Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018)

Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018) Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018) Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018) Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018) Preview A Complete Resource Book in Chemistry for JEE Main 2019 by Atul Singhal (2018)

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About Pearson

A.K Singhal U.K Singhal

A Complete Resource Book in

chemistry

my grandparents, parents and teachers

Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128

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Preface vii

Acknowledgements viii

Chapter 1 Basics of Chemistry 1.1–1.32

Chapter 2 Solid State 2.1–2.34

Chapter 3 Gaseous State 3.1–3.36

Chapter 4 Atomic Structure 4.1–4.38

Chapter 5 Solutions 5.1–5.38

Chapter 6 Energetics 6.1–6.40

Chapter 7 Chemical Equilibrium 7.1–7.42

Chapter 8 Ionic Equilibrium 8.1–8.54

Chapter 9 Redox Reactions and Electrochemistry 9.1–9.50

Chapter 10 Chemical Kinetics 10.1–10.46

Chapter 11 Surface Chemistry 11.1–11.34

Chapter 12 Periodic Properties 12.1–12.26

Chapter 13 Chemical Bonding 13.1–13.38

Chapter 14 Chemistry of Representive Elements 14.1–14.26

Chapter 15 Chemistry of Non-Metals I 15.1–15.30

Chapter 16 Chemistry of Non-Metals II 16.1–16.60

Chapter 17 Chemistry of Lighter Elements 17.1–17.28

Chapter 18 Chemistry of Heavier Elements (Metallurgy) 18.1–18.34

Chapter 19 Transition Metals Including Lanthanides and Actinides 19.1–19.26

Chapter 20 Coordination Compounds 20.1–20.40

Chapter 21 Nuclear Chemistry 21.1–21.28

Chapter 22 Purifi cation and Characterization of Carbon Compounds 22.1–22.26

Contents

JEE Mains 2018 Paper xi

JEE Mains 2017 Paper xviii

Chapter 23 General Organic Chemistry I 23.1–23.54

Chapter 24 General Organic Chemistry II 24.1–24.52

Chapter 25 Hydrocarbons and Petroleum 25.1–25.58

Chapter 26 Organic Compounds with Functional Groups Containing Halogens (X) 26.1–26.36

Chapter 27 Alcohol Phenol Ether 27.1–27.56

Chapter 28 Organic Compounds Containing Oxygen-II 28.1–28.70

Chapter 29 Organic Compounds with Functional Groups Containing Nitrogen 29.1–29.44

Chapter 30 Polymers 30.1–30.20

Chapter 31 Biomolecules and Biological Processes 31.1–31.40

Chapter 32 Chemistry in Everyday Life 32.1–32.22

Chapter 33 Environment Chemistry 33.1–33.16

Chapter 34 Practical Chemistry 34.1–34.34

About the Series

A Complete Resource Book for JEE Main series is a must-have resource for students preparing for JEE Main examination

There are three separate books on Physics, Chemistry, and Mathematics; the main objective of this series is to strengthen

the fundamental concepts and prepare students for various engineering entrance examinations It provides class-tested

course material and numerical applications that will supplement any ready material available as student resource

To ensure high level of accuracy and practicality, this series has been authored by highly qualifi ed and experienced

faculties for all three titles

About the Book

A Complete Resource Book in Chemistry for JEE Main 2018 is an authentic book for all the aspirants preparing for the

joint entrance examination (JEE) This title is designed as per the latest JEE Main syllabus, where the important topics

are covered in 34 chapters It has been structured in user friendly approach such that each chapter begins with topic-wise

theory, followed by suffi cient solved examples and then practice questions

The chapter end exercises are structured in line with JEE questions; with ample number of questions on single choice

correct question (SCQ), multiple-type correct questions (MCQ), assertion and reasoning, column matching, passage based

and integer type questions are included for extensive practice Previous 15 years’ questions of JEE Main and AIEEE are also

added in every chapter Hints and Solutions at the end of every chapter will help the students to evaluate their concepts and

numerical applications Because of its comprehensive and in-depth approach, it will be especially helpful for those students

who prefers self-study than going for any classroom teaching

Series Features

• Complete coverage of topics along with ample number of solved examples.

• Large variety of practice problems with complete solutions.

• Chapter-wise Previous 15 years’ AIEEE/JEE Main questions.

• Fully solved JEE Main 2017 questions

• 5 Mock Tests based on JEE Main pattern in the book.

• 5 Free Online Mock Tests as per the recent JEE Main pattern.

Despite of our best eff orts, some errors may have crept into the book Constructive comments and suggestions to

further improve the book are welcome and shall be acknowledged gratefully Kindly share your feedback with me at

singhal.atul50@yahoo.com or singhal.atul1974@gmail.com.

A.K Singhal

The contentment and ecstasy that accompany the successful completion of any work would remain essentially incomplete

if I fail to mention the people whose constant guidance and support has encouraged me

I am grateful to all my reverend teachers, especially, late J.K Mishra, D.K Rastogi, late A.K Rastogi and my

honour-able guide, S.K Agarwala Their knowledge and wisdom has continued to assist me to present in this work

I am thankful to my colleagues and friends, Deepak Bhatia, Vikas Kaushik, A.R Khan, Vipul Agarwal, Ankit Arora

(ASO Motion, Kota), Manoj Singhal, Yogesh Sharma, (Director, AVI), Vijay Arora, (Director, Dronacharya), Rajneesh

Shukla (Allen, Kota), Anupam Srivastav, Rajeev Jain (M V N), Sandeep Singhal, Chandan Kumar (Mentor, Patna),

P.S Rana (Vidya Mandir, Faridabad)

I am indebted to my father, B.K Singhal, mother Usha Singhal, brothers, Amit Singhal and Katar Singh, and sister,

Ambika, who have been my motivation at every step Their never-ending aff ection has provided me with moral support and

encouragement while writing this book

Last but not the least, I wish to express my deepest gratitude to my wife Urmila and my little, —but witty beyond her

years, daughters Khushi and Shanvi who always supported me during my work

A.K Singhal

Acknowledgements

18 Chemistry of Heavier Elements (Metallurgy) – 1 – – – 2 – 1 1 2 – –

19 Transition Metals Including Lanthanides and Actinides – 1 2 – 2 – 2 – 1 1 – –

Purification and Characterization of Carbon Compounds

26

Organic Compounds with Functional Groups Containing Halogens (X)

28 Organic Compounds Containing Oxygen-II – – 1 – 2 1 1 1 1 1 1 1

29

Organic Compounds with Functional Groups Containing Nitrogen

31 Biomolecules and Biological Processes – 1 1 1 1 2 1 1 1 1 1

Total No of Questions 33 38 30 28 30 30 29 22 21 34 35 30

1 For 1 molal aqueous solution of the following

com-pounds, which one will show the highest freezing

2 Hydrogen peroxide oxidizes [Fe(CN) ]6 4−to[Fe(CN )]6 3−

in acidic medium but reduces [Fe(CN) ]6 3− to

[Fe(CN) ]6 4− in alkaline medium The other products

formed are, respectively:

(A) H O and (H O OH )2 2 + −

(B) (H O + O2 2) and H2O

(C) (H O2 +O2) and (H O2 +OH−)

(D) H O and (H O O )2 2 + 2

3 Which of the following compounds will be suitable for

Kjeldahl’s method for nitrogen estimation?

4 Glucose on prolonged heating with HI gives:

(A) 6-iodohexanal (B) n-Hexane

(C) l-Hexene (D) Hexanoic acid

5 An alkali is titrated against an acid with methyl

orange as indicator, which of the following is a correct

combination?

Base Acid End point

(A) Strong Strong Pink to colourless

(B) Weak Strong Colourless to pink

(C) Strong Strong Pinkish red to yellow

(D) Weak Strong Yellow to pinkish red

6 The predominant form of histamine present in human

(B)

NH2

N

N H

8 Which of the following lines correctly show the

tem-perature dependence of equilibrium constant K, for anexothermic reaction?

1 T(K)

(A) A and D (B) A and B (C) B and C (D) C and D

9 How long (approximate) should water be

electro-lyzed by passing through 100 amperes current so that

the oxygen released can completely burn 27.66 g of

(III) Only one isomer is produced if the reactant

com-plex ion is a trans-isomer

(IV) Only one isomer is produced if the reactant

com-plex ion is a cis-isomer

The correct statements are:

(A) (II) and (IV) (B) (I) and (II)

(C) (I) and (III) (D) (III) and (IV)

11 Phenol reacts with methyl chloroformate in the

pres-ence of NaOH to form product A A reacts with Br2 to

form product B A and B are respectively:

(A)

OH

O OCH3

OH and

O Br

OCH3

(B)

OH

O OCH3

OH and

O Br

O and

(D)

O O

O Br

O and

12 An aqueous solution contains an unknown

concentra-tion of Ba2+ When 50 mL of a 1 M solution of Na2SO4

is added, BaSO4 just begins to precipitate The final

volume is 500 mL The solubility product of BaSO4 is

1 × 10-10 What is the original concentration of Ba2+?

(A) 1.0 × 10-10 M (B) 5 × 10-9 M

(C) 2 × 10-9 M (D) 1.1 × 10-9 M

13 At 518°C the rate of decomposition of a sample of

gas-eous acetaldehyde initially at a pressure of 363 Torr,was 1.00 Torr s-1 when 5% had reacted and 0.5 Torr

s-1 when 33% had reacted The order of the reaction is:

14 The combustion of benzene (1) gives CO2 (g) and

H2O (l) Given that heat of combustion of benzene atconstant volume is -3263.9 kJ mol-1 at 25°C; heat ofcombustion (in kJ mol-1) of benzene at constant pres-

sure will be: (R = 8.314 JK-1 mol-1)(A) -3267.6 (B) 4152.6

15 The ratio of mass percent of C and H of an organic

compound (CxHyOz) is 6 : 1 If one molecule of theabove compound (CxHyOz) contains half as much oxy-gen as required to burn one molecule of compound

CxHy completely to CO2 and H2O, then empirical mula of compound CxHyOz is:

for-(A) C2H4O3 (B) C3H6O3(C) C2H4O (D) C3H4O2

16 The trans-alkenes are formed by the reduction of

alkynes with:

(A) Sn-HCl (B) H2-Pd/C, BaSO4(C) NaBH4 (D) Na/liq NH3

17 Which of the following are Lewis acids?

(A) BCl3 and AlCl3 (B) PH3 and BCl3(C) AlCl3 and SiCl4 (D) PH3 and SiCl4

18 When metal ‘M’ is treated with NaOH, a white

gelat-inous precipitate ‘X’ is obtained, which is soluble inexcess of NaOH Compound ‘X’ when heated stronglygives an oxide which is used in chromatography as anabsorbent The metal ‘M’ is:

19 According to molecular orbital theory, which of the

following will not be a viable molecule?

21 Phenol on treatment with CO2 in the presence of NaOH

followed by acidification produces compound X as the

major product X on treatment with (CH3CO)2O in the

presence of catalytic amount of H2SO4 produces:

(A)

O O

(C)

O O

CO2H

CH3

(D)

CH3OH

23 Which type of ‘defect’ has the presence of cations in

the interstitial sites?

(A) Metal deficiency defect (B) Schottky defect

(C) Vacancy defect (D) Frenkel defect

24 The major product of the following reaction is:

Br

MeOH NaOMe

(A)

OMe

25 The compound that does not produce nitrogen gas by

the thermal decomposition is:

(A) (NH4)2SO4 (B) Ba(N3)2 (C) (NH4)2Cr2O7 (D) NH4NO2

26 An aqueous solution contains 0.10 M H2S and 0.20 M HCl If the equilibrium constant for the formation of

HS- from H2S is1 0 10 × − 7 and that of S2− from HSions is1 2 10 × − 13 then the concentration of S2− ions in aqueous solution is:

(A) 5 10× − 19 (B) 5 10× − 8 (C) 3 10× − 20 (D) 6 10× − 21

27 The oxidation states of Cr in [Cr(H O) Cl2 6] 3,[Cr(C H )6 6 2]

Cr(H O) Cl2 6 3 Cr(C H )6 6 2

[ ] ,[ ] and K Cr(CN) (O) (O )(NH )2[ 2 2 2 3 ] re

-spectively are:

(A) +3, 0 and +4 (B) +3, +4 and +6 (C) +3, +2 and +4 (D) +3, 0 and +6

28 The recommended concentration of fluoride ion

in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting3Ca (PO ) Ca(OH)3 4 2 2

(A) 3 Ca(OH) CaF{ 2} 2

(B) [CaF2]

(C) [3(CaF2).Ca(OH)2]

(D) [3Ca (PO ) CaF3 4 2 2]

29 Which of the following salts is the most basic in

1 As DTf ∝ mi Here m is same so decide by using i.

The complex giving least number of ions will have lowest

depression in freezing point and therefore highest freezing

point Hence, option (A) is correct (Van’t Hoff factor = 1)

Hence, the correct option is (A).

2 In acidic medium, H O 2 2 acts as an oxidant as follows

Oxidation number decreases

Oxidation number increases

In alkaline medium, H O 2 2 acts as reducing agent as follows

Oxidation number increases

Oxidation number decreases

Hence, the correct option is (D).

3 Kjeldahl’s method cannot be not used in the case of nitro,

azo compounds and also to the compounds containing

nitrogen in the ring, e.g., Pyridine Hence, here only

C H NH6 5− 2 is suitable for the test.

Hence, the correct option is (C).

4 Glucose on heating with HI gives n-hexane as followed

Hence, the correct option is (B).

5 As methyl orange is weak organic base So it is used in the

titration of weak base (NH4OH) vs strong acid (HCl)

Unionized form (Yellow) Ionized form(Red)

unproton-Hence, the correct option is (A).

Answer keys

1 (A) 2 (D) 3 (C) 4 (B) 5 (D) 6 (A) 7 (D) 8 (B) 9 (D) 10 (C)

11 (D) 12 (D) 13 (B) 14 (A) 15 (A) 16 (D) 17 (A) or (C) 18 (D) 19 (A) 20 (A)

21 (B) 22 (D) 23 (D) 24 (C) 25 (A) 26 (C) 27 (D) 28 (D) 29 (C) 30 (D)

Hints and solutions

7 Order of basicity in increasing order is as follows

Here (III) is most basic because its conjugate acid is

stabi-lized by equivalent resonance.

Out of (I) and (IV), (IV) is more basic due to + I effect of

-CH 3 group (II) is less basic than (I) and (IV) because N

atom is sp 2 (more s%) hybridized.

Hence, the correct option is (D).

As according to the balanced equation:

Here 27 66 g B H 2 6 , i.e., 1 mole B H 2 6 requires 3 mole of

O2 Here this oxygen is produced by electrolysis of water as

follows.

2 H O 2  →4F 2 H 2 + O 2

As 1 mole O2 is produced by 4 F charge

So 3 mole O2 will be produced by 12 F charge

trans-isomer mer-isomer (only)

Br Br + Br

NH3

NH3

H3N Co

Br Br

NH3

NH3

Br

H3N Co

Br −

Br −

Hence, the correct option is (C).

11 The sequence of the reaction is as follows

OPh

O O

O CH3

OMe O

Br (Major product) (B)

o, p-directing group

(A) C

For BaSO4 Ksp is given as

K (BaSO ) = [Ba ][SO ] sp 4 +2 4 −

10 − 10 = [Ba ] + 2 × 0 1

[Ba ]+2 = 10 M (in 500 mL solution)−9

[SO ] 4 − in 500 mL solution will be

.

Hence, the correct option is (A).

15 As ratio of mass percent of C and H in CxHyOz is 6 : 1.

So, ratio of mole percent of C and H in CxHyOz will be 1 : 2.

Hence, x : y = 1 : 2, which is possible in options (A), (B) and (C).

Now oxygen needed to burn CxHy

Hence, the correct option is (A) (C 2 H4O3).

16 Na/liq NH3 reduces alkynes into trans alkene (trans or anti

addition).

R C C R

C = C R

R Na/liq NH3

trans alkene

H

H Hence, the correct option is (D).

17 Here BCl3 and AlCl3 are Lewis acids as both ‘B’ and ‘Al’ has vacant p-orbitals and are e- deficient SiCl4 is also a Lewis acid as silicon atom has vacant 3d-orbital, so it can accept e- Hence, the correct option is (A) or (C).

18 Here metal is Al and the reactions are as follows

Hence, the correct option is (D).

19 Species Bond order

2 2 1 0 5. (exists)

As H2− has bond order as zero, so it cannot exist.

Hence, the correct option is (A).

20 It follows SN1 reaction as follows.

O

O

Δ HI

CH3

O

CH3

CH3H H

CH3

⊕

OH + +

21 Here Asprin is formed as follows

Hence, the correct option is (B).

22 KCl contains only ionic bond between K+ and Cl− ions

( K Cl + − ) while rest compounds have covalent bonds as

follows:

H H

H H

H H

Hence, the correct option is (D).

23 As in Frenkel defect, smaller ion (cations) displaces from its

actual lattice site into the interstitial sites.

Hence, the correct option is (D).

24 It follows E2 mechanism as follows

Br H

+ NaBr + MeOH

MeOH NaOMe

Hence, the correct option is (C).

25 Here (NH4)2SO4 an heating gives Ammonia while rest give N2.

Hence, the correct option is (C).

27 The oxidation states of Cr are 3, 0, 6, respectively.

Compound Oxidation states of Cr

28 3Ca (PO ) Ca(OH) F

3Ca (PO ) CaF 2OH

Hence, the correct option is (D).

29 CH3COOK is most basic among the given salts as it gives strong base KOH and weak acid CH3COOH.

Hence, the correct option is (C).

30 I3− has 3 lone pair e- as follows

I I− I Hence, the correct option is (D).

1 Which of the following compounds will give

signif-icant amount of meta product during mono-nitration

2 ∆U is equal to:

(A) Isochoric work (B) Isobaric work

(C) Adiabatic work (D) Isothermal work

3 The increasing order of the reactivity of the following

halides, for the SN1 reaction is:

(I) CH3CHCH2CH3

Cl

(II) CH3CH2CH2Cl

(III) p—H3CO—C6H4—CH2Cl

(A) (III) < (II) < (I) (B) (II) < (I) < (III)

(C) (I) < (III) < (II) (D) (II) < (III) < (I)

4 The radius of the second Bohr orbit for hydrogen atom

is:

(Plank’s constant, h = 6.6262 × 10-34 Js; mass of

electron = 9.1091 × 10-31 kg; charge of electron,

e = 1.60210 × 10-19 C; permittivity of vaccum, ∈0=

8.854185 × 10-12 kg-1 m-3 A2)

5 pKa of a weak acid (HA) and pKb of a weak base

(BOH) are 3.2 and 3.4, respectively The pH of their

salt (AB) solution is:

6 The formation of which of the following polymers

involves hydrolysis reaction?

(C) Nylon 6, 6 (D) Terylene

7 The most abundant elements by mass in the body of a

healthy human adult are:

Oxygen (61.4%); carbon (22.9%); hydrogen (10.0%);

and nitrogen (2.6%) The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is:

(C) 7.5 kg (D) 10 kg

8 Which of the following, upon treatment with

tert-BuONa followed by addition of bromine water, fails todecolourize the colour of bromine?

(A)

O Br

9 In the following reactions, ZnO is respectively acting

as a/an:

(a) ZnO + Na2O → Na2ZnO2(b) ZnO + CO2→ ZnCO3(A) Base and acid (B) Base and base(C) Acid and acid (D) Acid and base

10 Both lithium and magnesium display several similar

properties due to the diagonal relationship; however,the one which is incorrect is:

(A) Both form basic carbonates(B) Both form soluble bicarbonates(C) Both form nitrides

(D) Nitrates of both Li and Mg yield NO2 and O2 onheating

11 3-Methyl-pent-2-ene on reaction with HBr in presence

of peroxide forms an addition product The number ofpossible stereoisomers for the product is:

12 A metal crystallises in a face centred cubic structure.

If the edge length of its unit cell is ‘a’, the closestapproach between two atoms in metallic crystal will be:

R1 and R2 respectively at 300 K, then ln(k2/k1) is equal

to (R = 8.314 J mol-1K-1):

14 The correct sequence of reagents for the following

conversion will be:

15 The Tyndall effect is observed only when following

conditions are satisfied:

(a) The diameter of the dispersed particles is much

smaller than the wavelength of the light used

(b) The diameter of the dispersed particle is not much

smaller than the wavelength of the light used

(c) The refractive indices of the dispersed phase

and dispersion medium are almost similar in magnitude

(d) The refractive indices of the dispersed phase and

dispersion medium differ greatly in magnitude

(A) (a) and (d) (B) (b) and (d)

(C) (a) and (c) (D) (b) and (c)

16 Which of the following compounds will behave as a

reducing sugar in an aqueous KOH solution?

(A) CH2OH

OCOCH3HO

OH

(B) CH2OH

HO OH

(C) CH2OH

OCH3HO OH

(D)

CH2OH

OH OH OH

C(graphite)+2H g2( )→CH g4( ) will be:

(A) +74.8 kJ mol-1 (B) +144.0 kJ mol-1 (C) -74.8 kJ mol-1 (D) -144.0 kJ mol-1

18 Which of the following reactions is an example of a

redox reaction?

(A) XeF4+O F2 2→XeF6+O2 (B) XeF2+PF5→[XeF PF]+ 6− (C) XeF6+H O2 →XeOF4+ 2HF (D) XeF6+2H O2 →XeO F2 2+4HF

19 The products obtained when chlorine gas reacts with

cold and dilute aqueous NaOH are:

(A) ClO and ClO− 3− (B) ClO and ClO2− 3− (C) Cl and ClO− − (D) Cl and ClO− 2−

20 The major product obtained in the following reaction

is:

Δ BuOK

Br H

C6H5

C6H5

(A) ( )± C H CH O Bu CH CH H6 5 ( t )

2 6 5 (B) C H CH=CHC H6 5 6 5

21 Sodium salt of an organic acid ‘X’ produces

efferves-cence with conc H2SO4.‘X’ reacts with the acidified

aqueous CaCl2 solution to give a white precipitate

which decolouriszes acidic solution of KMnO4.‘X’ is:

(A) C6H5COONa (B) HCOONa

(C) CH3COONa (D) Na C O2 2 4

22 Which of the following species is not paramagnetic?

23 The freezing point of benzene decreases by 0.45°C

when 0.2 g of acetic acid is added to 20 g of benzene

If acetic acid associates to form a dimer in benzene,

percentage association of acetic acid in benzene will

25 On treatment of 100 ml of 0.1 M solution of COCl3

6H2O with excess AgNO3; 1.2 × 1022 ions are

precipi-tated The complex is:

(D) Only SO4−

28 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2 The molar mass of M2CO3 in g mol-1 is:

(D) O , F , Na , Mg− − + 2+

1 Nitration is carried out in presence of concentrated HNO3

and concentrated H2SO4 Here, aniline acts as base, in

pres-ence of H2SO4 its protonation takes place and anilinium ion

Anilinium ion is strongly deactivating group and meta

directing in nature so it gives meta nitration product in

sig-nificant amount (≈ 97%).

NH2

NO2

+conc.HNO3conc.H2SO4

Hence work involved in adiabatic process is at the expense

of change in internal energy of the system.

3 For SN1 reaction, reactivity is decided by ease of

dissocia-tion of alkyl halide

−

= 7+ (3.2)1 (3.4)

2

1 2

−

= 6 9

6 Formation of Nylon6 involves hydrolysis of its monomer

(caprolactum) in initial state as follows:

Hydrogen = 10.0% and Nitrogen = 2.6%

Total weight of person = 75 kg

Mass due to 1H is = 75 × 10 =

100 7 5 kg

1 H atoms are replaced by 2 H atoms.

Hence, mass gain by person = 7.5 kg

Answer keys

1 (C) 2 (C) 3 (B) 4 (D) 5 (B) 6 (A) 7 (C) 8 (A) 9 (D) 10 (A)

11 (D) 12 (D) 13 (D) 14 (A) 15 (B) 16 (A) 17 (C) 18 (A) 19 (C) 20 (B)

21 (D) 22 (B) 23 (D) 24 (D) 25 (D) 26 (B) 27 (C) 28 (B) 29 (A) 30 (A)

Hints and solutions

JEE Mains 2018 Papers xiii

8

Br

(fails to decolorize the colour of bromine) (no unsaturation)

tert-BuONa

O

O-tBu O

C6H5

(it decolorizes bromine solution)

to unsaturation)

tert-BuONa

Br O

(it decolorizes bromine solution due

(a) ZnO + Na2O → Na 2 ZnO2

acid base salt

(b) ZnO + CO2 → ZnCO 3

base acid salt

10 Mg can form basic carbonate like,

2

3

(4 stereo isomers possible due to 2 chiral centres as molecule is asymmetric )

H

(II)

CH3

Et H

H Br

H3C

CH3

CH3

H Br

H

CH3

H

Br H

H3C

(I) Et

CH3

Br H

H

(II) Et H

H Br

H

(IV)

CH3

Et H

Br H

(E E RT

(E E RT

2 1

CH3MgBr

[Ag(NH3)2]OH Tollens reagent

O

C O

CH3

CH3

H3C OH

HO

C

15 It is factual.

16 (i) Ester in presence of aqueous KOH solution give SNAE

reaction so following reaction takes place.

O—C—Me O

O CH2OHHOCH2

O

OH O

O CH2OHHOCH2

− + Me—C—O−

SN AE

Aq KOH

Reagent Tollen’s

Ring opening Hemiketal

O

OH

CH2OH HOCH2 O−

⊕ve silver mirror test

(ii) In above compound, in presence of aq KOH (SNAE)

reaction takes place and a-Hydroxy carbonyl compound

is formed which gives ⊕ve Tollen’s test So this compound

behaves as a reducing sugar in an aqueous KOH solution.

Here, xenon undergoes oxidation while oxygen undergoes

reduction (Redox Process)

19 Cl2 + 2OH − Cl + ClO − + H2O

[cold and dilute]

Hypochlorite

20 Elimination reaction is highly favoured if:

(a) Bulkier base is used

(b) Higher temperature is used

Hence in given reaction biomolecular ellimination reaction

provides major product.

C6H5

C6H5+ t BuOH + Br ⊕

molecular orbital (Paramagnetic)

CO(14e-) ⇒ No unpaired electron is present (Diamagnetic)

O2 (16) ⇒ Two unpaired electrons are present in ⋅π *

Molecular orbitals (Paramagnetic)

B2(10) ⇒ Two unpaired electros are present in π

bonding molecular orbitals (Paramagnetic)

23 In benzene acetic acid dimerises as follows:

(C)

N pyridine is aromatic (6πe - )

25 Moles of complex =Molarity volume ml× ( )

1000

=100 0 1× =

1000 0 01. . mole

Moles of ions precipitated with excess of AgNO3

.

= 0 02 moles

Number of Cl−present in ionization sphere =

Moleof ion precipitated with exess AgNO

Mole of complex

3 =0 02

0 0

11 =2

It means 2Cl− ions present in ionization sphere.

Hence, complex is [CO(H2O)5Cl]Cl2 H2O

26 DIBAL–H is electrophilic reducing agent which reduces

cynide, esters, lactone, amide, carboxylic acid into

corre-sponding aldehyde (partial reduction).

27 NO3− : The maximum limit of nitrate in drinking water can

be 50 ppm Excess nitrate in drinking water can cause

dis-eases like methemoglobinemia, etc.

SO4− : above 500 ppm of SO4− ion in drinking water causes

laxative effect otherwise at moderate levels it is normally

harmless.

F−: Above 2 ppm concentration of F− in drinking water

causes brown mottling of teeth.

Hence, the concentration given in question of SO 4 − and

NO 3 − in water is suitable for drinking but the concentration

of F− (i.e., 10 ppm) makes water unsuitable for drinking

Matter and its nature, Dalton’s atomic theory, Concept of atom, Molecule, Element, Compound, Precision,

Significant figures, SI units, Derived units, Laws of chemical combinations, Mole, Mass, Molecular mass,

Equivalent mass, Chemical equation, Stoichiometry of chemical equation and various levels of multiple-choice

questions

MATTER

Any species having mass and occupying space is known

as matter It can exist in the three physical states, namely,

solid, liquid and gas

Figure 1.1 Classification of Matter

Pencil, air, water, justify the physical states and are all

composed of matter

• At the bulk level or macroscopic level, we can

fur-ther classify matter as mixtures or pure substances

Mixture

A mixture is composed of two or more substances which

are known as its components or constituents (in any ratio)

The components of the mixture can be separated with the help of physical separation methods like filtration, crystal-lization, distillation

• A mixture is further classified into two categories—

homogeneous and heterogeneous

• In a homogeneous mixture all the components under-

go complete mixing forming a uniform composition

as, air or sugar solution

• In a heterogeneous mixture the composition formed due to the mixing of components is not entirely uniform like in the case of grains mixed with dust etc

Pure Substance

Pure substances have fixed compositions and their uents cannot be separated by using simple physical meth-ods of separation

constit-• A pure substance can be further classified into an ment or a compound

ele-• An element is composed of one type of particle

which could either be atoms or molecules Na, Cu,

Ag have only one type of atoms

• A compound is formed by the combination of two or

more atoms or different elements For example, H2O,

Dalton’s Atomic Theory

An atom is the smallest particle of an element which is

neutral in nature, retains all the properties of the element

and takes part in a chemical reaction The word atom was

introduced by Dalton (alamos means undivided)

The Dalton’s atomic theory was proposed by Dalton on

the basis of laws of chemical combination

Main assumptions

• Matter (of any type) is composed of atoms

• An atom is the smallest, fundamental, undivided

particle (Building block material)

• An atom can neither be created nor destroyed

• Atoms of an element have similar size, energy and

properties while atoms of different element differ in

these aspects

• Atoms combine in whole number ratios to form a

molecule, therefore, a molecule is the smallest

iden-tity that exists individually

Modern view about atom

According to the modem view:

• An atom is divisible into other smaller particles

which are known as subatomic particles It can also

combine in non-whole number ratio as in the case

of non- stoichiometric compounds (Berthollide

com-pounds) like Fe0.93O

• Atoms of same element also differ in mass and mass

related properties as in the case of isotopes

• Chemical reactions involve rearrangement of atoms

Molecule

The term molecule was introduced by Avogadro It is the

smallest particle (identity) of matter that can exist

inde-pendently and retains all the properties of the substance

Normally the diameter of the molecules is in the range of

4–20 Å and the molecular mass is between 2–1000

• In case of macromolecules, the diameter is in the

range of 50–250 Å and the molecular weight may be

in lakhs

Berzelius Hypothesis

According to the Berzelius hypothesis, “Equal volumes

of all the gases contain same number of atoms under the

similar conditions of temperature and pressure.”

This hypothesis on application to law of combining

volume confirms that atoms are divisible which is in

con-trary to Dalton’s theory

PHYSICAL QUANTITIES AND THEIR MEASUREMENTS

In order to describe and interpret the behaviour of ical species, we require not only chemical properties but also few physical properties Physical properties are mass, length, temperature time, electric current etc

Further, to express the measurement of any physical quantity we require its numerical value as well as its unit

Hence, the magnitude of a physical quantity can be given as Magnitude of physical quantity 5 Its numerical value 3 Unit

Precision and Accuracy

• The measurements are considered accurate when the age value of different measurements is closer to the actual value An individual measurement is considered more accurate when it differs slightly from the actual value

aver-• When the values of different measurements are close

to each other as well as to the average value, such measurements are called precise

• In fact, precision is simply the measurement of reproductability of an experiment

Significant Figures

These are some uncertainties in values during ment of matter In order to make accurate measurements

measure-we use these figures

The total number of digits in a number including the last digit with uncertain value is known as the number of significant figures, for example, 14.3256 6 0.0001 has six significant figures

Rules to determine significant numbers

• All non-zero digits as well as the zeros present between the non-zero digits are significant, for exam-ple, 6003 has four significant figures

• Zeros to the LHS of the first non-zero digit in a given number are not significant figures, for example, 0.00336 has only three significant figures

• In a number ending with zeros, if the zeros are ent at right of the decimal point then these zeros are also significant figures, for example, 33.600 has five significant figures

pres-• Zeros at the end of a number without a decimal are not counted as significant figures, for example,

12600 has just three significant figures

• The result of division or multiplication must be reported to the same number of significant figures

as possessed by the least precise term, for example, 3.331 3 0.011 5 0.036641 ≈ 0.037

• The result of subtraction or addition must be reported

to the same number of significant figures as

pos-sessed by the least precise term, for example, 5.1 1

7.21 1 8.008 5 20.318 ≈ 20.3

Rounding-off non-significant figures

Rounding-off non-significant figures means dropping of

the uncertain or non-significant digits in a number It is

possible as follows:

• If the rightmost digit to be rounded-off is 5, then

the preceding number is increased by one, for

exam-ple, 3.17 is rounded off to 3.2

• If the rightmost digit to be rounded-off is, 5, then the

preceding number is kept unchanged, for example,

5.12 is rounded off to 5.1

• If the rightmost digit to be rounded-off is equal to

5, the preceding number is kept as such in case of

an even value However, in case of an odd value it is

increased by one, for example, 4.45 is rounded-off to

4.4; 5.35 is rounded off to 5.4

Exponential notation or scientific notation

In case a number ends in zeros that are not to the right of

decimal point it is not essential that zeros are significant

For example, 290 has 2 or 3 significant figures and 19500

has 3, 4 or 5 significant figures

This confusion can be removed when the values are

expressed in terms of scientific notations, for example, 19500

can be written as 1.95 3 104 (3 significant figures), 1.950 3

104 (4 significant figures), 1.9500 3 104 (5 significant

fig-ures) In this kind of notation, every number can be written as

N 5 Number with non-zero digit

to the left of the decimal point

For example, 0.00069 can be expressed as 6.9 3 1024

2 2

5 kg m 2 1 sec 2 2 5 Pascal (Pa)

5 kg m 2 sec 2 2 5 Joule

 Plane angle (Radian, that is, ‘rad’)

 Solid angle (Steradian, that is, ‘str’)

LAWS OF CHEMICAL COMBINATIONS

Law of Conservation of Mass

• Law of conservation of mass was proposed by Lavoisier

in 1774

• It was verified by Landolt

• According to this law, “In a chemical change the

total mass of the products is equal to the total mass

of the reactants, that is, mass is neither created nor

destroyed.” For example, when a solution with

calcu-lated weight of AgNO3 and NaCl is mixed, white

pre-cipitates of AgCl are formed while NaNO3 remains

in solution The weight of the solution remains the

same before and after this experiment

• It is not applicable to nuclear reactions

Law of Constant Composition or Law of

Definite Proportion

• Law of constant composition was proposed by Proust

in 1779

• It was verified by Star and Richards

• According to this law, “A chemical compound always

contains same elements combined together in same

proportion by mass.” For example, NaCl extracted

from sea water or achieved from deposits will have

23 g Na and 35.5 g of chlorine in its one mole

• It is not applicable to non-stoichiometric compounds

like Fe0.93 O

Law of Multiple Proportion

• Law of multiple proportion was proposed by Dalton

in 1804

• It was verified by Berzilius

• According to this law, “Different weights of an ment that combine with a fixed weight of another element bear a simple whole number ratio.” For example, in case of CO, and CO2 weight of oxygen which combines with 12 g of carbon is in 1 : 2 ratio

ele-• It is applicable when same compound is prepared from different isotopes of an element For example,

H2O, D2O

Law of Reciprocal Proportion

• Law of reciprocal proportion was proposed by Richter in 1792

• It was verified by Star

• According to this law, “When two different elements undergo combination with same weight of a third element, the ratio in which they combine will either

be same or some simple multiple of the ratio in which they combine with each other.”

• It is also known as Law of equivalent proportion

which states “Elements always combine in terms of their equivalent weight.”

Law of Combining Volume

• Law of combining volume was proposed by Gay-Lussac

• It applies to gases

• According to this law, “When gases react with each other they bear a simple whole number ratio with one another as well as the product under conditions of same temperature and pressure.”

AVOGADRO’S LAW

• Avogadro’s law explains law of combining volumes

• According to this law “Under similar conditions

of temperature and pressure equal volume of gases contain equal number of molecules.”

• It is used in:

1 Deriving molecular formula of a gas

2 Determining atomicity of a gas

3 Deriving a relation Molecular mass 5 2 3 Vapour Density (M 5 2 3 V.D.)

4 Deriving the gram molecular volume

• Avogadro number (N0 or NA) 5 6.023 3 1023.

• Avogadro number of gas molecules occupies

22.4 litre or 22400 mL or cm3 volume at STP

• The number of molecules in 1 cm3 of a gas at STP is

equal to Loschmidt number that is, 2.68 3 1019

• Reciprocal of Avogadro number is known as

avogram

MOLE

• Mole is a unit which represents 6.023 3 1023

par-ticles, atoms, molecules or ions etc., irrespective of

their nature

• Mole is related to the mass of substance, the volume

of gaseous substance and the number of particles

• Volume of one mole of any gas is equal to 22.4 litres

or 22.4 dm3 at STP It is known as molar volume

• Mole 5W

M

5 Molar mass of substance (g.m.m.)Wt of substance in g _

Here, g.m.m 5 Gram molecular mass

Mole 5 Vol of substance in litre 22.4 L

Mole Concept: An Example

A mole of any substance (like N2) stands for:

• 6.023 3 1023 molecules of N2

• 2 3 6.023 3 1023 atoms of nitrogen

• 28 g of nitrogen

• 22.4 litre of N2 at STP

To find total number of identities

⇒ Total Number of Molecules = mole (n) × NA

⇒ Total Number of Atoms in = mole (n) × NA ×

No of atoms present in one molecule

⇒ Total Number of Electrons = mole (n) × NA ×

No of electrons in one molecule

⇒ Total Charge on Any Ion = mole (n) × NA × charge on

• The atomic mass of any element expressed in grams

is called g.a.m (gram atomic mass) or gram atom

• A gram atom has number of atoms of the element

Atomic mass 5 E 3 V Here, E 5 Equivalent weight

Average atomic mass

Molecular Mass

Molecular mass represents the total mass of a molecule, that is, number of times a molecule is heavier than 1

12weight of C–12 atom or 1

16 weight of an oxygen atom

• It is non-variable

Determination of molecular mass

Vapour density method

Mol mass 5 2 3 V.D

V.D 5 Volume at STP (in mL) W 3 22400 _

Here, W 5 Weight of substance in g V.D = Vapour Density

Graham’s diffusion method

r

r =

MM

1 2

2 1









Here r1, r2 are rates of diffusion/effusion for two species

while M1, M2 are their molecular masses respectively

Colligative properties method

pV 5 W _ m RT Here, p 5 Osmotic pressure in atm

• Equivalent weight is the weight of an element or a

compound which will combine with or displace

1.008 part by weight of H2 or 8 part by weight of O2

or 35.5 part by weight of Cl2

• Equivalent weight is a number and when it is denoted

in grams, it is called gram equivalent.

• It depends upon the nature of chemical reaction in

which substances take part

Methods to Find Equivalent Weight

For acids E = protocity or basicity of acid Molecular weight _

For example, for H3PO4, E = M 3

For H2SO4, E = M 2 (Also for H2C2O4 · 2H2O)

For bases E = Acidity or number of OHMolecular weight 2 ions

For example, for Ca(OH)2, E = M 2

For Al(OH)3, E = M3

or

Fe(OH)3

For ions E = Molecular weightCharge on ion _

For example, for SO 42 2, E = M 2

For P O 43 2

, E = M3

For compounds

E = Valency of cation or anion Molecular weight

For example, for CaCO3, E = M 2 For AlCl3, E = M 3

Na3PO4, E = M 3

For redox reactions

=

E Molecualr weightTotalchange in oxidation number Let’s take KMnO4 as an example

2KMnO4 1 2KOH 2K2MnO4 1 H2O 1 [O]

one unit change in oxidation number

3 In neutral medium, E 5 M3

2KMnO4 1 H2O 2KOH 1 2MnO2 1 3[O]

3 units change in oxidation number

For acidic salts

E 5 Number of replaceable H-atoms Molecular weight

For H3PO4, for example, Ca(OH)2 1 H3PO4 CaHPO4 1 2H2O

E 5 Wt of metal Wt of oxygen 3 8 Weight of oxygen 5 Weight of metal oxide

2 Weight of metal

(c) Chloride formation method

E 5 Wt of metal Wt of chloride 3 35.5 Weight of chloride 5 Weight of metal chloride

2 Weight of metal (d) Double decomposition method

Eq wt of salt taken

MOLE FRACTION

• Mole fraction is t he ratio of moles of one component

to the total number of moles present in the solution

It is expressed by X, for example, for a binary

solu-tion of two components A and B

and solute respectively

• Mole fraction does not depend upon temperature as

both the solute and the solvent are expressed by weight

CHEMICAL EQUATION AND

STOICHIOMETRY OF CHEMICAL

REACTIONS

1 A balanced chemical reaction represents a

stoichio-metric equation

2 In a stoichiometric equation, the coefficient of reactants

and products represents their stoichiometric amounts

3 The reactant which is completely used up during an

irre-versible reaction is called the limiting reagent while the

reactant left is called the excess reagent, for example, 20

g of calcium is burnt in 32 g of O2, then Ca is the limiting

reagent while O2 is the excess reagent

4 Stoichiometric calculations help in finding whether

the production of a particular substance is

economi-cally feasible or not

5 These stoichiometric calculations are of following

four types:

(a) Calculations based on weight–weight relationship

(b) Calculations based on weight–volume relationships

(c) Calculations based on volume–volume relationships

(d) Calculations based on weight–volume–energy relationships

6 If the amount of the reactant in a particular reaction is

known, then the amount of the other substance needed

in the reaction or the amount of the product formed in the reaction can be calculated

7 For stoichiometric calculations the following steps

must be considered:

(a) A balanced chemical equation using cal formulas of reactants and products must be written

(b) Here, the coefficients of balanced chemical tion provide the mole ratio of the reactants and products

(c) This mole ratio is convertible into weight–

weight (w/w) ratio, weight–volume (w/v) ratio

or volume–volume (v/v) ratio These are called percentage by weight, percentage by volume and percentage by strength respectively

SOLVED EXAMPLES Mole Concept

1 If 1 Faraday was to be 60230 coulombs instead of 96500

coulombs, what will be the charge on an electron?

The volume of H2 at NTP given by Zn

5 (1.67 2 A) 22.465.4 L (ii)From (i) and (ii)

8 How many years would it take to spend Avogadro

num-ber of rupees at the rate of 10 lac rupees per second?

Solution

Avogadro number 5 6.023 3 1023

Total rupees 5 6.023 3 1023 RsRate of spending 5 10 lac rupees/s 5 106 Rs/sNumber of years to spend all the rupees

5 106 3 60 3 60 3 24 3 365 Rs/year6.023 3 1023 Rs

5 1.90988 3 1010 years

9 Oxygen is present in a one litre flask at a pressure of

7.6 3 10210 mm of Hg Calculate the number of gen molecules in the flask at 0 °C

oxy-Solution

Since, PV 5 nRT 7.6 3 10210 atm 3 1 L _ 760

5 n 3 0.0821 L atm K21 mol21 3 273 K

n 5 7.6 3 10210 L atm

760 3 0.0821 L atm K21 mol21 3 273 K

5 10 _ 22.41 212 mol Number of molecules 5 (6.02 3 1023 mol21) 3 10 _ 22.41 212 mol

5 2.68 3 1010

3 Calculate the number of atoms of oxygen present in

88 g of CO2 What would be the mass of CO having the

same number of oxygen atoms?

Solution

Number of moles of CO2 5 44 g mol88 g 21

5 2 moles

1 mole of CO2 contains 2 moles of oxygen atoms,

2 moles of CO2 will contain 4 moles of oxygen atoms

Number of oxygen atoms 5 4 3 6.023 3 1023

5 2.5092 3 1024

1 mole oxygen atom is present in 1 mole of CO,

4 moles oxygen atoms are present in 4 moles of CO

As one molecule of CH4 contains (6 1 4) 5 10

ele-ctrons, 6.02 3 1022 molecules of CH4 will have

10 3 6.02 3 1022 5 6.02 3 1023 electrons

5 How many atoms of carbon has a young man given

to his bride-to-be if the engagement ring contains 0.5

carat diamond? (1 carat 5 200 mg)

Solution

Mass of diamond (C) 5 0.5 3 200 mg

5 100 mg 5 100 3 1023 g 5 0.1 g Number of mole of C 5 12 g mol0.1 g21

5 1/120 mole Number of C atoms 5 1 120 3 6.023 3 1023

5 5.02 3 1021

6 A mixture of aluminium and zinc weighing 1.67 grams

was completely dissolved in acid and the evolved 1.69

litres of hydrogen gas was measured at 273 K and one

atmosphere pressure What was the mass of

alumin-ium in the original mixture?

10 2.68 3 1023 moles of a solution containing an ion An1

required 1.61 3 1023 moles of Mn O 42 for the oxidation

of An1 to A O 32 in an acidic medium What is the value

of 5 N HNO3 are mixed and the volume of the mixture

is made 1000 mL by adding water Find the normality

of the resulting solution

12 How many millilitres of 0.5 M H2SO4 are needed to

dissolve 0.5 g of copper (II) carbonate?

13 50 litres of water containing Ca(HCO3)2 when

con-verted into soft water required 22.2 g Ca(OH)2

Calculate the amount of Ca(HCO3)2 present per litre

of hard water

Solution

Reaction Ca(HCO3)2 1 Ca(OH)2 2CaCO3 1 2H2O

14 The formula weight of an acid is 82 In a titration,

100 cm3 of a solution of this acid containing 39.0 g

of the acid per litre were completely neutralized

by 95.0 cm3 of aqueous NaOH containing 40.0 g of NaOH per litre What is the basicity of the acid?

Solution

Normality of NaOH 5 1 Normality of acid 5 1 3 95100 5 0.95 Suppose the equivalent mass of the acid is E

39 _ E 5 0.95

E 5 41 Therefore, basicity 5 82/41 5 2

15 One g of impure Na2CO3 is dissolved in water and the solution is made upto 250 mL To 50 mL of this made up solution, 50 mL of 0.1 N HCl is added and the mixture after shaking well required 10 mL of 0.16 N sodium hydroxide solution for complete neutralization

Calculate the per cent purity of the sample of Na2CO3

Solution

Strength of the Na2CO3 solution 5 4 g L21

Suppose the normality of Na2CO3 solution 5 Nx

As after mixing Na2CO3 and HCl solution, NaOH solution is added, so, according to the normality equation

50 3 Nx 1 0.16 3 10 5 50 3 0.1

Nx 5 0.068 N Strength (g L21) 5 Normality 3 Equivalent mass

5 0.068 3 53 5 3.6 g L21

So, purity of Na2CO3 5 3.6 3 100 _ 4

16 Calculate the volume of water to be added to a 100 mL

of 5N solution to make it 0.01 N

Solution

According to normality equation,

N1 V1 5 N2 V2 0.01 3 V1 5 5 3 100

V1 5 5 3 100 _ 0.01 5 50000 mL

So, volume of water to be added 5 50000 2 100

17 A small amount of CaCO3 completely neutralizes

525 mL of 0.1 N HCl and no acid is left in the end

After converting all calcium chloride to CaSO4, how

much plaster of paris can be obtained?

Calculations Based on Reactions

18 Metallic tin in the presence of HCl is oxidized by

K2Cr2O7 solution to stannic chloride What volume of

decinorrnal dichromate solution would be reduced by

The reaction is as follows:

6CaO 1 P4O10 2Ca3(PO4)2

852 g P4O10 3 mol P4O10

1 mole of P4O10 neutralizes 6 moles of CaO

3 moles of P4O10 will neutralize 18 moles of CaO Mass of CaO 5 18 3 56 5 1008 g

20 Find the weight of iron which will be converted into its

oxide by the action of 18 g of steam

Solution

The reaction is 3Fe 1 4H2O Fe3O4 1 4H2

4 moles steam reacts with 3 moles Fe

1 mole (18 g) steam reacts with 3/4 moles Fe

22 What should be the weight of NaNO3 to make 50 mL

of an aqueous solution so that it contains 70 mg Na

= 13935 mg

= 13.935 g

CONCEPTS AT A GLANCE

 Giorgi introduced MKS system

 π has infinite number of significant numbers

 20 carat gold is a mixture having 20 parts by weight of

gold and 4 parts by weight of copper

 Some substances like CuSO4.5H2O, Na2CO3.10H2O have a

tendency to lose water in air These are called efflorescent

substances and this tendency is called efflorescence.

 Some solid substances like NaOH, KOH, which have a

tendency to absorb moisture greatly from air and to get

wet are called deliquescent and this tendency is called

deliquescence

 Hygroscopic substances like quicklime (CaO) anhydrous

P2O5 etc., absorb moisture from air

 Compounds having similar chemical composition in the

same crystalline form are called isomorphs

For example, all alums [M2SO4 M2 (SO4)3 24H2O]

Here, M = Monovalent (K)

M = Trivalent (Al) FeSO4.7H2O (Green vitriol) and ZnSO4.7H2O

 Different crystalline forms of a substance are called

polymorphs and this phenomenon is called polymorphism.

For example, ZnS → Zinc blende

1

2

2 1

Here rl, r2 are rates of diffusion for two species while

M1, M2 are their molecular masses respectively

1 2 1 2

 Specifi gravity = Mass of Liquid

EXERCISES

Single Option Correct Type

1 The number of significant figures for the three

(d) All have same no.of O-atoms

4 A mole of any substance is related to

(a) number of particles

(b) volume of gaseous substances

(c) mass of a substance

(d) all of these

5 What is the weight of oxygen required for the

com-plete combustion of 2.8 kg of ethylene?

8 Boron has two stable isotopes, B10 (19%) and B11

(18%) Find the average atomic weight of boron

12 A bivalent metal has an equivalent mass of 32, the

molecular mass of metal nitrate is?

(a) 168 (b) 182 (c) 184 (d) 188

13 How many moles of potassium chlorate should be

decomposed completely to obtain 67.2 litres of gen at STP?

14 How many grams of phosphoric acid is required to

complete neutralize 120 g of sodium hydroxide?

(a) 0.98 (b) 98

15 The hydrated salt Na2CO3n H2O undergoes 63% loss

in mass on heating and becomes anhydrous The value

of n is

16 The vapour density of a mixture having NO2 and N2O4

is 27.6 The mole fraction NO2 in the mixture is (a) 1.6 (b) 0.8

(c) 2.4 (d) 0.6

17 Among the following pairs of compounds, the one that

illustrates the law of multiple proportions is (a) Cu and CuSO4 (b) CuO and Cu2O (c) H2S and SO2 (d) NH3 and NCl3

18 How many grams of KCl must be added to 75 g of

water to produce a solution with a molality of 2.25 (a) 1.257 g (b) 125.7 g

(c) 12.57 g (d) 25.14 g

19 Normality of 0.04 M H2SO4 is

(a) 0.02 N (b) 0.01 N

(c) 0.04 N (d) 0.08 N

20 Which among the following is the heaviest?

(a) one mole of oxygen

(b) one molecule of sulphur trioxide

(c) 100 amu of uranium

(d) 44 g of carbon dioxide

21 The empirical formula of a commercial ion exchange

resin is C8H7SO3Na The resin can be used to soften

water according to the reaction Ca+2 + 2C8H7SO3Na →

(C8H7SO3)2 Ca + 2Na+ What would be the maximum

uptake of Ca+2 by the resin expressed in mole/g resin?

(a) 0.0024 (b) 0.0246

(c) 0.246 (d) 24.6

22 A boy drinks 500 mL of 9% glucose solution The

number of glucose molecules he has consumed are

[mol wt of glucose 5 180]

(a) 0.5 3 1023 (b) 1.0 3 1023

(c) 1.5 3 1023 (d) 2.0 3 1023

23 The pollution of SO2 in air is 10 ppm by volume The

volume of SO2 per litre of air is

25 The number of grams of a dibasic acid (molecular

weight 200) present is 100 mL of its aqueous solution

to give decinormal strength is

28 The molecular weight of O2 and SO2 are 32 and

64 respectively At 15 8C and 150 mm Hg

pres-sure, one litre of O2 contains ‘N’ molecules The

number of molecules in two litres of SO2 under

the same conditions of temperature and pressure

30 50 gram of calcium carbonate was completely burnt in

air What is the weight (in grams) of the residue?

33 How many moles of acidified FeSO4 can be completely oxidized by one mole of KMnO4?

34 A compound possess 8% sulphur by mass The least

molecular mass is (a) 200 (b) 400 (c) 155 (d) 355

35 The vapour density of ozone is

37 1000 g calcium carbonate solution contains 10 g

car-bonate The concentration of solution is (a) 10 ppm (b) 100 ppm (c) 1000 ppm (d) 10,000 ppm

38 One mole of CH4 contains (a) 4.0 g atoms of hydrogen (b) 3.0 g atom of carbon (c) 6.02 3 1023 atoms of hydrogen (d) 1.81 3 1023 molecules of CH4

39 The amount of O2 for me a at N.T.P by the complete combastion of 1 kg coal is?

(a) 22.4 L (b) 2240 L (c) 1866 L (d) 100 L

40 The maximum number of molecules is present in

42 The incorrect statement for 14 g of CO is

(a) it occupies 2.24 litre at NTP

44 Which of the following statement is correct?

(a) 1 mole of electrons weighs 5.4 mg

(b) 1 mole of electrons weighs 5.4 kg

(c) 1 mole of electrons weighs 0.54 mg

(d) 1 mole of electrons has 1.6 3 10219 C of charge

45 Which of the following pairs of gases contain equal

number of molecules?

(a) CO2 and NO2 (b) CO and (CN)2

(c) NO and CO (d) N2O and CO2

46 The samples of NaCl are produced when Na combines

separately with two isotopes of chlorine Cl35 and Cl37

Which law is illustrated?

(a) Law of constant volume

(b) Law of multiple proportions

(c) Law of reciprocal proportions

48 A breakfast cereal in advertised to contain 110 mg of

sodium per 100 g of the cereal The per cent of sodium

in the cereal is (a) 0.110 % (b) 0.01 10 % (c) 11.0 % (d) 0.22 %

49 Express 145.6 L of chlorine in terms of gram moles.

(a) 6.5 g moles (b) 4.5 g moles (c) 0.65 g moles (d) 9.5 g moles

50 The number of significant figures in 306.45 and 40440

are respectively (a) 4, 5 (b) 5, 5 (c) 5, 4 (d) 4, 6

51 The quantity of PV K

BT represents the (a) molar mass of a gas

(b) number of molecules in a gas (c) mass of gas

(d) number of moles of a gas

52 Which is the correct order of micro, nano, femto and

pico here?

(a) micro < nano < pico < femto (b) pico < femto < nano < micro (c) femto < pico < nano < micro (d) femto < nano < micro < pico

53 Find the number of atoms present in 0.016 g of methane.

(a) 0.5 N0 (b) 0.05 N0 (c) N0 (d) 1.6 N0

54 15 litre atmosphere is equal to

(a) 1.515 3 108 erg (b) 15.15 3 109 erg (c) 1.515 3 1010 erg (d) 15.15 3 1012 erg

55 If equal moles of water and urea are taken in a

ves-sel what will be the mass percentage of urea in the solution?

(a) 22.086 (b) 11 536 (c) 46.146 (d) 23.076

56 Mixture X 5 0.02 mol of [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4 was prepared in 2 litre of solution

1 litre of mixture X 1 excess AgNO3 Y

1 litre of mixture X 1 excess BaCl2 Z Number of moles of Y and Z are

(a) 0.02, 0.01 (b) 0.01, 0.01 (c) 0.01, 0.02 (d) 0.02, 0.02

57 At STP the density of CCl4 vapour in g/L will be nearest to

(a) 8.67 (b) 6.87 (c) 5.67 (d) 4.26