A complete resource book in chemistry for jee main 2019 atul singhal

A Complete Resource Book in Chemistry for JEE Main 2019 Atul Singhal

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Dedicated to my grandparents, parents and teachers

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Preface

vii

Acknowledgements viii

Chapter 1 Basics of Chemistry

Chapter 2 Solid State

Chapter 3 Gaseous State

Chapter 4 Atomic Structure

Chapter 5 Solutions

Chapter

Energetics

Chapter 7 Chemical Equilibrium.

Chapter 8 Ionic Equilibrium

Chapter

Chapter

Chapter

Preface

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 qualified 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 sufficient 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 efforts, 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

Acknowledgements

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 honourable 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 affection 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

Chemistry Trend Analysis (2007 to 2018)

1. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

(A) [Co(HO)ClO 23 3 ]. 2

(B) [Co(HO)] Cl 26 3

(C) [Co(HO)Cl] Cl .H 25 22 O

(D) [Co(HO)Cl] Cl 24 22.2HO

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) HO and (HO OH ) 22 +

(B) (H O+ 22OO 2 )andH

(C) (H O) and(HO ) 22 2 ++OOH

(D) HO and (HO O) 22 2 +

3. Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation?

(A) N2Cl

(B) N

(C) NH2 (D) NO2

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?

BaseAcidEnd point

(A)StrongStrongPink to colourless

(B)WeakStrongColourless to pink

(C)StrongStrongPinkish red to yellow

(D)WeakStrongYellow to pinkish red

6. The predominant form of histamine present in human blood is (pKa, Histidine = 6.0)

(A) NH3 ⊕ N N H

(B) NH2 N N H

(C) ⊕ ⊕ NH3 N N H H

(D) ⊕ NH2 N N H H

7. The increasing order of basicity of the following compounds is:

(I) NH2 (II) NH

(III) NH NH2 (IV) NHCH3

(A) (IV) < (II) < (I) < (III)

(B) (I) < (II) < (III) < (IV)

(C) (II) < (I) < (III) < (IV)

(D) (II) < (I) < (IV) < (III)

8. Which of the following lines cor rectly show the temperature dependence of equilibrium constant K, for an exothermic reaction?

(A) A and D

(B) A and B

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

9. How long (approximate) should water be electrolyzed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane?

(Atomic weight of B = 10.8 u)

(A) 1.6 hours (B) 6.4 hours

(C) 0.8 hours (D) 3.2 hours

10. Consider the following reaction and statements:

[C Br ][Co(NH) Br ]NH o(NH 34)Br23 33 3 +−+ → +

(I) Two isomers are produced if the reactant complex ion is a cis-isomer.

(II) Two isomers are produced if the reactant complex ion is a trans-isomer.

(III) Only one isomer is produced if the reactant complex ion is a trans-isomer.

(IV) Only one isomer is produced if the reactant complex 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 chlorofor mate in the presence of NaOH to form product A. A reacts with Br2 to for m product B. A and B are respectively:

(A)

(B)

(C)

12. An aqueous solution contains an unknown concentration 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 gaseous 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:

(A) 0 (B) 2

(C) 3 (4) 1

14. The combustion of benzene (1) gives CO2 (g) and H2O (l). Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol-1 at 25°C; heat of combustion (in kJ mol-1) of benzene at constant pressure will be: (R = 8.314 JK-1 mol-1)

(A) -3267.6 (B) 4152.6

(C) -452.46 (D) 3260

15. The ratio of mass percent of C and H of an organic compound (CxHyOz) is 6 : 1. If one molecule of the above compound (CxHyOz) contains half as much oxygen as required to burn one molecule of compound CxHy completely to CO2 and H2O, then empirical formula of compound CxHyOz is:

(A) C2H4O3 (B) C3H6O3

(C) C2H4O (D) C3H4O2

16. The trans-alkenes are for med 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 gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an absorbent. The metal ‘M’ is:

(A) Fe (B) Zn

(C) Ca

(D) Al

19. According to molecular orbital theory, which of the following will not be a viable molecule?

(A) H 2 2

(C) He 2 +

(B) He 2 2 +

(D) H 2

20. The major product formed in the following reaction is:

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)

(C) (D)

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 is 10 10 7 . × and that of S2 from HSions is 12 10 13 × then the concentration of S2 ions in aqueous solution is:

(A) 510 19 × (B) 510 8 ×

(C) 310 20 × (D) 610 21 ×

27. The oxidation states of Cr in Cr(H O) Cl Cr(C H) 26 36 62 [] [] , Cr(H O) Cl Cr(C H) 26 36 62 [] [] , and KCr(CN) (O)(O )(NH) 22 22 3 [] respectively are:

(A) +3, 0 and +4

(B) +3, +4 and +6

(C) +3, +2 and +4

(D) +3, 0 and +6

22. Which of the following compounds contain(s) no covalent bond(s)?

KCl, PH3, O2, B3H6, H2SO4

(A) KCl, B3H6 (B) KCl, B2H6, PH3

(C) KCl, H2SO4 (D) KCl

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

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 converting 3Ca(PO ).Ca(OH) 34 22 [] to:

(A) 3Ca(OH) .CaF 22{}    

(B) CaF2 []

(C) 3 2 ()CaFCa(OH) 2 []

(D) 3Ca(PO ). CaF 34 22 []

29. Which of the following salts is the most basic in aqueous solution?

(A) Pb(CHCOO)32 (B) Al(CN) 3

(C) CH COOK 3

(D) FeCl 3

30. Total number of lone pair of electron in I 3 ion is:

(A) 12

(C) 6

(B) 3

(D) 9

Answer keys

Hints and solutions

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, HO22 acts as an oxidant as follows

Oxidation number decreases

Oxidation number increases

In alkaline medium, HO22 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 CH NH 65 2 is suitable for the test.

Hence, the correct option is (C).

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

CH3

CHO (CHOH)4 (CH2)4

CH2OH

Glucose n-hexane Δ HI

CH3

Hence, the correct option is (B).

7. Order of basicity in increasing order is as follows

5. As methyl orange is weak organic base. So it is used in the titration of weak base (NH4OH) vs strong acid (HCl)

MeOH Me OH

Unionized form (Yellow)

Ionized form (Red)

In basic medium, equilibrium lies in backward direction and therefore it shows yellow colour.

In acidic medium, equilibrium shifts in forward direction and therefore, colour changes from yellow to red.

Hence, the correct option is (D).

6. NH3 + N H N

In it, the N-atoms present in the ring will have same pKa values (6.0), while N atom outside the ring will have different pKa value (pKa > 7.4).

Hence, two N-atoms inside the ring will remain in unprotonated from in human blood as their pKa(6.0) < pH of blood (7.4), while the N-atom outside the ring will remain in protonated from because its pKa > pH of blood (7.4).

Hence, the correct option is (A).

Here (III) is most basic because its conjugate acid is stabilized by equivalent resonance.

Out of (I) and (IV), (IV) is more basic due to + I effect of -CH3 group (II) is less basic than (I) and (IV) because N atom is sp2(more s%) hybridized. Hence, the cor rect option is (D).

8. DG° = -RT ln K

DH° -TDS° = -RT ln K

° + ° ∆∆ H

RT S R ln K

Here ln K vs 1 T graph will be a straight line with slope equal to °∆ H R . As reaction is exothermic, so°∆ H itself will be negative resulting in positive slope. Hence, the cor rect option is (B).

9. BH OB OH O 26 22 3332+ → +

As according to the balanced equation:

Here 2766 26 ., gB H i.e., 1 mole BH26 requires 3 mole of O2. Here this oxygen is produced by electrolysis of water as follows.

22 2 4 22 HO HO F →+

As 1 mole O2 is produced by 4 F charge

So 3 mole O2 will be produced by 12 F charge

11. The sequence of the reaction is as follows H2O

Hence, the cor rect option is (D).

12. Ba Na SO BaSO Na mL 2 mL,1M 4 ++ + → + 2 450 4 50 2

For BaSO4 Ksp is given as

K(BaSO ) = [Ba ][SO] sp 4 2 4 2+−

10 01 10 2 −+=× [Ba]

[Ba]+− = 29

10 M(in 500 mL solution)

As Q = It

So, 12 × 96500 C = 100 × t(s)

On solving, we get t = × × 1296500 1003600 hours

t = 3.2 hours

Hence, the cor rect option is (D).

Br NH3 H3N H3N Co

Br NH3 NH3 H3N H3N Co Br Br + Br NH3 NH3 H3N Co Br Br NH3 NH3 Br H3N Co H3N mer-isomer Br Br Br NH3 H3N H3N Co Br Br

10. Br Br

cis-isomer trans-isomermer-isomer (only) fac-isomer

Hence, the cor rect option is (C).

o, p-directing group (A) C

Br (Major product) (B)

[SO] 4 2 in 500 mL solution will be 50 × 1 = M × 500

M = 0.1

Now, calculate [Ba+2] in original solution (450 mL) as follows

10-9 × 500 = 450 × M

Option (B), x = 3, y = 6

MM = × =× =× 10500

9 99

450 10 9 10 11110

Hence, the correct option is (D).

13. CHCHOCOCH

PTorr i 3 363 4 = →+

As we know that, rPRn ∝ (1)

Here PR = reactant pressure

n = order of reaction

Now rate of reaction is 1.00 torr s-1, when reactant pressure is 363363 5 100 −×       torr = 344.85 torr.

Similarly rate of reaction is 0.5 torr s-1, when reactant pressure is 363363 33 100 −×       torr = 243.21 torr.

So, applying equation (1)

1 05 34485 24321 = 

2 = (1.418)

⇒ 21 418 1 n = ∴ n ≈ 2

Hence, the correct option is (B).

14. C6H6 (l) + 15 2 O2 (g) → 6CO2 (g) + 3H2O (l)

Dng = 6 15 2 = 3 2

DH = DU + DngRT

DH = ××       3263900 3 2 8314298.J = -3267616 J = -3267.616 kJ

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

CxHy + x y +       4 O2 → xCO2 + y 2 H2O

Now z is half of oxygen atoms needed to burn CxHy here.

y

∴

Putting on the values of x and y from the given options:

Option (A), x = 2, y = 4

z =+

= 2 4 4 3

z =+

 = 3 6 4 45 .

Hence, the correct option is (A) (C2H4O3).

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

26 23 22 3 Al HO AlOH H + x +→ () ↓ ()

AlOHNaOHNa Al(OH (Soluble) () () ) 34 +→ []

Hence, the correct option is (D).

19. Species Bond order

(A) H 2 2 B.O. =−[] = 1 2 22 0 (does not exists)

(B) He 2 2 + B.O. =−[] = 1 2 20 1 (exists)

(C) He 2 + B.O. = 1 2 21 05 [] = (exists)

(D) H 2 B.O. =−[] = 1 2 21 05 (exists)

As H 2 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

Hence, the correct option is (A).

21. Here Asprin is formed as follows

Acetyl salicylic acid (Asprin) (Analgesic) OC OH

(i) NaOH

(ii) CO2, 4 7 atm, 125°C

H3C 2 C O/H+ (catalyst)

(iii) H+ (Kolbe’s reaction) (acetylation)

Hence, the correct option is (B).

22. KCl contains only ionic bond between K + and Cl ions () KCl+− while rest compounds have covalent bonds as follows:

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

(NH) SO 2NH(g) +H SO

Ba(N )(s) Ba +3N(g) (NH) Cr 42 43 24 32 2 42 2

Hence, the correct option is (A).

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

Cr(H O) Cl 26 3 [] +3

Cr(C 66H) 2 [] 0

KCr(CN) (O)O NH 22 22()() 3 [] +6

20 ++ −+−+ [] = X( 2) (2)( 4)

so, X = +6

Hence, the correct option is (D).

28. 3Ca(PO ). Ca(OH) F 3Ca(PO ). CaF2OH 34 22 34 22 [] + → [] + 2

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. I 3 has 3 lone pair e- as follows

II I

Hence, the correct option is (D).

26. HS aq.H S 2 +0.20) ()

(. ) ( 01 2 2 01 02

12 2 1 + +

=× S On solving

= x x x aa a KK K . (. ) 210 12 10 02 01 310 20 20 22 ×

S2220

Hence, the correct option is (C).

JEE Mains 2017 PaPErs

1. Which of the following compounds will give significant amount of meta product during mono-nitration reaction?

(A) OH (B) OCOCH3

(C) NH2 (D) NHCOCH3

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)

(A) 1.65Å (B) 4.76Å

(C) 0.529Å (D) 212Å

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:

(A) 7.2 (B) 6.9

(C) 7.0

(D) 1.0

6. The formation of which of the following polymers involves hydrolysis reaction?

(A) Nylon 6 (B) Bakelite

(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:

(A) 15 kg (B) 37.5kg

(C) 7.5 kg

(D) 10 kg

8. Which of the following, upon treatment with ter tBuONa followed by addition of bromine water, fails to decolourize the colour of bromine?

(A) O Br (B) C6H5 Br

(C) Br O (D) Br O

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 proper ties 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 on heating

11. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is:

(A) Six (B) Zero

(C) Two (D) Four

12. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest approach between two atoms in metallic crystal will be:

(A) 2a (B) 22a

(C) 2a (D) a 2

13. Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol-1. If k1 and k2 are rate constants for reactions

R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to (R = 8.314 J mol-1K-1):

(A) 8 (B) 12

(C) 6 (D) 4

14. The correct sequence of reagents for the following conversion will be:

O CHO CHO HO CH3 CH3 HO

(A) AgNHOH H/CH OH CH MgBr + 33 3 2 ()     + ,,

(B) CH MgBr H/CH OHAgNHOH 3 + 3 ,, 3 2 ()     +

(C) CH MgBr AgNHOH H/CH OH 3 + 3 ,, 3 2 ()     +

(D) AgNHOH CH MgBr H/CH OH 3 + 3 3 2 ()    

,,

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 g reatly 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 OCOCH3 HO OH HOH2C O

(B) CH2OH HO OH HOH2C O

(C) CH2OH OCH3 HO OH HOH2C O

(D) CH2OH OH OH OH HOH2C O

17. Given

CO gCOg grahite 22 () + () → () ;

∆° =− r HkJmol 1 3935

Hg Og HO 1 22 () + () → () 1 2 2 ;

∆° =− r . HkJmol 1 2858

C; Og HO lCHg Og 22 42 () + () → () + () 22

∆° =+ r 1 HkJm ol8903.

Based on the above ther mochemical equation, the value of DrH° at 298 K for the reaction

CH gCHg graphite () () () +→ 2 24 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) XeFO FXeF O 42 26 2 +→ +

(B) XeFPFXeF PF 25 6 +→ []+

(C) XeFH OXeOFHF 62 4 +→ + 2

(D) XeFH OXeO F4HF 62 2 +→ + 2 2

19. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are:

(A) ClOand ClO3

(B) ClOand ClO 23

(C) Cl andClO

(D) Cl andClO 2

20. The major product obtained in the following reaction is: Δ BuOK Br H C6H5 C6H5

(A) () () ± CH CH OBuCHCHH 65 t 26 5

(B) CH CH=CHCH 65 65

(C) + () CH CH OBuCHH 65 t 25 ()

(D) () () CH CH OBuCHC H 65 t 26 5

21. Sodium salt of an organic acid ‘X’ produces effervescence 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 22CO 4

22. Which of the following species is not paramagnetic?

(A) NO (B) CO

(C) O2 (D) B2

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 be:

(Kf for benzene = 5.12 K kg mol-1)

(A) 64.6% (B) 80.4%

(C) 74.6% (D) 94.6%

24. Which of the following molecules is least resonance stabilized?

(A) (B)

(C) N (D) O

25. On treatment of 100 ml of 0.1 M solution of COCl3 6H2O with excess AgNO3; 1.2 × 1022 ions are precipitated. The complex is:

(A) [CO(H2O)4 Cl2]Cl 2H2O

(B) [CO(H2O)3Cl3] . 3H2O

(C) [CO(H2O)6]Cl3

(D) [CO(H2O)5Cl]Cl2 H2O

26. The major product obtained in the following reaction is: O O COOH DIBAL–H

(B) OH CHO CHO

(C) COOH CHO

(D) CHO CHO

27. A water sample has ppm level concentration of following anions:

F10; SO 100; NO 50 4 2 3 == =

the

anion/anions that make/makes the water sample unsuitable for drinking is/are:

(A) Only NO3

(B) Both SO 4 2 and NO3

(C) Only F

(D) Only SO 4 2

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:

(A) 1186

(C) 118.6

29. Given

(A)

(B) 84.3

(D) 11.86

E1.36V,E 0.74V Cl /ClCr/Cr 2 3+ °= °=

E1.33V,E 1.51V. Cr O/Cr MnO/Mn 27 23+ 4 2+ °= °=

Among the following, the strongest reducing agent is:

(A) Cr

(C) Cr3+

(B) Mn2+

(D) Cl-

30. The group having isoelectronic species is:

(A) O, F, Na ,Mg 2+ 2+

(B) O, F, Na,Mg +

(C) O, F, Na,Mg 22+

(D) O, F, Na ,Mg +2+

Answer keys

Hints and solutions

1. Nitration is carried out in presence of concentrated HNO3 and concentrated H2SO4. Here, aniline acts as base, in presence of H2SO4 its protonation takes place and anilinium ion is formed.

NH2 H2SO4

H⊕

NH3HSO4 ⊕

Anilinium ion is strongly deactivating group and meta directing in nature so it gives meta nitration product in significant amount. (≈ 97%).

NH2

2. Using 1st law:

∆U = q + w

NH3HSO4 ⊕

conc.H2SO4

NO2 + conc.HNO3

For adiabatic process:

q = 0

So, ∆U = w

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 dissociation of alkyl halide

R − X R⊕ + X

Higher the stability of R⊕ (carbocation) higher would be reactivity of SN1 reaction.

As stability of cation follows order.

⊕

CH3——CH2——CH2

< CH3——CH——CH2 ——CH3

⊕ ⊕

< p—— H3CO——C6H4——CH2

Hence, the correct order is

4. Radius of nth Bohr orbit in H-atom = 0.53 n Z 2 Å

Radius of second Bohr orbit = 0532 1 2 .( ) ×

= 2.12 Å

5. pKa(HA) = 3.2

pKb (BOH) = 3.4

As given salt is of weak acid and weak base.

So, pH = 7+pKpKab 1 2 1 2

= 7+(3.2)(3.4)

6. Formation of Nylon6 involves hydrolysis of its monomer (caprolactum) in initial state as follows:

NH

Hydrolysis

H3N——CH2——CH2——

Caprolactum Nylon6

Δ /Polymerization O

7. Mass in the body of a healthy human adult has:

Oxygen = 61.4%, carbon = 22.9%,

Hydrogen = 10.0% and Nitrogen = 2.6%

Total weight of person = 75 kg

Mass due to 1H is =× = 75 10 100 75.kg

1H atoms are replaced by 2H atoms.

Hence, mass gain by person = 7.5 kg

8. Br

tert-BuONa O O-tBu O

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

tert-BuONa

C6H5 Br O

tert-BuONa Br O

C6H5

(it decolorizes bromine solution)

(it decolorizes bromine solution due to unsaturation)

Br O O

tert-BuONa

(it decolorizes bromine solution due to unsaturation)

9. ZnO is an amphoteric oxide but in given reaction it acts as follows:

(a) ZnO + Na2O → Na2ZnO2 acid base salt

(b) ZnO + CO2 → ZnCO3 base acid salt

10. Mg can form basic carbonate like, 56 7

(I) Et CH3 H (II) Et H H3C

(III) CH3 Et CH3 H Br H (IV) CH3 Et H Br H H3C

12. In FCC unit cell atoms are in contact along face diagonal.

So, 24aR =

Hence, closest distance (2R) = 2 2 2 aa =

13. Using arrhenius equation,

KA e Ea RT =⋅

KA e 1 ERT a 1 =⋅ (1)

KA e 2 ERT a 2 =⋅ (2)

So, equation (2)/(l) ⇒= K K e 2 1 (E E RT a1 a 2 )

(As Pre-exponential factors of both reactions in same)

Hence, ln ) , K K (E E RT 2 1 aa 1       = = × = 2 10000 8314300 4

14. O O CHO CO2H

[Ag(NH3)2]OH Tollens reagent

45 2 2 3 2 2

32 23 Mg CH O MgCO Mg OH HO HCO +−++ →⋅ ⋅↓ + ()

While Li can form only carbonate (Li2CO3) not basic carbonate.

11. +H2O2

2 3 4 Br

1 5

HBr

3-methyl pent-2-ene Anti markownikov product (4 stereo isomers possible due to 2 chiral centres as molecule is asymmetric )

O C O

OCH3 H+/CH3OH (esterification) CH3MgBr

CH3 Br H H

CH3 Et

(I)

(II) CH3 Et H H Br H3C CH3

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.

22. Using molecular orbital theory

NO(15e-) ⇒ One unpaired electron is present in π * molecular orbital. (Paramagnetic)

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

O2 (16) ⇒ Two unpaired electrons are present in π * Molecular orbitals. (Paramagnetic)

⊕ ve silver mirror test

Hemiketal O OH CH2OH HOCH2 O

Reagent Tollen’s Ring opening

(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.

17. C; ? r Og HO lCHg Og H H 22 42 f () + () → () + () ∆° = ∆° 22 8903 3935285 80

∆∆ ∆ rr products rreactant HH H °= °− ° ∑ ∑ ()()

B2(10) ⇒ Two unpaired electros are present in π bonding molecular orbitals. (Paramagnetic)

23. In benzene acetic acid dimerises as follows:

2

2 CH COOH CH COOH 33()

i =+    

1 1 2 1 α

i =− 1 2 α Here a is degree of association

∆=TiKm ff

0451

512

890 31 20 1393 52 2858 ..

fCH H 4 ∆° () =− =− f CH HkJ/mol 4 8903965 1748

18. XeFO FX eF O 42 26 2 ++ + +→ + 41 6 0

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

BrH HH

C6H5

C6H5 + tBuOH + Br ⊕ OtBu

C6H5

21. Na2C2O4 + H2SO4 Na2SO4 + CO↑ + CO2↑ + H2O conc.

Na2C2O4 + CaCl2 CaC2O4↓ + 2NaCl (white ppt.)

5CaC2O4↓ + 2KMnO4 + 8H2SO4

K2SO4 + 5CaSO4 2MnSO4 + 10CO2 + 8H2O (purple) (colourless)

1 2 0527 −= α .

α = 0945.

Percentage deg ree of association = 94.5%

24. (D) O is nonaromatic (non-planar) and hence least

reasonance stabilized.

(A) Benzene is aromatic (6πe -)

(B) O furan is aromatic (6πe -)

(C) N pyridine is aromatic (6πe -)

25. Moles of complex Molarity volumeml = × ()

1000

= × = 100 01 1000 001 . .mole

Moles of ions precipitated with excess of AgNO3

= × × 12 10 60210

22 23 .

= 002.moles

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 corresponding aldehyde (partial reduction).

27. NO3 : The maximum limit of nitrate in drinking water can be 50 ppm. Excess nitrate in drinking water can cause diseases like methemoglobinemia, etc.

SO 4 2 : above 500 ppm of SO 4 2 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 2 and NO3 in water is suitable for drinking but the concentration of F (i.e., 10 ppm) makes water unsuitable for drinking purpose.

28. Given chemical equation

MCO2HCl2MClH OCO 23 22 +→ ++

1 gm 0.01186 mol

⇒ from the balanced chemical equation.

1 M 0.01186 =

⇒=M84.3 gm/mol

29.

°=E 1.51V MnO/Mn 4 +2 (1)

°= E1.36V Cl /Cl 2 (2)

°= E1.33V Cr27O/Cr2+3 (3)

°= E0.74 Cr /Cr +3 (4)

As Cr+3 is having least reducing potential, so Cr is the best reducing agent here.

30.

ions O-2 F- Na + Mg+2

Atomic number = 8 9 11 12

Number of e = 10 10 10 10

O, F, Na ,Mg 2+ +2 are isoelectronic species here.

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Basics

Chapter Highlights

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.

BASIC CONCEPTS

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.

Matter

Pure substances Mixtures

Homogeneous mixtures

Heterogeneous mixtures Elements Compounds

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 further 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, crystallization, distillation.

• A mixture is further classified into two categories— homogeneous and heterogeneous.

• In a homogeneous mixture all the components undergo 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 constituents cannot be separated by using simple physical methods of separation.

• A pure substance can be further classified into an element or a compound.

• 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, CO2

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 identity 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 compounds) 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 independently 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 contrary to Dalton’s theory.

PHYSICAL QUANTITIES AND THEIR MEASUREMENTS

In order to describe and interpret the behaviour of chemical 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 average 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.

• 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 measurement of matter. In order to make accurate measurements 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 example, 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 present at right of the decimal point then these zeros are also significant figures, for example, 33.600 has five significant figures.

• 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 possessed by the least precise term, for example, 5.1 1 7.21

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 example, 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 figures). In this kind of notation, every number can be written as N 3 10n

Here,

n 5 Integer, Table 1.1 SI Units

Table 1.2 Derived Units

1 Exa (E) 5 1018

1 Zetta (Z) 5 1021

1 Yotta (Y) 5 1024

1 litre 5 10 3 m3 5 1 dm3

1 atmosphere 5 760 mm or torr 5 101.325 Pa or Nm 2

1 bar 5 105 Nm 2 5 105 Pa

1 calorie 5 4.184 joule

1 eV (electron volt) 5 1.602 3 10 19 joule

1 joule 5 10 7 erg

So, 1 eV 5 1.602 3 10 12 erg

1 cal . 1 J . 1 erg . 1 eV

‘Barn’ is a unit of area to measure the cross section of nucleus.

1 Barn 5 10 28 m2 ≈ 10 24 cm2

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 calculated weight of AgNO3 and NaCl is mixed, white precipitates 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 element 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.

• 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 particles, 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 5 W M

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

Here, g.m.m. 5 Gram molecular mass Mole 5

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 one ion × 1.6 × 10–19C

MASS

Mass can be expressed in terms of atoms or molecules as follows:

Atomic Mass

Atomic mass is the relative mass of an atom which shows the number of times an atom is heavier than 1 12 mass of C-12.

• 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

V 5 Valency

• Atomic mass 5 6.4

Specific heat in calories

This is known as Dulong Petit’s Law

Atomic mass unit

The quantity of 1 12 mass of an atom of C-12 represents it and it is abbreviated as a.m.u

1 A.m.u = 1.9910 12 1.6610

11 24 × =×

Mass ofone atom of anelement

Atomic mass

1A.m.u. =

Here 1. 99 × 10–23 g is wt. of one C–12 atom

Average atomic mass

(At. mass)AV

ma mb mc ab c 12 2 = ×+ ×+ × ++

Here m1, m2, m3 are masses of isotopes and a, b, c are their percentage ratio.

Molecular Mass

Molecular mass represents the total mass of a molecule, that is, number of times a molecule is heavier than 1 12 weight 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. 5

V.D.

W 3 22400

Volume at STP (in mL)

Here, W 5 Weight of substance in g

V.D. = Vapour Density

Graham’s diffusion method

r r = M M 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

V 5 Volume in litre

W 5 Weight in gram

R 5 Universal gas constant

T 5 Given temperature

m 5 Molar mass

EQUIVALENT WEIGHT

• 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 = Molecular weight protocity or basicity of acid

For example, for H3PO4, E = M 3

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

For bases

E = Molecular weight Acidity or number of OH ions

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

For Al(OH)3, E = M 3 or Fe(OH)3

For ions

E = Molecular weight Charge on ion

For example, for SO4 2 , E = M 2

For PO4 3 , E = M 3

For compounds

E = Molecular weight

Valency of cation or anion

For example, for CaCO3, E = M 2

For AlCl3, E = M 3

Na3PO4, E = M 3

For redox reactions

Molecualrweight

= E

Totalchange in oxidation number

Let’s take KMnO4 as an example.

1. In acidic medium, E = M 5 +7 +2

2KMnO4 1 3H2SO4 K2SO4 + 2MnSO4 + 3H2O 1 5 [O]

5 units change in oxidation number

2. In basic medium, E 5 M 1

17

one unit change in oxidation number

3. In neutral medium, E 5 M 3

1

3 units change in oxidation number

For acidic salts

E 5

Molecular weight

Number of replaceable H-atoms

For H3PO4, for example,

Ca(OH)2 1 H3PO4 CaHPO4 1 2H2O

E 5 M 2

Other methods

(a) Hydrogen displacement method

E 5 W 3 11200

Volume of H2 at NTP

(b) Oxide formation method

E 5 Wt of metal

Wt of oxygen 3 8

Weight of oxygen 5 Weight of metal oxide

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

Weight of metal

(d) Double decomposition method

Eq.wt. of salt taken

Eq.Wt. of salt ppt. = Wt.ofsalttaken Wt.ofsalt ppt.

(e) Metal displacement method

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 solution of two components A and B.

XA 5 nA

nA 1 nB

XB 5 nB nA 1 nB

XA 1 XB 5 1

• Mole fraction of solute (X2 ) 5 n2 n1 1 n2

Here, n1 and n2 represent number of moles of solvent 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 stoichiometric 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 irreversible 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 economically 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 chemical formulas of reactants and products must be written.

(b) Here, the coefficients of balanced chemical equation 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.

Mole Concept

1. If 1 Faraday was to be 60230 coulombs instead of 96500 coulombs, what will be the charge on an electron?

Solution

One mole electron carries 1 Faraday charge. 6.023 3 1023 electrons carry 5 60230 C

1 electron carries 5 60230 C 6.023 3 1023

51 3 10 19 C.

2. If a piece of copper weighs 0.635 g, how many atoms does it contain?

Solution

Number of moles of Cu in 0.635 g

5 0.635 g

63.5 g mol 1 5 10 2 mol

1 mole Cu contains 6.023 3 1023 atoms of Cu

10 2 mole Cu contains 6.023 3 1023 3 10 2

5 6.023

3 1021 atoms of Cu.

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

88 g

44 g mol 1

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

Its mass is 4 (12 1 16) 5 112 g.

4. Calculate the total number of electrons present in 1.6 gram of methane.

Solution

Molecular mass of methane

5 16 g mol 1

16 g CH4 contains 6.02 3 1023 molecules of CH4 1.6 g CH4 contains 6.02 3 1022 molecules of CH4

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

The volume of H2 at NTP given by Zn

5 (1.67 A) 22.4 65.4 L (ii)

From (i) and (ii)

3 3 22.4 3 A 54 1 (1.67 2 A) 22.4 65.4 5 1.69

142.2 3 A 5 176.26

A 5 1.248 g

7. Find the equivalent mass of H3PO4 in the reaction:

Ca(OH)2 1 H3PO4 CaHPO4 1 2H2O

Solution

As in this reaction only two hydrogen atoms are replaced so its equivalent mass will be given by the following expression:

Equivalent mass of H3PO4

5 Molecular mass of H3PO4 2

5 98 2 5 49.

8. How many years would it take to spend Avogadro number 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 Rs

Rate of spending 5 10 lac rupees/s 5 106 Rs/s

Number of years to spend all the rupees

5 6.023 3 1023 Rs 106 3 60 3 60 3 24 3 365 Rs/year 5 1.90988 3 1010 years

9. Oxygen is present in a one litre flask at a pressure of 7.6 3 10 10 mm of Hg. Calculate the number of oxygen molecules in the flask at 0 °C.

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 aluminium in the original mixture?

Solution

Let the mass of aluminium in the sample be ‘A’ g. The mass of Zn 5 (1.67 2 A) g

The volume of H2 at NTP given by Al (i)