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I SBN-93-87444-84-3

CL MEDIA (P) LTD.

Edi t i on : 2018

Revi sed Ed i t i on : 2019

A-45, Mohan Cooperative Industrial Area,Near Mohan Estate Metro Station,New Delhi - 110044

I SBN : 978-93-89310-53-5Typeset by : CL Media DTP Uni t

ABOUT THE AUTHOR

Dr. Manu Kaushal is a renowned faculty of Chemistry and hasteaching experience of more than 10 years. He has worked as asenior faculty with institutes like Aakash Institute (New Delhi) andHelix Institute (Chandigarh). Through this book, he wants to sharehis vast experience with each and every student in India and helpthem prepare the best for their most important exam.

PREFACEStudents appearing for NEET cannot rely on conventional books and measuresto crack the entrance examination. Keeping in mind the cut-throat competition,it becomes mandatory for all the aspirants to have a thorough exposure of alltypes of problems, especially in Chemistry (Part I).

The objective of this book is to help students get rid of any phobia aboutChemistry.

The questions have skillfully been set along with solutions to help studentsdevelop an all round problem solving methodology.

I am thankful to my parents, son 'Yuvraj', daughter 'Neyhal' and my wife whoencouraged me to write this book.

Finally, I would like to thank GK Publications (P) Ltd. for giving me anopportunity to write this book.

Regards

Dr. MANU KAUSHAL

CONTENTS 1. Some Basic Concepts in Chemistry 1.1 – 1.36

2. Structure of Atom 2.1 – 2.24

3. Classification of Elements and Periodicity in Properties 3.1 – 3.18

4. Chemical Bonding and Molecular Structure 4.1 – 4.22

5. States of Matter 5.1 – 5.34

6. Chemical Thermodynamics 6.1 – 6.26

7. Chemical Equilibrium 7.1 – 7.30

8. Ionic Equilibrium 8.1 – 8.26

9. Redox Reactions 9.1 – 9.28

10. Hydrogen 10.1 – 10.14

11. s-block Elements 11.1 – 11.18

12. Some p-block Elements 12.1 – 12.18

13. Organic Chemistry-Some Basic Principles and Techniques (GOC) 13.1 – 13.38

14. Isomerism 14.1 – 14.32

15. Hydrocarbons 15.1 – 15.62

16. Environmental Chemistry 16.1 – 16.6

Questions Based on Calculation of Mole,Molecules, Atomic Mass& Laws ofChemical Combination

1. Match the fol lowing

2

2

2

3

Column I Column I I

a 0.55 mole i 4.48 L of SO at STP

b 0.1 mole i i 12.32 L of H at STP

c 0.2 mole i i i 22 g of COd 0.5 mole iv 2.24 L of NH at STP

(a) a - i , b - i i , c - i i i , d - iv

(b) a - i , b - i i i , c - iv, d - i i

(c) a - i , b - iv, c - i , d - i i i

(d) a - i i i , b - iv, c - i i , d - i

2. H ow many moles of an elect r on weight oneki logram?

(a) 236 10

(b) 311

109.1

(c) 546.02

109.1

(d)

81 109.1 6.02

3. I f water sample are taken from r ivers or lake,then hydrogen & oxygen wil l be in approximaterat io1:8 (by wt) . This statement is in accordancewith which of the fol lowing laws?

(a) Law of reciprocal propor t ion

(b) Law of constant composit ion

(c) Law of mult iple propor t ion

(d) Both 1 & 3

4. Hydrogen and oxygen combine to form H 2O &H 2O containing 5.93% & 11.2% hydrogen (by wt)respect ively. This date i l lust rates–

(a) Law of mult iple propor t ions

(b) Law of constant composit ion

(c) Law of reciprocal propor t ion

(d) Both 1 & 2

5. Which of the following combinat ions il lustrate thelaw of reciprocal propor t ions?

(a) 2 3 2CO,CO ,C O (b) 2 3 2SO ,SO ,H O

(c) 2 2 2H S,SO ,H O (d) Both (a) & (b)

6. At STP, 5.6 l i t res of a gas weighs 60 g. The vapourdensity of the gas is -

(a) 30 (b) 60

(c) 120 (d) 240

Some Basic Conceptsin Chemistry

1CHAPTER

1.2 Some Basic Concepts in Chemistry

7. Let atomic mass of an element be ‘A' grams. Thenthe mass of 10 atoms of element ‘A' in amu is –

(a)A10

(b) 23A

6.02 10

(c) 10 A (d) 2310A

6.02 10

8. Arrange the fol lowing samples in correct order ofno. of molecules –

(1) 100 ml of di-oxygen at STP

(2) 100 ml of di-hydrogen at STP

(3) 100 ml of H 2O (l iquid)

(4) 100 ml of Ozone at STP

(a) 1 = 2 = 3 = 4 (b) 2 > 3 > 1 > 4

(c) 3 > 4 > 1 > 2 (d) 3 > 4 = 1 = 2

9. Which of the fol lowing is cor rect ?

(a) 1gm atom of two different elements weighs1gm each.

(b) 1gm molecule of two different molecules mayhave same weight .

(c) 1amu is reciprocal of Avogadro's No.

(a) Both a & b

(b) Both b & c

(c) a, b & c

(d) Both a & c

10. xgm of element P (at . wt = 15) contains y atoms.How many atoms are contained in 2x g of elementQ (at . wt = 30)

(a) 2y (b) 4y

(c)y2

(d) y

11. 10 moles of CO2 do not contain

(a) 120g of C

(b) 246.02 10 atoms of O

(c) 10NAmolecules of CO2(d) 20gm atoms of O

12. The number of molecules in 16g of CH 4 is–

(a) 233 10 (b) 236 10

7 (c) 23A

16N (6.02 10 ) (d)

2316 103

13. The total number of atoms in 0.1 mole of a

tr iatomic gas is 23A(N 6.02 10 )

(a) 221.8 10 (b) 236 10

(c) 231.8 10 (d) 233.66 10

14. I f 1021 molecules are removed form 200 mg of CO2,then the number of moles of CO2 left are–

(a) 32.88 10 (b) 328.8 10

(c) 30.28 0 (d) 21.66 10

15. Calculate the number of the gram atoms of anelement in 40 kg, i f weight of 1 atom of an

element is 236.644 10 g .

(a) 102 g atoms (b) 10 g atom

(c) 103 g atoms (d) None of these

16. The number of atoms in 558.5 g of Fe (at . wt =55.85 g) is –

(a) Twice that in 60 g C

(b) 226.02 10

(c) Half in 8 g He

(d) A558.5 N

17. Which of the fol lowing contains the gr eatestnumber of atoms?

(a) 1 g of C4H 10 (b) 1g of N2(c) 1g of Ag (d) 1g of H 2O

18. The number of neutr ons in 1.8g of H 2O wi l lapproximately be –

(a) 234.216 10 (b) 238.4 10

(c) 244.8 10 (d) None of these

19. The numbers of atoms present in 52 amu of Heare–

(a) 13 (b) 13 NA

(c) N A (d)AN

6

20. I dent i fy t he cor r ect st at ement among t hefollowing–

(a) 0.5gm atom of oxygen contains AN4

oxygen

molecules.

Some Basic Concepts in Chemistry 1.3

(b) 0.5gm molecules of oxygen contains AN4

oxygen molecules.

(c) Total elect rons present in 0.5g atom of oxygenare 4 NA.

(d) Total elect rons present in 0.5g atom of 2O is

A7N4

.

(a) a & b (b) a & d

(c) b & d (d) a & c

21. Which one of the fol lowing is cor rect for 2 moles

of 3N ?

(a) The mass of 2 moles is 84g.

(b) I t contains 32 NA valence elect rons.

(c) I t contains 21 NA total elect rons.

(d) I t contains 42 NA total protons.

(a) a & c (b) b, c, d

(c) a, b & d (d) Al l of these

22. 11.2L of a gas at STP weighs 14g. The gas couldbe –

(a) N2O (b) NO2(c) N2 (d) NH 3

23. Which of the fol lowing weighs most?

(a) 50g of Fe (b) 5g atoms of nit rogen

(c) 0.1g atom of Ag (d) 1023 atoms of C

24. The number of gram molecules of oxygen in COmolecules is –

(a) 10g molecules (b) 5 g molecules

(c) 1 g molecule (d) 0.5 g molecules

25. What mass of H 2 wi l l contain same number ofatoms as that of 1.6g of O2?

(a) 2gm (b) 0.2g

(c) 0.1gm (d) None of these

26. What is the mass of 1 molecule of oxygen?

(a) 225.314 10 g (b) 23 g3.5 10

(c) 2353.14 10 g (d) 235.314 10 g

27. The number of molecules of H2S contained in 0.40mole of H 2S are–

(a) 232.408 10 (b) 2324.08 10

(c) 202.4 10 (d) 2224.8 10

28. An organic compound on analysis was found tocontain 0.032% . I f i ts molecule contains 2 atomsof S, molecular mass of the compound is-

(a) 200 amu (b) 2000 amu

(c) 20000 amu (d) 200000 amu

29. Which one of the fol lowing is not cor rect for 1mole of CH 4 ?

(a) I t contains 236.02 10 molecules

(b) I t occupies 22.4L at STP

(c) I t weighs 16 gms

(d) I t weighs 16 amu

30. The percentage of Se in peroxide anhydrousenzyme is 0.5% by weight (at . wt 73.4). Then whatis the minimum molecular weight of peroxideanhydrous enzyme?

(a) 14680 amu (b) 29360 amu

(c) 58720 amu (d) None of these

31. The vol ume at STP occupi ed by 14g of N 2, 1.5 moles of CO2 & 1021mol ecul es of O2 gas respectively is–

(a) 11.2 L, 33.6 L, 0.037 L

(b) 33.6 L, 44.8 L, 11.2 L

(c) 44.8 L, 44.8 L, 11.2 L

(d) None of these

32. The rat io of oxygen atoms in 1g O, 1g O2 & 1g O3 is–

(a) 1: 2 : 3 (b) 2: 3 : 1

(c) 3: 2 : 1 (d) 1: 1 : 1

33. The abundances of isotopes 54 56 57Fe, Fe & Fe

are 5%, 90% & 5% respect ively. Then, the at . wt .of Fe is –

(a) 58 (b) 55.95

(c) 68 (d) 59.21

34. How many moles of 3 4 2Mg (PO ) wil l contain 0.25

mole of oxygen atoms?

(a) 22.5 10 (b) 0.02

(c) 23.125 10 (d) 21.25 10


35. The masses of oxygen, which combine with fixedmass of hydr ogen t o for m H 2O & H 2O2respect ively, bears the simple rat io 1:2.

The above statement i l lust rates the –

(a) law of constant composit ion

(b) law of mult iple propor t ion

(c) law of reciprocal propor t ion

(d) law of conservat ion of energy

36. 2g of an oxide of M contains 1g of M, 8g of anotheroxide of M contain 3.2g of M. This date i l lust ratesthe law of –

(a) Reciprocal propor t ions

(b) Constant propor t ion

(c) Conservat ion of energy

(d) mult iple proport ion

37. The total number of elect rons in 180mg of glucoseare –

(a) 2 A9.6 10 N (b) 96 NA

(c) 3 A9.6 10 N (d) 19 A96 10 N

38. P & Q are two elements which form 2 3 2P Q & PQ

molecules. I f 0.15 mole of 2 3 2P Q & PQ weighs

15.9g & 9.3g, respect ively, the atomic masses ofP & Q are, respect ively

(a) 26 & 18

(b) 20 & 32

(c) 28 & 56

(d) 56 & 28

Questions based on limiting Reagent &Stoichiometry

39. I f 1.5 moles of oxygen combine with Al to formAl2O3, then the mass of Al (at mass = 27) used inthe react ion is-

(a) 45 gm (b) 54 gm

(c) 35 gm (d) 22 gm

40. Assuming ful ly decomposed, the volume of CO2released at STP on heat ing 9.85 g of BaCO3 (Ba =137, C = 12, O = 16) wi l l be-

(a) 0.84 L (b) 2.24 L

(c) 4.06 L (d) 1.12 L

41. The mass of C anode consumed (giving only CO2)in the product ion of 270 kg of aluminium metalfrom bauxite by Hall 's process is–

2 3 22Al O 3C 4Al + 3CO

(at mass of Al = 27 gm)

(a) 90 kg (b) 540 kg

(c) 180 kg (d) 270 kg

42. When a mixture of 10 mole of SO2, 15 mole of O2was passed over a catalyst , 8 mole of SO3 wasformed. How many moles of SO2 & O2 did not enterinto combinat ion?

(a) 2 & 11 (b) 3 & 14

(c) 8 & 12 (d) 2 & 5

43. A mixture of N 2 and H 2 is al lowed to react in aclosed container to form NH 3.The react ion ceasesbefore either reactant has been totally consumed.At this stage, 4 moles each of N2, H 2 and NH 3 arepresent .

Which of the fol lowing statements are cor rectregarding above exper iment?

(1) Init ial moles of N 2 were 6.

(2) Init ial moles of H 2 were 10.

(3) Init ial moles of H 2 were 8

(4) For complete react ion to occur, H 2 gas wouldhave been the l imit ing reagent .

(a) 1, 2 & 3 (b) 2, 3 & 4

(c) 1, 2 & 4 (d) 1, 3 & 4

44. Cort isone is a molecular substance containing 21atoms of C per molecule. The mass percentage ofcarbon in cor t isone is 69.98%. The molecularmass of cor t isone (in amu) is–

(a) 176.5 amu (b) 252.2 amu

(c) 287.6 amu (d) 360.1 amu

45. Air contains near ly 20% oxygen by volume. Thevolume of air needed for complete combust ion of100 cc of acetylene wil l be-

(a) 1.25 cc (b) 12.5 cc

(c) 125 cc (d) 1250 cc

46. I f 0.5 mole of BaCl2 is mixed with 0.2 moles ofNa3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed are–

(a) 0.70 (b) 0.50

(c) 0.30 (d) 0.10


47. The volume (in l i t res) of CO2 l iberated at STPwhen 10 g of 90% pure l ime stone (CaCO3) isheated completely is-

(a) 2.016

(b) 20.16

(c) 2.24

(d) 22.4

48. A cer tain compound has the molecular formulaX4O6. I f 10 amu of compound contains 6.06 amuof X, the atomic mass of X is –

(a) 32 amu (b) 37 amu

(c) 42 amu (d) 48 amu

49. 146 g of HCl is neutral ized by x gms of NH 3. Theweight of ammonium sulphate required for theproduct ion of x gms of NH 3 is–

(a) 264 g

(b) 403 g

(c) 528 g

(d) 1056 g

50. Consider the react ions–

3 2 2 2 3Mg N 6H O 3Mg(OH) 2NH

Pt3 2 2

52 NH O 2NO 3H O

2

What amount of 3 2Mg N must be used to produce

sufficient NH 3 so that i t may give 400 g of NO?

(a)400 10

g6

(b)400 3

g10

(c)100 3

g400

(d)400

g100 3

51. An element ‘A' reacts with the compound BO3 toproduce A3O4 & B2O3. The no. of moles of A3O4produced if 1 mole each of A & BO3 are al lowed toreact , is–

(a) 3 (b) 1

(c)13

(d)23

52. When xg C is burnt with y gm oxygen in a closedvessel, no residue is left . Which of the fol lowingis cor r ect r egar ding the r elat ive amounts ofoxygen & carbon?

(a)yx

must be less than 1.33

(b)yx

must be greater than 1.33

(c)yx

must be greater than 2.67

(d)yx

must be between 1.33 & 2.67

53. What volume of 20% by mass of HCl solut ion isrequired to obtain 213g of chlor ine gas from thefol l owing r eact i on (densi t y of HCl solu t ion

= gm

1.20ml

)

4 22KMnO 16HCl 2KCl 2MnCl

2 25Cl 8H O

(a) 2.22 l i t re (b) 3.75 l i t re

(c) 1.46 l i t re (d) None of these

54. Consider air as a 4 : 1 mixture of N2 & O2 byvolume. The mass of O2 in a hall with dimension

5m 5m 5m at STP

(a) 142.8 kg

(b) 160 kg

(c) 35.7 kg

(d) None of these

55. A compound contains 28% N2 & 72% of metal byweight . Three atoms of metal combine with twoatoms of N. The atomic weight of metal is

(a) 24 (b) 12

(c) 36 (d) 26

56. On heat ing 4.9 gm of impure KClO3, i ts mass isreduced by 0.384 gm & conver ted into a gas. Theper cent age of or iginal K ClO3 t hat has beendecomposed is

(At . wt . of K = 39, Cl = 35.5, O = 16)

(a) 20% (b) 30%

(c) 14.7% (d) 12.25%


57. 90 ml of pure dry O2 is subjected to Si lent elect r icdischarge. I f only 10% of i t is conver ted to O3, thevolume of O2 & O3 respect ively in the react ionmixture is

(a) 84 ml & 78 ml (b) 81 ml & 6 ml

(c) 71 ml & 9 ml (d) 90 ml & 6 ml

58. 2 2 3(g) (g) (g)2SO O 2SO

Above react ion is car r ied out by taking 10 molesof SO2 & 15 moles of O2 to give 8 moles of SO3.Then, Choose the cor rect opt ion

(a) O2 is excess reagent

(b) SO2 is l imit ing reagent

(c) Percentage yield of react ion is 80%

(d) All are cor rect .

59. N aBr, used t o pr oduce AgBr for use i nphotography can be self prepared as fol lows

2 2Fe Br FeBr ....(1)

2 2 3 83FeBr Br Fe Br ....(2)

3 8 2 3 2Fe Br 4Na CO 8NaBr 4CO

3 4Fe O ...(3)

I f the yield of 2nd react ions 60% & 3rd react ion is70% t hen mass of Fe r equ i r ed t o pr oduce

32.06 10 kg NaBr is

(a) 105 kg

(b) 107 kg

(c) 103 kg

(d) None of these

60. A mixture of formic acid & oxal ic acid is heatedwith concentrated H 2SO4. The gas produced iscol lected & on i ts t reatment with KOH solut ion,the volume of the gas decreases by one sixth.Then, the molar rat io of two acids in the or iginalmixture is

(a) 2 2 4HCOOH : H C O 4 : 1

(b) 2 2 4H C O : HCOOH 1 : 4

(c) 2 2 4HCOOH : H C O 3 : 1

(d) 2 2 4H C O : HCOOH 1 : 3

61. An oleum is labeled as 109%. The mass % age offree SO3 & H 2SO4 are respectively(a) 60 & 40 (b) 40 & 60

(c) 30 & 70 (d) 70 & 30

62. Consider air as a 4 : 1 mixture of N 2 & O2 byvolume. The mass of O2 in a hal l with dimension

5m 5m 5m at STP is –

(a) 142.8 kg (b) 160 kg

(c) 35.7 kg (d) None of these

63. Reaction 2 2 3N 3H 2NH is used to produce

ammonia. When 450g of H2 is reacted with (excess)N 2, then 1575g of NH 3 is produced.

What is the percentage yield of this react ion?

(a) 41.5% (b) 30.8%

(c) 61.8% (d) 20.7%

64. An ore contains 1.34% of Ag2S by mass. How muchgram of this ore would have to be processed inorder to obtained 1gm Ag?

(at . Mass of Ag = 108 & S = 32)

(a) 74.6 gm

(b) 85.7 gm

(c) 108 gm

(d) 1.14 gm

65. A metal oxide has formula M2O3. I t can be reducedby hydrogen to give free metal and water. I f 0.16g of metal oxide requires 6 mg of hydrogen forcomplete reduct ion, then the atomic wt . of metalis –

(a) 27.90

(b) 159

(c) 79.86

(d) 56

66. 1 Mole of N 2 & 4 mole of H 2 are al lowed to reactin a vessel and 50% of expected moles of NH 3 areformed. Fur ther, H 2O is added to this react ionmixture that dissolves al l NH 3 produced. What isthe mole fract ion of H 2 in the gaseous mixtureafter react ion?

(a)16

(b)56

(c)13

(d)23


67. 3 2CaCO CaO CO

2 2 3 22KOH CO K CO H O

The decomposit ion of a cer tain mass of CaCO3gave 11.2 dm3 of CO2 gas at STP. The mass ofKOH required to completely neutral ize the gasis – (K = 39, Ca = 40)

(a) 28g (b) 20g

(c) 56g (d) 42g

68. What mass of phosphor ous pent oxi de wi l lbe obtained on heat ing 1.33 g phosphorous and5.07 g of O2?

(a) 3.05 g (b) 7 g

(c) 4 g (d) 9 g

69. A bangle made up of Cu – Zn al loy weighs 10g. I tis accident ly dropped into a beaker containingsulphur ic acid & 224 ml of H 2 gas is evolved atSTP. The percentage of zinc (by weight) in thebangle, i f only zinc reacts with , is –

2 4 4 2Zn H SO ZnSO H

(At . wt . of Zn = 65, At . wt . of Cu =63.5)

(a) 22.4%

(b) 6.5%

(c) 65%

(d) None of these

70. 4 2CaSO x H O looses 6.2% of i t by weight uponheat ing. The value of x wi l l be–

(M. wt . of CaSO4 = 136)

(a) 4

(b)12

(c) 2

(d) 3

71. Hydrated salt 2 3 2Na CO x H O undergoes 63% lossin mass on heat ing and becomes anhydrous. Thevalue of x is –

(Na2CO3 = 106)

(a) 10

(b) 12

(c) 8

(d) 18

72. Potassium chlorate decomposed on heat ing as–

3 22KClO 2KCl 3O

How much wil l mass of 80% pure KClO3 produce48g of oxygen? (at . mass of K = 39g)

(a) 98g sample (b) 153.12g sample

(c) 245g sample (d) 122.5g sample

Equivalent Weight & Molecular Mass

73. The equivalent weight of H 3PO4 in the fol lowingreact ion is–

3 4 2 4 2H PO Ca (OH) CaHPO 2H O

(a) 98 (b) 49

(c) 32.66 (d) 40

74. An oxide of metal has 20% oxygen (by wt .), theequivalent wt . of metal and metal oxide are,respect ively

(a) 32, 40

(b) 30, 46

(c) 38, 42

(d) 38, 38

75. When a metal is burnt , i ts weight is increased by24%. Calculate the equivalent wt . of the metal.

(a) 67.7

(b) 33.3

(c) 12.25

(d) 69.91

76. Equivalent weight of a bivalent metal is 37.2. Themolecular weight of i ts chlor ide is-

(a) 15.87 (b) 145.4

(c) 22.35 (d) 120.7

77. The equivalent weight of an element is 4. I t schlor ide has a vapour density 59.25. Then thevalency of the element is –

(a) 4 (b) 3

(c) 2 (d) 1

78. 0.16 g of dibasic acid required 25 ml of decinormalNaOH solut ion for complete neutral izat ion. Themolecular weight of the acid wi l l be –

(a) 32

(b) 64

(c) 128

(d) 256


79. Two oxides of a metal M contain 40% & 50% ofthe metal respect ively. I f the formula of the firstoxide is M 2O3, then the formula of the secondoxide is–

(a) MO (b) MO2(c) M3O4 (d) M2O5

80. The chlor ide of a metal has 52.8% chlor ine bymass. The equivalent wt . of metal is -

(a) 31.7 (b) 35.2

(c) 63.5 (d) 20

Questions Based on Concentration Terms

81. What is normali ty of 1M solut ion of H 3PO4?

(a) 3N (b) 2N

(c) 4N (d) 1N

82. A mixture of ethanol & water contains 54% waterby mass. Calculate the mole fract ion of alcohol inthis solut ion.

(a) 0.25 (b) 0.95

(c) 1.65 (d) 3.25

83. 214.2g of (sucrose) sugar syrup contains 34.2g ofsugar. The molal i ty of the solut ion is -

(Formula of sucrose is C12H 22O11)

(a) 0.55 (b) 1.5

(c) 6.6 (d) 4.72

84. How much H 2O should be added to 200 c.c. ofsemi-normal solut ion of NaOH to make is exact lydecinormal ?

(a) 200 cc (b) 400 cc

(c) 600 cc (d) 800 cc

85. 25 ml of 3M HNO3 is mixed with 75 ml of 4MHNO3. Then, the molar i ty of the final mixtureis–

(a) 3.75M (b) 2.62M

(c) 4.31M (d) 5M

86. Molar i ty of 20% of H 2SO4 by mass whose densityis 1.02g /cc is –

(a) 8.4M (b) 2.08M

(c) 7.28M (d) 1.7M

87. What is the mass % of C in CO2?

(a) 0.034% (b) 27.27%(c) 3.4% (d) 28.7%

88. The density of 3 molal solut ion of NaOH is 1.110g/ml. The molar i ty is

(a) 2.97 M (b) 5 M

(c) 6 M (d) 9 M

89. Which of the fol lowing terms are unit less?

(1) molali ty (2) molar ity

(3) mole fract ion (4) mass percent

(a) 1 & 2 (b) 3 & 4

(c) 1 & 4 (d) 2 & 3

90. I f 250 ml of a solut ion contains 24.5g H 2SO4 ,calculate the molar ity & normality of the solut ion.

(a) 1M, 2N (b) 2M, 1N

(c) 3M, 3N (d) 3M, 1N

91. Find the density of an aqueous H 3PO4 solut ionwhose weight % age is 9.8% & molar i ty is 1.25.

(a) 1.25 g/ml (b) 2.5 g/ml

(c) 1.5 g/ml (d) 2 g/ml

92. When 644g of 2 4 2Na SO H O is dissolved in 4000

ml of aqueous solut ion, the molar i ty of solut ionwas found to be 0.5 M. The value of x is –

(a) 5 (b) 10

(c) 8 (d) 2

93. Which of the fol lowing graphs is NOT correct lyrepresented?

(a)

Moles of

solute

Temperature

(b)

Temperature

Molal ity

(c)

Dilution

Molefraction

of solvent

(d) Al l are cor rect


94. The mole fr action of a solut e in a solut ion is 0.1. At 298 K, molar i ty of the solut ion is same as i ts molal i ty. Densi ty of thi s solut ion is 2g cm– 3. The rat io of molecular weight s of solut e & solvent is-

(a) 5 : 1

(b) 7 : 2

(c) 9 : 1

(d) None of these

95. What i s t he volume r equi r ed of a 20% H Cl

solut ion of density 1.20 g/ml to prepare 363 g of

AsCl3, according to the fol lowing equat ions–

(As =75, Cl = 3.55)

2 KMnO4 +16HCl →2KCl +2MnCl 2

2 25Cl 8H O

2 32As 3Cl 2 AsCl

(a) 2.56 L

(b) 0.73 L

(c) 1.46 L

(d) 2.9 L

96. I n 1200g solut ion, 12g urea is present . I f Densityof solut ion = 1.2 g/ml, then molar i ty of solut ionof

(a) 0.2 M (b) 10 M

(c) 0.167 M (d) 12 M

97. Upon heat ing a l i t re of M2

HCl solut ion, 2.675g

of HCl is lost due to evaporat ion & the volume ofthe solut ion shr inks to 750 ml. Then, the molar i tyof result ing solut ion is X Molar & the number ofmil l imoles of HCl in 100 ml of final solut ion arey. Then X & Y are, respect ively

(a) 0.569, 56.9

(b) 0.769, 76.9

(c) 0.70 & 7

(d) 0.23 & 23

98. How many grams of Conc. HNO3 solut ion shouldbe used to prepare 250 ml of 2 M HNO3?

Conc. Acid is 70% (w/w) HNO3(a) 90g (b) 70g

(c) 54g (d) 45g

99. What is the concentrat ion of nit rate ions i f equalvolumes of 0.1M AgNO3 & 0.1M NaCl are mixedtogether?

(a) 0.1M

(b) 0.2M

(c) 0.05M

(d) 0.25M

100. I f the molal i ty of an aqueous solut ion is 0.111,then the mole fract ion of solute wi l l be -

(a) 0.02 (b) 0.002

(c) 0.2 (d) 0.1

101. A compound contains 28% N2 & 72% of a metal byweight . 3 atoms of metal combined with 2 atomsof N. The atomic weight of metal is -

(a) 24

(b) 12

(c) 36

(d) 48

Empirical Formula & Molecular Formula

102. Empir ical formula for CH 3COOH is same as thatfor –

(a) C5H5O5(b) C12H22O11(c) C6H12O6(d) None of these

103. A compound contains C = 40%, H = 6.67% & restis oxygen. The empir ical formula of the compoundis –

(a) C12H22O11 (b) CH2O

(c) C2H2O (d) CH3O

104. The empir ical formula of an organic compoundcontaining carbon & hydrogen is CH 2. The massof 1 l i t re of this organic gas is exact ly equal tothat of 1 l i t re of N2 at STP. So, molecular formulaof this gas –

(a) C2H4 (b) C3H6(c) C6H12 (d) C4H8

105. A gaseous hydrocarbon gives upon combust ion0.72 g of water & 3.08 g of CO2. The empir icalformula of the compound is

(a) C3H4 (b) C6H5(c) C9H8 (d) C2H4


106. 0.2012g of an organic compound containingcarbon, hydrogen & oxygen, gave on completecombust ion, 0.4431g CO2 & 0.1462g of H 2O. Themolecular weight of the compound is 100g. Then,the molecular formula of the compound is–

(a) CaCO3(b) C5H8O2(c) C4H20O2(d) None of these

Miscellaneous

107. Consider the given react ion

4 2 7 2 2 2 7 2H P O 2NaOH Na H P O 2H O

I f 534 gm of H 4P2O7 i s reacted wi th 233 10

mol ecu l es of N aOH , t hen t ot al number ofmolecules produced in the product are

(a) A2.5 N

(b) A5 N

(c) A7.5 N

(d) A75 N

108. Equivalent weight of a metal is 20. I f the metalforms a t r iposit ive ion, i ts atomic weight wi l l be

(a) 20

(b) 40

(c) 60

(d) None of these

109. 2g of a metal carbonate on decomposit ion gave1.5 g of metal l ic oxide. The equivalent mass ofmetal is

(a) 58 (b) 29

(c) 5.8 (d) 2.9

110. The vapour density of a tet rabasic acid is x. Theequivalent mass of the acid is

(a)2x

(b)3x

(c)23x

(d)32x

111. 0.36 gm of a metal was dissolved in 100 ml of 1NH 2SO4 & the volume of solut ion was made upto1000 ml by adding water. 40 ml of this solut ionwas neutral ized completely by 28 ml of 0.1 NNaOH. The equivalent mass of the metal is

(a) 3 (b) 6

(c) 12 (d) 24

112. V1 ml of NaOH of molar ity X & V2 ml of of Ba(OH)2molar i ty Y are mixed together. The mixture iscompletely neutral ized by 100 ml of 0.1 N HCl. I f

1

2

V 1V 4

& X

4Y

. Then, what fract ion of acid is

neutralized by Ba(OH)2?

(a) 0.5

(b) 0.25

(c) 0.33

(d) 0.67

113. What volume of 75% H 2SO4 by mass is requiredto prepare 1.5 l i t res of 0.2 M H 2SO4? (Density ofthe sample is 1.8 g/ml)

(a) 21.7 ml (b) 33 ml

(c) 92 ml (d) 56 ml

114. ‘X' grams of CaCO3 was completely burnt in air.The weight of sol id residue formed is 28g. Whatis the value of ‘X'? (in grams)

(a) 50 (b) 75

(c) 60 (d) 95

115. 12.5 ml of a solut ion containing 6 grams of dibasicacid in one l i t re was found to be neutral ized by10 ml of decinormal solut ion of NaOH, then themolecular mass of acid is

(a) 150 g (b) 257 g

(c) 267 g (d) 277 g

116. When NA is avogadro's number then the numberof oxygen atoms in one gm. equivalent of oxygenis

(a) N A

(b) AN2

(c) AN2

(d) A2 N


117. What is the col lect ive wt . (in gms) of 236.02 10

molecules of N 2 & 233.10 10 molecules of SO2?

(a) 30g

(b) 32g

(c) 34g

(d) 60g

118. ‘X' l i t res of Carbon monoxide is present at STP. I tis completely oxidized to CO2. The volume of CO2formed is l i t res of STP. What is the value of ‘X'in l i t res?

(a) 11.207

(b) 34.4

(c) 19.9

(d) 27.7

119. The density of a solut ion containing x % by massH 2SO4 is y (g/ml). Then i ts normali ty is

(a)10

98xy

(b)20

98 xy

(c)100 1

98 98

xy

(d) Both (b) & (c)

120. 0.45 N and 0.6 N NaOH solut ion are mixed in 2 :1 by volume. The amount of solute present in 1l i t re of this solut ion is

(a) 20g (b) 75g

(c) 45g (d) 40 g

121. The density of 3M aq. 2 2 3Na S O solut ion is 1.25

g/ml. Then, ident i fy the cor rect statements

(a) % age by wt . of 2 2 3Na S O is 37.92

(b) The moles fract ion of 2 2 3Na S O is 0.065

(c) Molar i ty of Na+ is 6 & that of 22 3S O is 3

(d) Al l of these

122. A metal M of equivalent mass E forms an oxide ofmolecular formula M xOy. The atomic mas of metalis given by

(a) 2E

yx (b) E

xy

(c)Ey

(d) None of these

123. 0.1 mole of a carbohydrate with empir ical formulaCH 2O has 1g of hydrogen. What is i ts molecularformula?

(a) C5H10O5 (b) C6H12O6(c) C12H22O11 (d) None of these


1. (c) 2. (d) 3. (d) 4. (a) 5. (c) 6. (c) 7. (c) 8. (d) 9. (b) 10. (d)

11. (b) 12. (b) 13. (c) 14. (a) 15. (c) 16. (a) 17. (a)* 18. (d) 19. (a) 20. (d)

21. (c) 22. (c) 23. (b)* 24. (b) 25. (c) 26. (d) 27. (a) 28. (d) 29. (d) 30. (a)

31. (a) 32. (d) 33. (b) 34. (c) 35. (b) 36. (d) 37. (a) 38. (a) 39. (b) 40. (d)

41. (a) 42. (a) 43. (c) 44. (d) 45. (d) 46. (d) 47. (a) 48. (b) 49. (a) 50. (a)

51. (c) 52. (b) 53. (c) 54. (c) 55. (a) 56. (a) 57. (b) 58. (* ) 59. (c) 60. (a)

61. (b) 62. (c) 63. (c) 64. (b) 65. (d) 66. (b) 67. (c) 68. (a) 69. (b) 70. (b)

71. (a) 72. (b) 73. (b) 74. (a) 75. (b) 76. (b) 77. (b) 78. (c) 79. (a) 80. (a)

81. (a) 82. (a) 83. (a) 84. (d) 85. (a) 86. (b) 87. (b) 88. (a) 89. (b) 90. (a)*

91. (a) 92. (b) 93. (c) 94. (b) 95. (c) 96. (a) 97. (a) 98. (d) 99. (c) 100. (b)

101. (a) 102. (c) 103. (b) 104. (a) 105. (c) 106. (b) 107. (c) 108. (c) 109. (a) 110. (a)

111. (c) 112. (d) 113. (a) 114. (a) 115. (a) 116. (b) 117. (d) 118. (a) 119. (d) 120. (a)

121. (d) 122. (a) 123. (a)


1. (c) Moles of 24 48

SO 0 222 4

Moles of 212 32

H 0.5522.4

Moles of 222

CO 0.544

Moles of 32.24

NH 0.122.4

2. (d) 319.1 10 kg weight of 1 elect ron

311

1 kg elect rons109.1

311

109.1 elect ron means 31

A

1109.1 moles

N

31 231 1

10 109.1 6

81

6 109.1

10

6.02

819.1

3. (d)

4. (a) H 2O2 has 5.93% hydrogen by wt . So, 5.93 gmof hydrogen is present in 100 gm H 2O2.

Oxygen present in 100 gm 2 2H O 94 07g .

5.93g H combined with O = 94.07

1g(fixed wt.of H) H combined with O

94 075 93

….. (i )

H 2O has 11.2% hydrogen by wt . So, 11.2 gmhydrogen is present in 100 gm H 2O

So, 11.2 gm hydrogen is present in 100 gm H 2O.

Oxygen is present in 100 gm = 100 – 11.2

= 88.8 g

H O

11 2g 88 8g88 8

1g g11 2

....(i i )

(Fixed wt . of H)

Rat io of weights of oxygen 94 075 9388 811 2

94 07 11 22 : 1

5 93 88 8

5. (c)

6. (c) At STP

5.6 L of gas weighs = 60g

22.4 L of gas weighs 605.6

4

22.4

= 240g

M. mass of gas = 240g

Now, 2 V.D. M.mass

240V.D 120

2

7. (c) 236.02 10 atoms of element weigh = A gms

1 atoms of element weigh 23A

gms6.02 10

10 atoms of element weigh

23A

10gms6.02 10

Now, 1 amu A

1gms

N

So, mass of 10 elements

2323

A10 6.02 10 amu 10 A amu

6.02 10


8. (d) 100 ml O2 (at STP)

No. of moles 100

22400

Molecules AN224

100 ml O3 (at STP)

No. of moles 100

22400

Molecules AN224

100 ml H 2 (at STP)

No. of moles 100

22400

Molecules AN224

H2O

of l iquid H 2O = 1g/ml

1ml 1g

100 ml 100g

Now,

2 A18g H O N molecules

A2 A

N100g H O 100 5.55N

18

9. (b) N 2O = 44gm, CO2 = 44gm

1gm molecule of both these gases have sameweight .

10. (d) Moles of P =15x

Moles of Q 230 15

x x

So, both wil l contain same no. of atoms.

11. (b) CO21 mole of CO2 has C = 12g

10 mole of CO2 has C =12 10 120g

CO21 mole of CO2 has O atoms = 2 moles

10 mole of CO2 has O atoms 2 10 20 moles

2320 6.20 10

2412 10

CO21 mole CO2 has molecules = NA

10 mole CO2 has molecules A10 N

Also, 1 mole CO2 has O = 2 mole atoms of O

10 mole CO2 has O = 20 mole atoms of O

= 20gm atoms of O

12. (b) 16gm (molar mass of CH 4 ) has molecules

236 10

13. (c) 1 mole of a t r iatomic gas, contains molecules

= NA0.1 mole of a t r iatomic gas, contains molecules

AN 0.1

0.1 mole of a t r iatomic gas, contains atoms

AN 0.1 3

230.3 6 10

231.8 10

14. (a) Conver t ing 200 mg CO2 into moles

Numbers of molesW 200M

3

10010

44

22

1110 0.045 10

22

Conver t ing 1021 molecules into moles

212

2310

n 0.166 106 10

1 2No,of molesleft 0.045 10 0.166 10

1 110 (0.045 0.166 10 )

110 (0.045 0.0166)

10.028 10

1 328 10 2.8 101000


15. (c) Weight of 1 atom of an element

236.644 10 g

Weight of NA atom of an element

23 236.644 10 6 10 g

23 2340 10 10 g

= 40 g

atomic weight = 40 g

No. of gm atoms (No. of moles of atoms)

340000 10 g atoms40

16. (a) No. of atoms in 55.85 g Fe = NA

No. of atoms in 558.5 g Fe AN

558.555.85

AN 10

246.02 10

Atoms in 60 g C

A12g C N atoms

AN60gC12

5

60 A5 N

Now, twice of 24A5 N 6.02 10

17. (a) 4 10C H

Molar mass = 58g

58g 4 10C H , contains molecules = NA

58g 4 10C H , contains atoms = AN 14

1g 4 10C H , contains atoms

AA

N 140.241 N

58

(b) N2Molar mass = 28g

28g N 2, contains molecules = NA28g, N 2, contains atoms = NA × 2

1g, N 2, contains atoms AN 228

AA

N0.0714 N

14

(c) Ag

M. mass = 108g

108g Ag, contains atoms = NA

1g Ag, contains atoms A AN

0.009 N108

(d) H2O

M. mass = 18g

18g H 2O, contains molecules = NA

18g H 2O, contains atoms = AN 3

1g H 2O, contains atoms A

AN 3

0.166 N18

18. (d) 18g H 2O has molecules = NA

1.8g H 2O has molecules AN 1.8

18

226.02 10 molecules

Now,

1 H 2O molecule has neutrons = 8

2226.02 10 H O molecule has neutrons

22=8 6.02 10

2248 10

19. (a) 4amu 1atom of He

152amu 52 13atomsof He

4

20. (d) 0.5g atom of oxygen means 0.5 moles of atomsof oxygen.

1 atom of oxygen 21

moleculesO2

A A1

0.5 N atom of oxygen 0.5 N2

A2

Nmolecules of O

4

A0.5gm atom ( 0.5 N atoms)

1 atom of O 8 elect rons

A A0.5 N atom of O 8 0.5 N

A4 N electrons


21. (d) 1 mole of 3N weighs = 42g

2 mole of 3N weighs 42 2 84g

1 molecule 3N , contains valencee e– = 16

A2 N

(2mole)

molecule 3N

, contains valence e–

A A16 2 N 32 N

1 molecule 3N , contains total e– = 22

A2 N molecule 3N , contains total e–

A A22 2 N 44 N

1 molecule 3N , contains total protons = 21

A2 N molecule 3N , contains total protons

A21 2 N

A42 N

22. (c) 11.2 L gas weighs = 14g

22.4 L gas weighs 214

22.4 28g 11.2

So, molar mass of gas is 28g

So, it is N2.23. (b) (a) 50g of Fe

(b) 1g atom (1 mole of atoms) of nit rogen weighs= 14g

5g atom (5 mole of atoms) of nit rogen weighs

14 5 = 70g

(c) 1g atom (1 mole of atom) of Ag weighs = 108g

0.1g at om (0.1 mol e of at om) of Ag wei ghs108

0.11

= 10.8g

(d) 236 10 atoms of C, weigh = 12g

1023 atoms of C, weigh 23

2312

106 10

= 2g

24. (b) 1 CO molecule has O = 1 atom

1 CO molecule has oxygen molecule 12

246.02 10 CO molecule has oxygen molecule

241 6.02 10 molecule2

To form gram molecules

No. of moles of molecules

24

23

6.02 102

6.02 101

= 5g molecules

25. (c) ATOMS IN 1.6gm of O2

32gm O2 = AN molecules of O2

32gm O2 = A2 N atoms of oxygen

1.6gm O2 A2 N 1.6

322

10

226.02 10 of oxygenatoms

A 22 N atomsarein H 2g

2226.02 10 atomsarein H

2

2 6.02

236.02

10

2210 0.1g

26. (d) NA molecules of O2, weigh = 32g

1 molecules of O2, weigh 2332

g6 10

27. (a) 1 moles of H 2S = NA molecules of H 2S

0.40 mole of H 2S = NA × 0.4 molecules of H 2S

= 6 × 1023 × 0.4 molecules of H 2S

= 2.4 × 1023

28. (d) 0.032 amu S is present in molecule = 100 amu

2 32 amu S is present in molecule

1002 32 1000

0.032

= 200000 amu

29. (d) 1 mole of CH 4 weighs 16 gms

1 molecule of CH 4 weighs 16 amu


30. (a) Minimum molecular weight of enzyme wil l berecorded when the amount of Se is minimum init i .e. 1 atom of Se.

0.5 amu of Se is present in enzyme = 100 amu

73.4 amu of Se is present in enzyme

10073.4amu

0.5

31.1428

(a) 14g N2 means moles = =0.5mole

1 mole occupied volume at STP = 22.4 L

0.5 mole occupied volume at STP 22.4 0.5

= 11.2 L

1.5 moles of CO21 mole of CO2 occupies vol. (at STP) = 22.4 L

1.5 mole of CO2 occupies vol. (at STP) 22.4 1.5

= 33.6 L

1021 Molecules of O2No. of moles of O2

212

2310

0.166 10 0.001666 10

1 mole of O2 = Vol.

22.4 L

0.00166 mole of O2

= 22.4 0.00166 0.037 li t res

32. (d) 16g O has atoms = NA

1g O has atoms AN

atoms16

32 gm O2 has molecules = NA

32 gm O2 has atoms A=2 N O atoms

1 gm O2 has atoms A2 N

32

A

NO atoms

1616

48g O3 has molecules of O3 = NA48g O3 has atoms of O3 = A3 N

1g O3 has atoms of O3 A3 N

48

AN O atoms16

So, rat io = 1: 1 : 1

33. (b) Av. Atomic mass 1 1 2 2 3 3A x A x A x

100

54 5 56 90 57 5100

270 5040 285100

34. (c) 8 oxygen atom are in 1 molecule of

3 4 2Mg (PO )

A0 25 N Oxygen atom are in

A1

0 25 N8

molecules

moles of A3 4 2

A

0 25 NMg (PO )

8N

A

A

0 25 N 18 N

0 258

20 03125 3 125 10

35. (b)

36. (d)

37. (a) Glucose has the formula C6H 12O6Now, each C has e– = 6

each H has e– = 1

each O has e– = 8

Total e– in one molecule C6H 12O6= 36 + 12 + 48 = 96

Total e– in 180g glucose A96 N

Total – in 3180 10 g glucose

3A96 N 180 10180

A0.096 N

2A9.6 10 N


38. (a) 0.15 moles of P2Q3 weight = 15.9 gm

1 moles of P2Q3 weight 15 9

106gm0 15

2 3So, M. Mass o P Q 1 mf 06g

0.15 moles of PQ2 weight = 9.3 gm

1 moles of P2Q3 weight 9.3

62 gm0.15

2So, M. Mass PQ mof 62g

Now,

2× At.wt. of P + 3 × At . wt . of Q = 106At .wt . of P + 2 × At. wt . of Q = 62

Let at .wt . of P = x

& at .wt . of Q = y

2x + 3y = 106 ....(1)

x + 2y = 62 ....(2)

From 2, x = 62 – 2y

Putt ing value in 1 –

2 (62 – 2y) + 3y = 106

124 – 4y + 3y = 106

124 – y = 106

18 y

26x

39. (b) 2 2 33

2Al O Al O2

2O Al

1.5 moles of O2 combine with Al = 2 moles

2 27 54gm

40. (d) 3 2BaCO BaO + CO

BaCO3, M.Mass = 197 gm

197 gm BaCO3 l iberates CO2 = 1 mole (= 22.4 L atSTP)

9.85 gmBaCO3 l iberates CO2 22.4

9.85L197

= 1.12 L

41. (a) 2 3 22Al O 3C 4Al + 3CO

To produce 4 mole Al, C consumed = 3 moles

To produce 4 27 gm Al, C consumed

= 3 12 gm

To produce 270000 gm Al, C consumed

3 12

4

4 2793

90000gm270000

= 90 kg

42. (a) 2 2 32SO O 2SO

g g(g) ( ) ( )

At t 0 10 15 0

At t t 10 152

x

x x

= 8 x Moles of SO3 formed

No. of moles of SO2 which did not react

= 10 – x = 10 – 8 = 2 molesNo. of moles of O2 which did not react

x15

2

815

4

211moles

43. (c)2 2 3N 3H 2NH

(g) (g) (g)

At t 0 a b 0

At t t a b 3 2

x x x

Moles of N 2 at t ime, t = a – x = 4_____ (1)

Moles of H 2 at t ime, t = b – 3x = 4_____ (2)

Moles of NH 3 at t ime, t = 2x = 4______ (3)

From (3),

2x = 4

x = 2


Putt ing value of x in (1)

a – x =4

a – 2 = 4

a = 6

Put t ing value of x in equat ion (2)

b – 3x = 4

b – 6 = 4

b = 10

44. (d) 69.98 amu of C is present in cor t isone

= 100 amu

21 12 amu of C is present in cor t isone

10021 12

69.98

= 360.1 amu

45. (d)

2 2 2 2 2

Acetylene

(g) (g) (g) ( )l

5C H O 2CO H O

2

To burn 1 vol. acetylene, O2 required = 2.5 vol.

OR, To burn 1 cc acetylene, O2 required =2.5 vol.

To burn (100 vol.) acetylene, O2 required

2.5 100

= 250 vol

= 250 ml

(or 250 cc)

Now,

20 vol. O2 comes from air = 100 vol.

250 vol. O2 comes from air

100

5

20250 1250 vol.

46. (d) 43 242 33BaCl 2Na Ba )PO O(P

3 4Na PO 2BaCl

2 moles react with 3 moles of BaCl20.2 moles react with

So, BaCl2 left = 0.5 – 0.3 = 0.2 moles

BaCl2 is excess reagent

Na3PO4 is LIMITING REAGENT

Now,

3 4Na PO 43 2PB Oa ( )

2 moles react to form 1 mole

0.2 moles react to form 1

0.22

= 0.1 mole of 43 2PB Oa ( ) is formed.

47. (a)3 2CaCO CaO + CO

(s) (s) (g)

CaCO3 taken = 10g

But i ts percentage pur i ty = 90%

Pure CaCO3 90

=10100

9gm

moles of pure CaCO3 9

0.09mole100

So, moles of CO2 l iberated = 0.09

Volume of CO2 l iberated (at STP)

=0.09 22.4 = 2.016L

48. (b) Let atomic mass of X = y amu

4 6(4y 96)amu X O hasX 4y amu

4 6

4y 1010amu X O hasX

4y +96

4y 106.06

4y +96

40y = 6.06(4y +96)

40y 24.24y + 581.76

40y 24.24y = 581.76

15.76y = 581.76

581.76y 36.91 37amu

15.76


49. (a) 3 4HCl NH NH Cl

146moles 4

36 5

So, moles of NH 3 reqd. = 4

wt of NH3 reqd. 4 17 68g

4 2 4 3 2 4(NH ) SO 2NH H SO

Now, 32 17g NH is produced by 4 2 4(NH ) SO

= 132 g

68g NH 3 is produced by

4 2 4

132(NH ) SO 68 264 g

34

50. (a) Pt3 2 25

2NH O 2NO 3H O2

n (moles) of NO 40030

So, NH 3 reqd. 403

moles

Now, 3 2 2 2 3Mg N 6H O 3Mg (OH) 2NH

2 mole NH 3 are produce from 3 2Mg N 1mole

3

40 1 40 20mole NH moles

3 2 3 3

3 2Mg N 24 3 28

= 72 + 80 = 100g

wt. of 3 220

Mg N reqd. 1003

51. (c)(1) (1)

3 3 4 2 39A 8BO 3A O 4B O

3BO

9 moles 8 m les

A

o

81 moles moles 0 88 moles

9

‘A' is the l imit ing Reagent .Now,9 moles of A gives A3O4 moles = 3

1 moles of A gives 3 43 1

A O moles moles9 3

52. (b) 2 2gm gmC O CO

x y

2OC12gm 32gm

32gm 2 63gm

12 x x

y = 2.67 x

1 33 yx

53. (c) Now,

4 22KMnO 16HCl 2KCl 2MnCl

2 25Cl 8H O

2Cl HCl

5 71gm 16 36 5g

16 36 5213gm 213 350 4g

5 71

HCl avai lable = 20% (w/w)

20gm HCl is obtained from solut ion = 100 gm

350.4 gmHCl is obtained from solut ion

100350 4

20

wt. of soln. = 1752 gm

reqd.

Now, of soln. = 1.20 g/ml

1.20g soln. has vol. = 1 ml

1752 g soln. has vol. 1

17521 20

= 1.46 l i t res

54. (c) Let volume occupied by O2 = n

Volume occupied by N 2 = 4n

Total volume of the hal l = 125 m3

Now,

n + 4n = 125 m3

5n = 125 m3

3n 25 m


10– 3 m3 = 1L

33

125m 25 25000L

10

At STP,

222.4L 32g O

3225000L 25,000g

22.4

= 35714 g

= 35.71 kg

55. (a) Formula of metal nit r ide wil l be 3 2M N .

Let atomic wt . of metal be yamu.

So, % age by wt . of metal 3y

100 723y 28

300y = 216y + 72 × 28

72 2824amu

84

y

56. (a) 3 22 KClO 2KCl + 3O

Whatever KCI & O2 is generated, i t wi l l comefrom pure KClO3.

Mass of KClO3 conver ted into 2O 0.384 gm

Moles of O2 formed 0.384

= 0.012 moles32

According to Stoichiometry,

3 moles O2 are formed from KClO3 = 2 moles

0.012 moles O2 are formed from

32

KClO 0.0123

moles of pure 32

KClO 0.0123

Wt. of pure 32

KClO 0.012 122.53

= 0.98 g

% age of pure 30.98

KClO 100 20%4.9

57. (b) 2 3(g) (g)3O 2O

At t = 0 90 vol 0

At t = t1090

90100

2 10

903 100

Vol. of O2 1090

= 90 90 9 81 ml100

Vol. of 32 10

O = 90 6 ml3 100

58. 2 2 3(g)(g) (g)15moles

2SO O 2SO

10moles

According to Stoichiometry, if 2 moles of SO2 reactthen moles of O2 reacted = 1

I f 10 moles of SO2 react then moles of

21

O2

105

reacted = 5

O2 left = 10 moles

O2 is excess regent

SO2 is l imit ing reagent

I f 2 moles SO2 react , moles of SO3 formed = 2

I f 10 moles SO2 react , moles of SO3 formed = 10

Maximum moles of SO3 formed = 10

According to quest ion, moles of SO3 formed = 8

So % age yield 8

100 80%10

59. (c) Moles of NaBr reqd. 3 32.06 10 10

=104

44206 10= 1.98 10 moles

104

3 9Fe Br & NaBr are in the rat io 1 : 8 (by moles)

To for m 41 98 10 moles of NaBr, mol es of

Fe3Br 8 reqd 41 1 98 10

8

= 40 247 10 moles


But yield of react ion is 70%, So moles of Fe3Br 8

reqd. wi l l be more than 40 247 10 .

Let x moles of Fe3Br 8 are reqd.

470 0 247 10100

x

40.247 10 10070

x

60 247 1070

60 0035 10 moles

Now, from react ion 2 –

To form 60 0035 10 moles of 3 8Fe Br moles of

2FeBr required

63 0 0036 10

60 0106 10 moles

But its yield is 60%, so actual FeBr2 moles requiredwil l be y –

660 0 0106 10100

y

60 0106 10100

60

y

80 000176 10 moles

I f FeBr 2 required is 80 000176 10 moles , Fe

required Wil l be 80 000176 10 moles

So, Fe reqd. 8=0 000176 10 moles

wt. of Fe reqd. 8=0 000176 10 56

899 10 g10000

499 10 g

4100 10 g

610 g

3106 10 kg1000

60. (a) Let moles of HCOOH & H 2C2O4 are a & brespect ively

H SO2

42HCOOH H O CO

a 0 0

0 a a

H SO2 2 4 2 2

42H C O H O CO CO

b 0 0 0

0 b b b

H 2O is absorbed by H 2SO4 & CO2 is absorbed bykOH

So, 2

2

vol.of COTotal vol.of (CO CO )

1 b6 a b b

a4

b

Molar rat io 2 2 4HCOOH & H C O 4 : 1

61. (b) I f oleum is labeled as 109%.

I t means 9gm water is added in 100g oleum. Letus calculate how much SO3 can be conver ted intoH 2SO4 by 9gm of water

3 2 2 4SO H O H SO

80g 18g 98g

i .e., 9g H 2O can dissolve 40 g SO3 to form 49 gH 2SO4.

mass of SO3 in 100g oleum = 40g

% by mass of 3SO 40%

% by mass of 2 4H SO 60%

62. (c) Let volume occupied by O2 = x

Volume occupied by N 2 = 4x

Total volume of the hal l = 125m3

Now, x + 4x =125m3

5x = 125n3

x = 25m3

3 310 m 1L

3

31

25m 25 25000L10

01 final (1-36) (a).p6502 final (1-24) (a).p6503 final (1-18) (a).p6504 final (1-22) (a).p6505 final (1-34) (a).p6506 final (1-26) ok07 final (1-30) ok.p6508 final (1-26) ok.p6509 final (1-28) ok.p6510 final (1-14) ok.p6511 final (1-18) ok.p6512 final (1-18) ok.p6513 final (1-38) ok.p6514 final (1-32) ok15 final (1-62) ok.p6516 final (1-6) ok

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