in the container will be -

(1) 6.022 × 1023 (2) 1.2046 × 1023 (3) 2 moles (4) 6.023 × 1024

Q.108 Find the volume of CO2 obtained at N.T.P. on heating 200 gm of 50% pure CaCO3 –

(1) 11.2 litre (2) 22.4 litre (3) 44.8 litre (4) None of these

Q.109 Asample ofAlF3 contains 3.0 × 1024 F– ions. The number of formula units in this sample are –

(1) 9.0 × 1024 (2) 3.0 × 1024 (3) 0.75 × 1024 (4) 1.0 × 1024

Q.110 The mass of 70% pure H2SO4 required for neutralisation of 1 mol of NaOH -

(1) 49 gm (2) 98 gm (3) 70 gm (4) 34.3 gm

Q.111 Assuming that petrol is iso-octane (C8H18) and has density 0.8 gm/ml. 1.425 litre of petrol on complete

combustion will consume oxygen -

(1) 50 L (2) 125 L (3) 125 mol (4) 50 mol

Q.112 What volume of air at STP containing 21% of oxygen by volume is required to completely burn sulphur(S

8) present in 200 g of sample, which contains 20% inert material which does not burn. Sulphur burns

according to the reaction :

1/8S8(s) + O

2(g) SO

2(g)

(1) 23.52 litre (2) 320 litre (3) 112 litre (4) 533.33 litre



Q.113 The percentagebymole of NO2 ina mixture ofNO2(g) and NO(g)havingaverage molecularmass 34 is :(1) 25% (2) 20% (3) 40% (4) 75%

Q.114 26 c.c. of CO2 are passed over red hot coke. The volume of CO evolved is -

(1) 15 c.c (2) 10 c.c. (3) 32 c.c. (4) 52 c.c.

Q.115 An iodized salt contains 0.5 % of NaI.Aperson consumes 3 gm of salt everyday. The number of iodideions going into his body everyday is(1) 10–4 (2) 6.02 ×10–4 (3) 6.02 × 1019 (4) 6.02 × 1023

Q.116 The mass of CO2 produced from 620 gm mixture of C2H4O2 & O2, prepared to produce maximumenergy is (Combustion reaction is exothermic)(1) 413.33 gm (2) 593.04 gm (3) 440 gm (4) 320 gm

Q.117 The mass of P4O10 produced if 440 gm of P4S3 is mixed with 384 gm of O2 isP4S3 + O2 P4O10 + SO2

(1) 568 gm (2) 426 gm (3) 284 gm (4) 396 gm

Q.118 Two oxides of Metal contain 27.6% and 30% oxygen respectively. If the formula of first oxide is

M3O4 then formula of second oxide is -

(1) MO (2) M2O (3) M2O3 (4) MO2

Q.119 Assuming complete precipitation ofAgCl, calculate the sum of the molar concentration of all the ions if 2lit of 2MAg2SO4 is mixed with 4 lit of 1 M NaCl solution is :(1) 4M (2) 2M (3) 3 M (4) 2.5 M

Q.120 What volumes should you mix of 0.2 M NaCl and 0.1 M CaCl2 solution so that in resulting solution theconcentration of positive ion is 40% lesser than concentration of negative ion.Assuming total volume ofsolution 1000 ml.(1) 400 ml NaCl , 600 ml CaCl2 (2) 600 ml NaCl, 400 ml CaCl2(3) 800 ml NaCl, 200 ml CaCl2 (4) None of these

Q.121 Excess of carbon dioxide is passed through 50 mL of 0.5 M calcium hydroxide solution. After thecompletion of the reaction, the solution was evaporated to dryness. The solid calcium carbonate wascompletely neutralised with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (At.mass of calcium = 40)(1) 200 cm3 (2) 500 cm3 (3) 400 cm3 (4) 300 cm3



EXERCISE-2

[PREVIOUS YEAR'S QUESTIONS]

Q.1 Pressure in a mixture of 4g of O2and 2g of H

2confined in a bulb of 1L at 0OC is (A.I.I.M.S. 1999)

(1) 15.210 atm (2) 25.215 atm (3) 31.205 atm (4) 45.215 atm

Q.2 12g of alkaline earth metal gives 14.8g of its nitrite.Atomic weight of metal is - (A.I.I.M.S. 2000)

(1) 12 (2) 20 (3) 40 (4) 14.8

Q.3 Volume of CO2obtained by the complete decomposition of 9.85 g BaCO

3is- (CPMT-2000)

(1) 2.24L (2) 1.12L (3) 0.84L (4) 0.56L

Q.4 2.5L NaOH of 1M solution is mixed with 3L NaOH of 0.5 M solution. What is the molarity of the

resultingsolution- (CBSE 2002)

(1) 0.80 M (2) 1.0 M (3) 0.73 M (4) 0.50 M

Q.5 Avogadro's number is the number of molecules present in : (AFMC 2003)

(1) 1L of molecule (2) 1g of molecule

(3) gram molecular mass (4) 1g-atom of molecules

Q.6 In Haber process, 30L of dihydrogen and 30L of dinitrogen were taken for reaction which yielded only

50% of the expected product. What will be the composition of gaseous mixture under the aforesaid

condition in the end? (CBSE AIPMT 2003)

(1) 20 L ammonia, 10 L nitrogen, 30 L hydrogen




Q.7 The number of sodium atoms in 2 moles of sodium ferrocyanides is : (BHU 2004)

(1) 12 × 1023 (2) 26 × 1023 (3) 34 × 1023 (4) 48 × 1023

Q.8 The amount of H2S required to precipitate 1.69 g BaS from BaCl

2solution is :

(Haryana PMT 2005)

(1) 3.4 g (2) 0.24 g (3) 0.34 g (4) 0.17 g

Q.9 The oxygen obtained from 72 kg of water is : (RPMT 2005)

(1) 72 kg (2) 45 kg (3) 50 kg (4) 64 kg

Q.10 Total number of protons in 10g of calcium carbonate is (NA

= 6.023 × 1023) : (RPMT 2005)

(1) 3.01 × 1024 (2) 4.06 × 1024 (3) 2.01 × 1024 (4) 3.24 × 1024



Q.11 Total number of atoms represented by the compound CuSO4· 5H

2O is : (BHU 2005)

(1) 27 (2) 21 (3) 5 (4) 8

Q.12 A gas is found to have a formula [CO]x. If its vapour density is 70, the value of x is : (AMU 2006)

(1) 2.5 (2) 3.0 (3) 5.0 (4) 6.0

Q.13 A gas mixture contains O2and N

2in the ratio of 1 : 4 by weight. The ratio of their number of molecules

is (AFMC 2006)

(1) 1 : 8 (2) 1 : 4 (3) 3 : 16 (4) 7 : 32

Q.14 2.76 g of silver carbonate on being strongly heated yield a residue weighing: (BHU 2006)

(1) 2.16 g (2) 2.48 g (3) 2.64 g (4) 2.32 g

Q.15 The weight of one molecule of a compound C60

H122

is : (BHU 2007)

(1) 1.3 × 10–20g (2) 5.01 × 10–21 g (3) 3.72 × 1013 g (4) 1.4 × 10–21 g

Q.16 The largest number of molecules is in : (AMU 2008)

(1) 34 g of H2O (2) 28 g of CO

2(3) 46 g of CH

3OH (4) 54 g of N

2O

5

Q.17 How many moles of lead (II) chloride will be formed from a reaction between 6.5g of PbO and

3.2 g of HCl? (CBSE AIPMT 2008)

(1) 0.044 (2) 0.333 (3) 0.011 (4) 0.029

Q.18 What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely 1L of

propane gas (C3H

8) measured under the same conditions? (CBSE AIPMT 2008)

(1) 7L (2) 6L (3) 5L (4) 10L

Q.19 What volume of CO2will be liberated at NTP, if 12g of carbon is burnt in excess of oxygen?

(AFMC 2009)

(1) 11.2 L (2) 22.4 L (3) 2.24 L (4) 1.12 L

Q.20 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water

produced in this reaction will be : (CBSE AIPMT 2009)

(1) 3 mol (2) 4 mol (3) 1 mol (4) 2 mol

Q.21 60 g of a compound on analysis produced 24 gm carbon, 4 gm hydrogen and 32 gm oxygen. The

empirical formula of the compound is : (BVP 2010)

(1) CH2O

2(2) CH

2O (3) CH

4O (4) C

2H

4O

2

Q.22 Volume of a gas at NTP is 1.12 × 10–7 cm3 . The number of molecules in it is : (Manipal 2010)

(1) 3.01 × 1012 (2) 3.01 × 1024 (3) 3.01 × 1023 (4) 3.01 × 1026

Q.23 The mass of one mole of electron is : (CPMT 2010)

(1) 9 × 10–28 g (2) 0.55 mg (3) 9.1 × 10–24 g (4) 6 × 10–12 g



Q.24 In an experiment, 4 g of M2O

xoxide was reduced to 2.8 g of the metal. If the atomic mass of the metal

is 56 g mol–1 , the number of O atoms in the oxide is : (AFMC 2010)

(1) 1 (2) 2 (3) 3 (4) 4

Q.25 Number of atoms present in 4.25 g of NH3is : (AFMC 2010)

(1) 6.023 × 1023 (2) 4 × 6.023 × 1023 (3) 1.7 × 1024 (4) 4.25 × 6.023 × 1023

Q.26 Which has the maximum numberof molecules amongthe following? [CBSE (Final) 2011]

(1) 8 g H2

(2) 64 g SO2

(3) 44 g CO2

(4) 48 g O3

Q.27 Mole fraction of the solute in a 1.00 molal aqueous solution is [CBSE (Pre) 2011]

(1) 1.7700 (2) 0.1770 (3) 0.0177 (4) 0.0344



EXERCISE-3

[ASSERTION & REASON TYPE QUESTIONS]

Each of the questions given below consist of Assertion and Reason. Use the following Key to choose

the appropriate answer.

(A) If both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.

(B) If both Assertion and Reason are correct, and Reason is not the correct explanation of Assertion.

(C) If Assertion is correct and Reason is incorrect.

(D) IfAssertion is incorrect but Reason is correct.

(E) If bothAssertion and Reason are incorrect.

Q.1 Assertion :Approximate mass of 1 atom of O16 in gms is (16 / NA)

Reason : 1 atom of O16 weighs 16 a.m.u & 1 a.m.u = (1 / NA) g.

(1)A (2) B (3) C (4) D (5) E

Q.2 Assertion : During a chemical reaction total moles remains constant.

Reason : During a chemical reaction total mass remains constant.

(1)A (2) B (3) C (4) D (5) E

Q.3 Assertion : For the reaction producing Fe & CO2 by the reaction of Fe2O3 and C the ratio

of stoichiometric coefficients of Fe2O3 : Fe is 1 : 2.

Reason : During a chemical a reaction atoms can neither be created nor be destroyed.

(1)A (2) B (3) C (4) D (5) E

Q.4 Assertion : 2A + 3B C

4/3 moles of 'C' are always produced when 3 moles of 'A' & 4 moles of 'B' are added.

Reason : 'B' is the liming reactant for the given data.

(1)A (2) B (3) C (4) D (5) E

Q.5 Assertion : Molality of pure ethanol is lesser than pure water.Reason :As density of ethanol is lesser than density of water.

[Given : dethanol = 0.789 gm/ml; dwater = 1 gm/ml](1)A (2) B (3) C (4) D (5) E



EXERCISE-4

[SUBJECTIVE BOARD QUESTIONS]

Very Short Answer Question :Q.1 What is the S.I. unit of density?

Q.2 What is one a.m.u. or one ‘u’?

Q.3 Which isotope of carbon is used for getting relative atomic masses?

Q.4 Write down the empirical formula of acetic acid.

Short Answer Question :Q.5 What will be the mass of one atom of C-12 in grams?

Q.6 Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate Ca3(PO4)2.

Q.7 If two elements can combine to form more than one compound, the masses of one element that combinewith a fixed mass of the other element, are in whole number ratio.(a) Is this statement true ?(b) If yes, according to which law ?(c) Give one example related to this law

Q.8 Calculate the average atomic mass of hydrogen using the following data :Isotope % Natural abundance Molar mass1H 99.985 12H 0.015 2

Long Answer Question :Q.9 A vessel contains 1.6g of dioxygen at STP (273.15K, 1atm pressure). The gas is now transferred to

another vessel at constant temperature, where pressure becomes half of the original pressure.Calculate : (i) volume of the new vessel (ii) number of molecules of dioxygen.

Q.10 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction?CaCO3 (s) + 2HCl (aq) CaCl2(aq) + CO2(g) + H2O (l)

What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with 100 g of CaCO3?Name the limiting reagent. Calculate the number of moles of CaCl2 formed in the reaction.

Q.11 A box contains some identical red coloured balls, labelled as A, each weighing 2 gm. Another boxcontains identical blue coloured balls, labelled as B, each weighing5 gm. Consider the combinationsAB,AB2,A2B andA2B3 and show that law of multiple proportions is applicable.



EXERCISE-1

Q.1 4 Q.2 2 Q.3 2 Q.4 2

Q.5 1 Q.6 2 Q.7 1 Q.8 2

Q.9 2 Q.10 1 Q.11 4 Q.12 1

Q.13 2 Q.14 3 Q.15 4 Q.16 1

Q.17 1 Q.18 2 Q.19 2 Q.20 4

Q.21 1 Q.22 3 Q.23 1 Q.24 2

Q.25 2 Q.26 4 Q.27 2 Q.28 3

Q.29 4 Q.30 2 Q.31 1 Q.32 2

Q.33 2 Q.34 1 Q.35 2 Q.36 2

Q.37 1 Q.38 4 Q.39 3 Q.40 2

Q.41 2 Q.42 4 Q.43 1 Q.44 3

Q.45 3 Q.46 3 Q.47 1 Q.48 4

Q.49 4 Q.50 1 Q.51 2 Q.52 1

Q.53 1 Q.54 2 Q.55 3 Q.56 1

Q.57 1 Q.58 4 Q.59 1 Q.60 2

Q.61 4 Q.62 4 Q.63 3 Q.64 2

Q.65 2 Q.66 1 Q.67 3 Q.68 4

Q.69 2 Q.70 3 Q.71 1 Q.72 2

Q.73 3 Q.74 1 Q.75 1 Q.76 1

Q.77 4 Q.78 2 Q.79 3 Q.80 2

Q.81 1 Q.82 1 Q.83 3 Q.84 4

Q.85 1 Q.86 4 Q.87 3 Q.88 4

Q.89 2 Q.90 4 Q.91 2 Q.92 1

Q.93 1 Q.94 2 Q.95 3 Q.96 3

Q.97 4 Q.98 4 Q.99 1 Q.100 2

Q.101 4 Q.102 4 Q.103 2 Q.104 1

Q.105 1 Q.106 2 Q.107 2 Q.108 2

Q.109 4 Q.110 3 Q.111 3 Q.112 4

Q.113 1 Q.114 4 Q.115 3 Q.116 3

Q.117 2 Q.118 3 Q.119 2 Q.120 4

Q.121 2



EXERCISE-2

Q.1 2 Q.2 3 Q.3 2 Q.4 3

Q.5 3 Q.6 4 Q.7 4 Q.8 3

Q.9 4 Q.10 1 Q.11 2 Q.12 3

Q.13 4 Q.14 1 Q.15 4 Q.16 1

Q.17 4 Q.18 3 Q.19 2 Q.20 2

Q.21 2 Q.22 1 Q.23 2 Q.24 3

Q.25 1 Q.26 1 Q.27 3

EXERCISE-3

Q.1 1 Q.2 4 Q.3 1 Q.4 4

Q.5 2

EXERCISE-4

Q.1 Kg m–3

Q.2 1 a.m.u.

Q.3 C–12

Q.4 CH2O

Q.5 1.992 × 10–23 gm

Q.6 Ca = 38.71%, P = 20.0%, O = 41.29%

Q.7 (a)Yes, the given statement is true(b) True, Law of multiple proportions.(c) Carbon and oxygen combine to form CO and CO2

2C + O2 2CO2 × 12g 32g carbon monoxideC + O2 CO212g 32g carbon dioxide

Q.8 1.00015u

Q.9 (i) 2.24L (ii) 3.011 × 1022

Q.10 HCl will be limiting reagent, 0.095 mol

Q.11 Combination AB AB2 A2B A2B3Mass ofA(g) 2 2 4 4Mass of B(g) 5 10 5 15

Masses of B which combine with fixed mass ofA(say 4 g) will be 10 g, 20 g, 5 g and 15 g i.e., theyare in the ratio 2 : 4 : 1 : 3 which is a simple whole number ratio. Hence, the law of multiple proportionsis applicable.



EXERCISE-1

1. Same ammount of oxygen combine with different % of H.

Law of multiple proportion.

2. Ratio of H and O is fixed which is law of definite proportions.

3. For nuclear reation law of conservation of mass does not hold good.

4. BaCl2 + H2SO4 BaSO4 + 2HCl

0.1 mole 0.1 mole 0 0

0 0 0.1 0.1

Mass of reactant = Mass of product

20.8 + 9.8 = 7.3 + x

x = 23.3 gm

5. % is fixed so it proves law of constant proportion.

6. % of Fe = MassMolar

protonsof.NoAtomFeof.wt × 100

=67200

56x× 100 =

3

1

x =3

1×

56

672= 4

7. Moles of Ca =M

W=

AN

N

N =40

2× 6.022 × 1023 = 3.0115 × 1022

8.AN

N=

M

W

N =12

102 12× 1023 × 6.022 = 1.004 × 1011

9. Mass of H2O = 65 × 0.8 = 52 Kg

N =18

1052 3× 6.022 × 1023 = 17.39 × 1026



10. atomNof.No

atomCof.No=

14/6.4

12/3.67=

1

17

11. Total number of atoms = 19 × 6 × 1023 atoms

12. ngas =4.22

2.11=

2

1mole

2

1=

M

W= M = 28 gm

13. n =666

333= 0.5

number of H2O = 0.5 × 18 × 6.02 × 1023 = 9 × 1023

14.40

x× NA = y

80

x2× NA = y

15. 1 mole (CO2) 6.023 × 1023 atom of C

16. C has max moles so will have maximum number of atoms.

17. 1 amu = 1.67 × 10–24 gm.

18. H2 has maximum moles so it will have maximum atoms.

19. P = 1 =V

M= M = 18 gm

OH2n =

18

18= 1

Total number of e– = 10 × NA

20. 1 mole of P4 contain 4 mole of P or 4 NA atoms of P.

21. nC =12

12= 1

Total number of e + p + n = 18 × 6.022 × 1023



22. Total number of Na atom = 2 × NA

23.2CaCln =

111

111= 1

Number of Ca2+ = NA

Number of Cl– = 2NA

24.xSOn =

4.22

6.5=

4

1=

x1632

16

=4

1=

x2

1

= n = 2

25. O moles are present in 1 mole Mg3 (PO4)3

4

1mole will be present in

8

1×

4

1=

32

1= 3.125 × 10–2

26. 9.1 × 10–31 Kg weight mean 1 e–

1 Kg weight mean = 31101.9

1

e–

Moles of e– =022.61.9

1010 2331

=023.61.9

108

27.2HV at NTP = 2 × 22.4 = 44.8 liter..

28.4CHn =

4.22

48.4=

16

W

W = 3.2 gm.

30. n =4.22

6.5=

VD2

8

VD = 16

31. O2 : H2 : CH4

32

32:

2

32:

16

32

1 : 16 : 2

32. VDA = 4 VDB2 × VDA = 4 × 2 (VD)B

MA = 4 × MB



33. PM = dRT

M =1

2730021.0293.1 = 28.9 gm

VD =2

M= 14.49.

34.3HNOn =

63

2.25= 0.4

2NOn =2

3× 0.4

2NOW =2

3× 0.4 × 46 = 27.6 gm

35. Using POSe

2 ×2Fn = 6 ×

6uFn

2Fn =352

1023 3

Molecules of F2

=352

1023 3× N

A= 1 × 1019

36. Ag + HNO3 AgNO3 + NaCl Ag Cl

108

x

108

x=

5.143

87.2

x =5.143

87.2× 108 = 2.16 gm.

37. POAC on Cl

CaCl2 AgCl

2 ×2CaCln =

5.143

35.14

2 ×111

w= 0.1

w = 5.55 gm.



38. NO +2

1O2 NO2

4.2 3.2

Mole1

14.0

5.0

1.0

0.14 0.2

NO is LR So

2NOn = 0.14

2NOw = 0.14 × 46 = 6.44 gm.

39. C H3 8

O23CO2

20 60 ml

CH4

O2CO2

x x

COO2 CO2

60 – x 80 – x

2COv = 60 + x – 180 – x = 140 ml.

40.ON

CO

2

2

w

w=

5

2

ON

CO

2

2

n

n=

544

442

=

5

2

41. urea NH – C – NH22

O

% of N =60

142× 100 = 46.66

42. Mass of C = 0.5 × 6 × 12 = 36 gm.

43. Mg +2

1O2 MgO

24

2.1= 1 × nMgO

nMgO = 0.5



44. Attleast one sulphur will be there in 1 molecule

3.4 = 100M

32

M = 941

45. % of N = %14100142132

66

46. 2Fe(NO3)

3+ 3Na

2CO

3 Fe

2(CO

3)

3+ 6NaNO

3

2

5.2

3

6.3

1.25 1.2

Na2CO

3is LR

% yield =6.32

3.6

× 100 = 87.5%

47. 2O2

3 O

3

4.22

2.67= 3mole 2 mole

mole3.015.02n3O

gm4.14483.0w3O

49. EMn

EM

massMolecular

massEmpirical

n

1

78

13

n = 6

MF = (CH)6 = C6H6

50. C(s) + 2Cl2 CCl4

1

mole3

2

mole2

4CCln = 1 mole = 12 0+ 1 + 35.5 × 4 = 154 gm.

51. 2SO2 (g) + O2 (g) + 2H2O () 2H2SO4

2

6.5

2

8.4

SO2 is LR

Mole of H2SO4 = 5.6



52. 2H2 (g) + O2 (g) 2H2O

50 mole 3.125

O2 is LR

OH2n = 2 × 3.125

OH2w = 18 × 2 × 3.125 = 112.5 gm

53. H2SO4 + Ca(OH)2 CaSO4 + 2H2O

0.5 0.2

Ca(OH)2 is LR

4CaSOn = 0.2

54. A + 2B C

1

5

2

8

B is LR

Moles of C = 4

55. 4A + 2B + 3C AuB2C3

4

1

2

6.0

3

72.0

0.25 0.3 0.24

C is LR

324 CBAn =3

72.0= 0.24

56. Zn + H2SO

4 ZnSO

4+ H

2

0.1 mole22400

224= 0.01 mole

wZn

= 0.1 0+ 65 = 0.65

57. 3BaCl2 + 2Na3PO4 24Cl POB

3

3

5.0= 1.66

2

2.0= 0.1

243 POBan = 0.1 mole



58. 1 gm H2O is present in 1 ml of water

M =18

1×

1

1000= 55.5 M

59. M =v

n NaOH× 100

1 =25040

w

× 100

w = 1 gm

60. M =180

36×

400

1000= 0.5 M

61. 15 gm is present in 100 ml solutions

d =V

M= Msolution = 1.1 × 100 = 110 gm

Mass of solvent = 110 – 15 = 95 gm

m =9598

15

× 1000 = 1.6 m

62. Conrains =1000

10× 106 = 104 ppm

63. Molarity =5.58

85.5×

200

1000= 0.54

64. C6H5CO2H + NaOH C6H5CO2Na + H2O

0.1 mole 0.1 mole = 4 gm

65. Molar concentration.

66. OH2x =

46

414

18

18

18/18

=

91

1

= 0.1

67. M =M

dx10=

98

9884.110 = 18.4 M



68. BaCl2 + H2SO4 BaSO410 mm 20 mm

LR 10 mm

69. Msolute =Total

solute

n

n=

80

20= 0.25

70. M1V1 + M2V2 = M3V3

M3 =5.5

5.0315.2 = 0.73

71. Mass of solution = 1.9 × 2000 = 3800 gm

Mass of acid = 0.98 × 3800 = 3724 gm

72. 2HCl + Ba(OH)2 BaCl2 + 2H2O

1 3

LR

mm OH– = 2 × 2.5

molerity of OH– =50

5.22= 0.1 M

73.32CONaM =

106

3.25×

250

1000= 0.955 M

NaM = 1.9 M

23CO

M = 0.955 M

74. NaM =

3NaNOH

23

101070 33

=85

10w 3

w = 12.934 gm

75. Number of ion = 5 × 0.6 × 6 × 1023 = 1.8 × 1024

76. OH2M =

18.0

360×

1000

1000= 20 M



77. 3 mole of NaOH in 1000 gm of water

XNaOH

=

18

10003

3

= 0.5

OH2X = 0.95

78. H2SO

4+ 2NaOH Na

2SO

4+ 2H

2O

X 8

4 = M ×42SOHV

42SOHV =1.0

4= 40 ml

79. M =342

171×

1000

1000= 0.5 M

80. AgNO3 + NaCl AgCl + NaNO30.1 × v 0.2 × v

3NO

M =v2

v1.0 = 0.05 M

81. HCl + NaOH

20 × v = 19.85 × 0.01

v = 0.0099 ml

82. Specific gravity (in terms of H) = Hydrogenofdensity

Gasofdensity

=

11200/1

1000/248.1= 13.98

83. H2SO4 required = 1 × 0.1 = 0.1 moles = 9.8 gr.

from information 98 gr. H2SO4 is present in = 100 gr. solution

=8.1

100mL .solution

9.8 gm H2SO4 is present in =98

8.9

8.1

100 mL

= 5.55 mL.



84. 1 gm ion ofAl3+ = 1 molAl3+ = NA ions

charge = NA × (3e) coulomb

85. Mav =2

130161 = 23 g/mol

d =4.22

23= 1.03 g/L

86. Mole fraction of H2 =

30

w

2

w

2/w

=

16

15

from dalton's Law,Total

H

P

P2

=2HX =

16

15

87. CS2 C S

6 gm 32 gm

CO2 C O

12 gm 32 gm

6 gm 16 gm

with fixed mass of C (= 6 gm), ratio of weight of S & O

O

S

w

w=

16

32= 2.

in SO2 S O

10 gm 10 gm

O

S

'w

'w=

10

10= 1.

O

S

w

w= 2 ×

O

S

'w

'w

Law of raciprocal proportions hold.

88. Volume of H2=

2

1× 22.4 = 11.2 L

89. d = 10.2 g/c.c.

V =3

4×

3

2

1

c.c.



weight of ball = V × d

number of Fe atoms =3

4× ×

3

2

1

×

100

84×

56

1× N

A

= 4.8 × 1022 atoms

90. Under same condition of tempanol pressure equal volume of different gases contans equal molecular of

each gas.

91. Under same condition of temperature and pressure equal volume of different gases contains same

numberof molecules.

2Hn = nHe

atoms of H2 = 2 × atoms of He

92. In both cases

Al + 3H+ Al3+ +2

3H2

equal moles of H2 will be evolved in both cases.

93. X Y

mole ratio a : b

mole fraction (Let) x : 1 – x

therefore mole ratio b : a

mole fraction 1 – x : x

Case I

Mav = 16 x + 28 (1 – x) = 20

x =12

8=

3

2

Case II

Mav = 16 (1 – x) + 28 x

= 16 ×3

1+ 28

3

2=

3

72= 24 g/mol ]

94. 12C + 11 H2 +2

11O2 C12H22O11

moles12

84= 6

2

12= 6

4.22

56= 2.5

forlimiting12

6

11

6

2/11

5.2=

11

5

reagent Limittign reagent O2



mole of sucrose produced =11

2× 2.5 =

11

5

mass of sucrose =11

5× 342 = 155.5

95. A4+ B2–

formula A B2

96. In B

with 32 part of X, mass of Y = 84

In C

with 32 part of X, mass of Y = 84 ×3

5= 140

with 16 part of X, mass of Y =2

140= 70 gm.

97. C2H5OH 2CO2

1 mole 2 mole

= 44 × 2 = 88 gm.

98. NH3 + O2 NO + H2O unbalanced

4NH3 + 5O2 4NO + 6H2O balanced

moles of NH3 (given) =17

8.6= 0.4

moles of O2 required =4

5× 0.4 = 0.5

99. weight of C–14 in 12 gm carbon = 12 ×100

2= 0.24 gm.

moles of C–14 =14

24.0

number of C–14 =14

24.0× 6.022 × 1023

= 1.032 × 10–22

100. Moles of H2SO4 left =98

1098 3– 23

20

1002.6

1001.3

= 0.5 × 10–3

101. dHg

= 13.6 g/cc

Let 'a' is edge length

volume = a3

mass of 1 mole = a3 × d × NA

= 200

a = 2.9 × 10–8 cm



102. Let volume of one molecule of H2O = V

weight of 1 mole = V × d × NA = 18 gm.

V × 1 × NA = 18

V = 2310022.6

18

= 3 × 10–23 cm3

103. CO2 + 2NaOH Na2CO3 + H2O

moles of NaOH =40

12= 0.3

moles of CO2 =2

3.0= 0.15

C6H12O6 + 6O2 6O2 + 6H2O

moles of C6H12O6 =6

15.0

weight of glucos =6

15.0× 180 = 4.5 gm.

104. moles ofA= mole ofAO

M

35.1=

16M

88.1

M = 40.75 g/mol

105. moles of H2O = 5 × moles of (CuSO4 . 5H2O)

= 5 ×5.249

1000

mass of H2O = 5 ×5.249

1000× 18

= 360.7 gm.

106. CO + CO2 CO2

0.8 mole 0.8 mole

CO2 + 2 NaOH Na2CO3 + H2O

moles of NaOH required = 0.8 × 2 = 1.6

extra NaOH required = 1.6 –40

40= 0.6 mole

107. moles in container =44

4.4+

4.22

24.2= 0.2 moles

molecules = 0.2 × 6.022 × 1023 = 1.204 × 1023



108. CaCO3 CaO + CO2

weight of CaCO3 in sample = 200 ×100

50= 100

moles of pure CaCO3 =100

100= 1

moles of CO2 = 1

volume (at NTP) = 22.4 L

109. F = 3.0 × 1024

noumber of formula unit ofAlF3 =3

F=

3

103 24

= 1.0 × 1024

110. 2 NaOH + H2SO4 Na2SO4 + H2O

mole 1 1/2

weight of H2SO4 (pure) required = 98 ×2

1= 49 gm

weight of H2SO4 (70 % pure) required = 49 ×70

100= 70 gm.

111. Weight of 1.425 L iso octane = 1425 × 0.8 = 1140 gm.

moles = 18128

1140

= 10

C8H18 + 12.5O2 8CO2 + 9H2O

moles of O2 required = 12.5 × 10 = 125 mole

112.8

1S

8+ O

2 SO

2(g)

weight of pure S = 200 ×100

80= 160 gm/

moles of S8=

832

160

moles of oxygen =832

160

× 8 = 5

volume of O2= 5 × 22.4 L

volume of air = 5 × 22.4 ×21

100= 533.3 L



113. NO2 NOmoles x 1 – x

46x + 30 (1 – x) = 34

x =4

1

mole percentage = 25 %

114. CO2 + C 2COvolume of CO2 = 26 c.c. volume of CO = 52 c.c.

115. Weight of NaI = 3 ×100

5.0= 1.5 + 10–2

moles of NaI = 3 ×100

5.0= 10–4

number of I– ions = 10–4 × 6.022 × 1023

= 6.022 × 1019

116. C2H4O2 + 2O2 2CO2 + 2H2Oformaximumenrgymole ratio should be 1 : 2 let moles of C2H4O2 = x mole of O2 = 2x 60 x + 32 × 2x = 620

x = 5 moles of CO2 produced = 5 × 2 = 10 mass of CO2 = 10 × 44 = 440 g/mole

117. P4S3 + 8O2 P4O10 + 3SO2

weight 440 gm 384 gm

moles220

440= 2

32

384= 12

for limiting reagent =1

2

8

12 O2 is limiting reagent

mass of P4O10 produced =8

1× 12 × (31 × 4 + 10 × 16) = 426 gm

118. In M3O4 oxygen percentage is 27.6 %

with 27.6 gm oxygen weight of metal conbined = 72.4 gm

with 16 × 4 gm oxygen weight of metal combined =6.27

4.72× 16 × 4 = 168

Atomic weight of metal =3

168= 56 g/mol

formula of 2nd oxide



M O

weight % 70 30

moles 1.25 1.875

mole ratio 1 : 1.5 2 : 3

formula M2O3

119. Ag2SO4 + 2NaCl 2 AgCl + Na2SO4moles 2 × 2 = 4 4 × 1 = 4 0after ppt 2 0 2

Agn = 2 × 2 = 4

24SO

n = 2 × 1 + 2 × 1 = 4

Nan = 2 × 2 = 4

total moles of ions = 12volume after mixing 4 + 2 = 6 L

total of ions =6

12= 2 M

120. [Na+] + [Ca2+] =100

60[Cl–]

i.e. Nan + 2Ca

n =100

60 Cln

no option match with requirement.

121. Ca(OH)2 + CO2 CaCO350 × 5 = 25 milimoles

25milimoles formed.CaCO2 + 2HCl CaCl2 + H2O + CO2

0.1 N = 0.1 M

2 ×3CaCOn = nHCl

2 × 25 = (0.1) × VV = 500 mL

EXERCISE-2

1. n =32

4+

2

2=

8

9

P =V

nRT=

1

2730821.08/9= 25.215 atm

2. M M(NO2)

2

M

12= 23214M

6.39

(POAC for metal)

M = 40 (error : use 39.6 at place of 14.8 g)



3. BaCO3

BaO + CO2

3COn =3BaCOn = 16312137

85.9

= 0.05

2COV = 0.05 × 22.4 = 1.12 L

error : Volume of CO2at STP

4. M1

V1

+ M2

V2

= MV

2.5 × 1 + 3 × 0.5 = (M) × (5.5)

M = 0.73 M

5. Gram molecularmass : mass of one mole of molecule.

6. N2

+ 3H2

2NH3

30 30 0

100 % yield 20 0 20

in case of 50 % yield 20+2

1×10 0+

2

3×10 10 L

error : Option 4 10 L NH3, 25 L N

2, 15 LH

2

7. Na4

[Fe (CN)6]

number of Na atom in 2 mole

= 2 × 4 × 6 × 1023 = 48 × 1023

8. BaCl2+ H

2S BaS + 2 HCl

SH2n = n

BaS=

32137

69.1

= 0.01

SH2m = 0.01 × 34 = 0.34 g

9. H2O (18 g H

2O has 16 gm O)

weight of O =18

16× 72 = 64 kg

10. CaCO

number of proton in one mole (100 gm) = 20 + 6 + 8 × 3 = 50

number of protons in 10 gm =100

50× 10 × N

A= 5 × 6.022 × 1023

11. CuSO4

. 5 H2O

number of atoms = 21



12. X =1612

270

= 5

13. O2

: N2

weight 1 : 4

mole32

1:

28

4

7 : 32

14. Ag2

CO3

2Ag + CO2

+2

1O

2

fromstiochiometry 276 gm 2 × 108 = 216 gm

with 2.76 gm 2.16 gm

15. Weight of one molecule C60

H12

= 60 × 12 + 122

= 842 amu

= 842 × 1.66 × 10–24 gm

= 1.4 × 10–2 gm

16. Moles of H2O =

18

34

17. PbO + 2HCl PCl2

6.5 3.2

molea168.2

5.6

= 0.029

5.36

2.3= 0.087

PbO is limiting reagent.

18. C3H

8+ 5O

2 3CO

2+ 4H

2O

with 1 L C3H

8, 5 L O

2is required.

19. C + O2

CO2

12 gm

= 1 mole 1 mole = 22.4 L

20. 2H2

+ O2

2H2O

weight 10 64

moles 5 2

O2is limitingreagent

moles of H2O formed = 2 × 2 = 4.

21. weight mole mole ratio

C 24 2 1

H 4 4 2

O 32 2 1

emperical formula = CH2O



22. Number of molecules =4.22

1012.1 7× 6.02 × 1023 = 3.01 × 1012

23. mass of 1 mole e–

= 9.1 × 10–31 × 6.022 × 1023 kg

= 0.55 mg

24. M2O (weight of M = 56 × 2 = 112 gm

weight of O = 16 x)

112

x16=

8.2

8.24

x = 3

25. Number of atoms =17

25.4× 4 × 6.022 × 1023

= 6.022 × 1023

26. Number of moles =.W.M

.wtGiven

27. 1 molal solution

1 mole in 1000 gm of H2O

mole fraction =

18

10001

1

= 0.0177

EXERCISE-3

1. mass number of oxygen is 16.

2. Mass remains constant but moles can be changed.

3. 2 Fe2O

3+ 3C 4Fe + 3CO

2

4. We cann't say reaction will be always 100 % complete.

5. Molality of pure ethanol is lesser because of higher M.W. of ethanol then pure water.

Bansal Quick Review Table

Instruction to fill(A) Write down the Question Number you are unable to solve in columnAbelow, by Pen.(B) After discussing the Questions written in columnAwith faculties, strikes off them in the manner

so that you can see at the time of Revision also, to solve these questions again.(C) Write down the Question Number you feel are important or good in the column B.

COLUMN : A COLUMN : B

Exercise I

Exercise II

Exercise III

EXERCISE NO. Question I am unable tosolve in first attempt

Good / Importantquestions

Exercise IV

Advantages

1. It is advised to the students that they should prepare a question bank for the revision as it is very difficultto solve all the questions at the time of revision.

2. Using above index you can prepare and maintain the questions for your revision.

7 inch

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