Practice Book of Chemistry for Jee Main

Atomic Number, Mass Number and IsotopesAll atoms can be identified by the number of protons and neutrons they contain. The

atomic number (Z) is the number of protons in the nucleus of each atom of an element. In

a neutral atom, the number of protons is equal to the number of electrons present in the

atom.

The chemical identity of an atom can be determined solely from its atomic number. For

example atomic number of oxygen is 8. This means neutral oxygen atom has 8 protons

and 8 electrons.

The mass number (A) is the total number of neutrons and protons present in the nucleus

of an atom of an element. Except for hydrogen (which has only one proton), all atomic

nuclei contain both protons and neutrons.

The mass number ( )A P N= + = +Z N = atomic number + number of neutrons

Thus, N (number of neutrons) = −A Z

mass number

ZA X

atomic number

Atomic number, number of neutrons, and mass number of all must be positive integers

(whole numbers).

Atoms of a given element do not have the same mass. Most elements have two or more

isotopes (atoms of same element that have the same atomic number but different mass

numbers).

e.g., there are three isotopes of hydrogen.

11H hydrogen 1 P 0 N

12H deuterium 1 P 1 N

13H tritium 1 P 2 N

Thus, isotopes have different number of neutrons.

The chemical properties of an element are determined primarily by the protons and

electrons in its atoms, neutrons do not take part in chemical changes under normal

conditions. Therefore, isotopes of the same element have similar chemical reactions.

Average mass number of natural occurring element containing two or more isotopes is

given by

AAX

X

A X A X

X X= = + +

+ +ΣΣ

1 1 2 2

1 2

K

K

Chapter

1 Atoms, Molecules

Ionsand

where A A1 2, ,K are mass numbers of the different isotopes

with percentages or ratios as X X1 2, ,K

If 1735Cl and 17

37Cl are two isotopes of chlorine in the ratio of

3 1: , then average mass number of Cl is

A (Cl) 35.5= × + ×+

=35 3 37 1

3 1

If there is formation of cations, electrons are lost but

number of protons remains unchanged. 2656Fehas 26 ,P 26e−

and 30 N ; 2656 2Fe + has 26P , 24e− and 30 N .

If there are 10 electrons in Na+ then Z = 11, P = 11.

The MoleThe mole (abbreviated as mol) is the SI base unit for a

amount of a chemical species. It is always associated with

a chemical formula and refers to Avogadro’s number of

particles It is designated as N 0 whose value is

N 0 = × −6.022 10 mol23 1

1 mole of every substance 6.022 1023= × species

Thus, 1 mole of oxygen atom 6.023 10 atoms23= ×

1 mole of oxygen (O )2 gas 6.022 1023= × molecules

The molar mass of a substance is the mass in grams of

1 mole of that substance.

Moles of a substance =mass in grams

molar mass

11

mol

molar mass of

X

X= ,

1

101

23

mol

6.022 atoms

X

X×=

Atomic Mass UnitAtoms are so tiny that even the smallest speck of dust

visible to the naked eye contains about 1019 atoms. To

express mass of one atom, unit called atomic mass unit

(amu) is used. amu is also called dalton (Da).

One amu is defined as exactly one-twelth the mass of an

atom 612C and is equal to 1.66054 10 g24× −

Mass of one 612C atom = 12 amu (exactly)

1 amu =mass of one C atoms

12

612

= × −1.66054 10 g24

and 1 g 6.022 10 amu23= ×

Thus, mass of one H-atom = =1

0N1 amu

mass of one O-atom = =16

0N16 amu

Laws of ChemicalCombinations

Law of Mass Conservation (Lavoisier, 1774)

When hydrogen gas burns and combines with oxygen to

yield water (H O),2 the mass of water formed is equal to the

mass of hydrogen and oxygen consumed. This is in

accordance with the law of mass conservation which is

defined as

“Mass is neither created nor destroyed in chemical

reactions”

2H ( ) O ( ) 2H O( )2 2 24 g 32 g 36 g

g g l+ →

CaCO ( ) CaO( ) CO ( )3100 g

256 g 44 g

s s g→ +

Law of Multiple Proportions

(John Dalton, 1803)

Nitrogen and oxygen can combine either in a 7 8: mass

ratio to make a substance denoted by NO or in a 7 16: mass

ratio to make a substance denoted by NO .2

4 Practice Book of Chemistry for JEE Main & Advanced

Grams of A

Moles of A

Atoms of A

Multiplyby N0

Divide byatomic mass

Given

Calculation of number of atom

Atomic mass of A

Atomic massunit of A

Divide by N0

Given

=Mass of

one atom of A

Calculation of a.m.u.

Grams of A

Moles of A

Molecules of A Find

Multiplyby N0

Divide bymolar mass

Calculation of number of moles

NO NO2

mass ratio of N and O 14 16: 14 32:

7 8: 7 16:

This is in accordance with law of multiple proportions

which states “If two elements combine in different ways to

form different substance, the mass ratio are small, whole

number multiples of each other.”

Thus,N : O mass ratio in NO

N : O mass ratio in NO

(7

2

= g N/ 8 gO)

(7 gN/16 gO)2=

Also 1 g of hydrogen can combine with 8 g of oxygen to

yield H O2 or with 16 g of oxygen to yield H O .2 2 Thus,

hydrogen reacts in a multiple of 2

H :O mass ratio in H O

H :O mass ratio in H O

(1 g H/8 gO)

(1 g H/2

2 2

=16 gO)

= 2

Law of Definite Proportions(Proust, 1779)

This states “different samples of a pure chemical substance

always contain the same proportions of elements by mass”.

Every samples of water (H O)2 contains 1 part hydrogen

and 8 parts oxygen by mass

H O2

↓ ↓2 16 1 8: : : :

Every sample of carbon dioxide (CO )2 contains 1 part

carbon and 2.67 parts oxygen by mass

CO2

↓ ↓12 32: : : 1 : 2.67

If formula is to be derived then take molar ratio in which

number of moles =mass

molar mass or atomic mass

Law of Reciprocal Proportions

“If two elements X and Y combine together and each also

combines with a third element Z, then the ratio by weight

in which X and Y combine together is either that ratio in

which they separately combine with a fixed weight of Z, or

simple multiple or fraction of that proportion”.

Hydrogen, carbon and oxygen combine to form three

compounds : H O,2 CO2 and CH .4

CH4 C and H 3 1:

H O2 O and H 8 1:

CO2 C and O 3 8:

Thus, weight ratio of C and O in CO2 is same as weight

ratio of C and O in CH4 and H O2 .

Gay Lussac’s Law of Combining Volumes

When gases react together at constant temperature and

pressure, they do so in volumes which bear a simple ratio to

each other and also the volume of gaseous products.

H + Cl 2HCl2 2 → Ratio

1 1 2 1 1 2: :

N + 3H 2NH2 2 3→ 1 3 2: :

1 3 2

Atoms, Molecules and Ions 5

C O

H

H O2CH4

CO2

1. A sample of CaCO3 is 50% pure. On heating 1.12 L of

CO2 (at STP) is obtained. Residue left (assumingnon-volatile impurity) is(a) 7.8 g (b) 3.8 g(c) 2.8 g (d) 8.9 g

2. In the decomposition of 10 g of MgCO ,3 0.1 mole CO2

and 4.0 g MgO are obtained. Hence, percentage purityof MgCO3 is(a) 50% (b) 60% (c) 40% (d) 84%

3. Consider the following pairs,I. CH ,4 C H2 6 II. CO, CO2

III. NO, NO2 IV. H O,2 H O2 2

In which cases, law of multiple proportion is followed?(a) I, II (b) I, II, III(c) I, III, IV (d) I, II, III, IV

4. Two substances I and II of carbon and oxygen haverespectively 72.73% and 47.06% oxygen. Hence, theyfollow(a) law of multiple proportion(b) law of reciprocal proportion(c) law of definite proportion(d) law of conservation of mass

5. In which case purity of the substance is 100%?(a) 1 mole of CaCO3 gave 11.2 L CO2 (at STP)(b) 1 mole of MgCO3 gave 40.0 g MgO(c) 1 mole of NaHCO3 gave 4 g H O2

(d) 1 mole of Ca(HCO )3 2 gave 1 mole CO2

6. Consider the following laws of chemical combinationwith examplesI. Law of multiple proportion : N O,2 NO, NO2

II. Law of reciprocal proportion : H O,2 SO ,2 H S2

Which is correct with examples?(a) I and II(b) I only(c) II only(d) None of the above

7. H S2 contains 94.11% sulphur; SO2 contains 50%oxygen and H O2 contains 11.11% hydrogen. Thus,(a) law of multiple proportion is followed(b) law of reciprocal proportion is followed(c) law of conservation of mass is followed(d) All of the above

8. Sodium combines with 1735 Cl and 17

37 Cl to give two

samples of sodium chloride. Their formation followsthe law of(a) gaseous diffusion (b) conservation of mass(c) reciprocal proportion (d) None of these

9. According to Dalton’s atomic theory, the smallestparticle in which matter can exist, is called(a) an atom (b) an ion(c) an electron (d) a molecule

10. The nucleus of an atom consists of(a) neutron (b) proton(c) electron (d) Both (a) and (b)

11. 1735 Cl and 17

37 Cl are two isotopes of chlorine. If average

atomic mass is 35.5 then ratio of these two isotopes is(a) 35 37: (b) 1 3:(c) 3 1: (d) 2 1:

12. Ionic mass of X3 − is 17. If it has 10 electrons, thennumber of neutrons are(a) 10 (b) 13(c) 7 (d) 17

13. M2 + ion is isoelectronic of SO2 and has ( )Z + 2

neutrons (Z is atomic number of M). Thus, ionic massof M2 + is(a) 70 (b) 66(c) 68 (d) 64

14. X +, Y 2 + and Z − are isoelectronic of CO .2 Increasing

order of protons in X +,Y 2 + and Z − is

(a) X Y Z+ + −= =2 (b) X Y Z+ + −< <2

(c) Z X Y− + +< < 2 (d) Y X Z2+ + −< <

15. X −,Y 2 − and Z3 − are isotonic and isoelectronic. Thus,increasing order of atomic number of X, Y and Z is(a) X Y Z< < (b) Z Y X< <(c) X Y Z= = (d) Z X Y< <

16. Number of atoms in increasing order in 1.6 g CH ,4 1.7 g

NH3 and 1.8 g H O2 is(a) H O = NH = CH2 3 4 (b) H O < NH < CH2 3 4

(c) CH < NH < H O4 3 2 (d) CH = NH < H O4 3 2

Format I MCQs with only Correct OptionONE

17. Which has maximum number of H-atoms per gram ofthe substance?(a) CH4 (b) CuSO 5H O4 2⋅(c) H O2 2 (d) H O2

18. If each O-atom has two equivalents, volume of oneequivalent of O2 gas at STP is(a) 22.4 L (b) 11.2 L(c) 5.6 L (d) 44.8 L

19. Each drop of H O2 has 0.018 mL at room temperature.

Number of H O2 molecules in one drop is(a) 1 10 3× − (b) 6.02 1020×

(c) 22.4 10 3× − (d) 6.02 3 102× ×

20. 1 g CH4 and 4 g of compound X have equal number of

moles. Thus, molar mass of X is(a) 16 g mol 1− (b) 32 g mol 1−

(c) 4 g mol 1− (d) 64 g mol 1−

21. Mass of one atom of an element is 6.64 10 g.23× − This

is equal to(a) 6.64 10 u23× − (b) 40.0 u

(c)1

40u (d) 6.64 u

22. If Avogadro’s number would have been 1 10 10× − ,

instead of 6.02 1023× then mass of one atom of Hwould be(a) 1 u (b) 1 1010× u

(c) 6 u (d) 6 1013× u

23. Mass of one atom of X is 2.66 10 g,23× − then its 32 g is

equal to(a) 32 2.66 10 mol23× × −

(b)32

2.66 10 6.02 10mol

–23 23× × ×

(c)32 2.66 10

6.02 10mol

–23

23

× ××

(d) None of the above

24. Number of mole of 612C in 1 amu is

(a)1

0N(b) N0

(c) N02 (d)

1

120N ×

25. Mass of one 714N-atom is

(a) 14 u (b) 7 u(c) 14 g (d) 7 g

26. If two compounds have same empirical formula butdifferent molecular formula, they must have(a) same viscosity(b) same vapour density (VD)(c) different molecular weight(d) different percentage composition

27. If the equivalent weight of an element is 32, then thepercentage of oxygen in its oxide is(a) 16 (b) 40 (c) 32 (d) 20

28. A hydrocarbon has 3 g carbon per gram of hydrogen,hence, simplest formula is(a) CH4 (b) C H6 6

(c) C H3 8 (d) CH2

29. Molar ratio ofNa SO2 3 andH O2 is1 7: inNa SO H O.2 3 2⋅xHence, their mass percentage is(a) 12.5 : 87.5 (b) 87.5 : 12.5(c) 50 50: (d) 75 25:

30. Which has the maximum percentage of chlorine?(a) C H Cl6 6 6 (b) C H Cl6 5

(c) CH Cl3 (d) CCl4

31. In which of the following pairs do the two speciesresemble each other most closely in chemicalproperties?(a) 1

1 H and 12 H (b) 8

16O and 816 2–O

(c) 1224 Mg and 12

24 2+Mg (d) 714 N and 7

14 3–N

32. One isotope of a non-metallic element has massnumber 127 and 74 neutrons in the nucleus. The anionderived from the isotope has 54 electrons. Hence,symbol for the anion is(a) 54

127 X − (b) 53127 X −

(c) 5374 X − (d) 54

74 X −

33. Which of the following is the richest source ofammonia on a mass percentage basis?(a) NH NO4 3 (b) NH CONH2 2

(c) NH Cl4 (d) HNC(NH )2 2

34. Which of the following substances contains greatestmass of chlorine?(a) 5.0 g Cl2 (b) 0.5 mol Cl2(c) 0.10 mol KCl (d) 30.0 g MgCl2

35. When 0.273 g of Mg is heated strongly in a nitrogen(N2 ) atmosphere, 0.378 g of the compound is formed.Hence, compound formed is(a) Mg N3 2 (b) Mg N3

(c) Mg N2 3 (d) MgN

36. A certain metal sulphide, MS ,2 is used extensively as a

high temperature lubricant. If MS2 is 40.06% by masssulphur, metal M has atomic mass(a) 160 u (b) 64 u (c) 40 u (d) 96 u

37. The molar mass of a compound if 0.372 mole of it has amass of 186 g, is(a) 200 g (b) 372 g(c) 500 g (d) 186 g

38. Which of the following has maximum number ofC-atoms?(a) 4.4 g CO2 (b) 3.0 g C H2 6

(c) 4.4 g C H3 8 (d) 1.3 g C H6 6


39. Mg C2 3 ( )X is decomposed by H O2 forming a gaseous

hydrocarbon ( )Y . 8.4 g of X gives ………… mol of Y.(a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4

40. Number of atoms in 20 g Ca is equal to number ofatoms in(a) 20 g Mg (b) 1.6 g CH4

(c) 1.8 g H O2 (b) 1.7 g NH3

41. Na SO H O2 3 2⋅ x has 50% H O2 by mass. Hence, x is

(a) 4 (b) 5 (c) 6 (d) 7

42. Mass of one atom of X is6.66 10 g23× − . Hence, number

of moles of atom X in 40 kg is(a) 10 mol3 (b) 10 mol3−

(c)40 10

6.66 10mol

3

23

×× − (d)

40 10

6.022 10mol

3

23

××

43. To make 0.01 mole which of the following hasmaximum mass?(a) NaHCO3 (b) Na CO2 3

(c) Na SO2 4 (d) Na C O2 2 4

44. In a glass-tube, there are 18 g of glucose. 0.08 mole ofglucose is taken. Glucose left in the glass-tube is(a) 0.10 g (b) 17.92 g(c) 3.60 mol (d) 3.60 g

45. Rest mass of an electron is 9.11 10 kg.31× − Molar mass

of the electron is(a) 1.50 10 kg mol31 1× − − (b) 9.11 10 kg mol31 –1× −

(c) 5.5 10 kg mol7 1× − − (d) 6.02 10 kg mol23 1× −

46. A sample of ammonium phosphate (NH ) PO4 3 4

contains 3.18 moles of H-atoms. The number of molesof oxygen atoms in the sample is(a) 0.265 (b) 0.795(c) 1.06 (d) 3.18

47. A sample ofCuSO 5H O4 2⋅ contains 3.782 g of Cu. How

many grams of oxygen are in this sample? (Cu = 63.5)

(a) 0.952 g (b) 3.80 g(c) 4.761 g (d) 8.576 g

48. Cortisone is a molecular substance containing21 atoms of carbon per molecule. The mass percentageof carbon in cortisone is 69.98%. Its molar mass is(a) 176.5 (b) 252.2(c) 287.6 (d) 360.1

49. If the dot under a question mark has a mass of1 10 6× − g, how many atoms are required to make such

a dot? (of carbon)(a) 5 1016× (b) 12 10 6× −

(c)10

12

6−(d)

1 10 6

0

× −

N

50. One equivalent of magnesium oxide weighs 20 g thenone equivalent of magnesium chloride weighs(a) 29.75 g (b) 47.5 g (c) 95.0 g (d) 20.0 g

51. A spherical ball of radius 7 cm contains 56% iron. Ifdensity is 1.4 g /cm3 , number of moles of Fe present

approximately is(a) 10 (b) 15(c) 20 (d) 25

52. In the following final result is …0.1 moleCH 3.014 + × 1023 molecules CH4 − =9.6 g CH4 x

mole H atoms(a) 0 mol H atom (b) 0.2 mol H atom(c) 0.3 mol H atom (d) 0.4 mol H atom

53. If we assume that N 0 = × −1.2 10 mol ,23 1 then molar

mass of O2 will be taken as

(a) 32 g mol 1− (b)32

6g mol 1−

(c) 32 1023× −g mol 1 (d)1 10

32

23× −g mol 1

54. CO, CO ,2 C O2 3 follow

(a) law of definite proportion(b) law of multiple proportion(c) law of conservation of mass(d) All of the above

55. In a gas S and O are 50% by mass, hence, their moleratio is(a) 1 1: (b) 1 2:(c) 2 1: (d) 3 1:

56. What mass of ammonium phosphate (NH ) PO4 3 4

would contain 14.0 g of nitrogen?(a) 50.0 g (b) 25.0 g(c) 12.5 g (d) 100.0 g

57. What mass of propane C H3 8 contains the same mass of

carbons as contained in 1.35 g of barium carbonate,BaCO3?(a) 1.35 g (b) 1.00 g(c) 0.10 g (d) 0.135 g

58. Which element has maximum percentage in iron (III)sulphate (IV)?(a) Iron (b) Sulphur(c) Oxygen (d) Equal

59. How many moles of oxygen are contained in one litreof air if its volume content is 21% in standardconditions?(a) 0.21 mol (b) 0.045 mol(c) 0.067 mol (d) 0.0094 mol

60. A compound with molecular mass 180 is acylated withCH COCl3 to get a compound with molecular mass 390.The number of amino groups present per molecule ofthe former compound is [JEE Main 2013]

(a) 2 (b) 5(c) 4 (d) 6


61. Mass of one atom of the element A is 3.9854 g× −10 23 .

How many atoms are contained in 1 g of the element A?(a) 2.5092 × 1022 (b) 6.022 × 1023

(c) 3.9854 × 1023 (d) 1.66 × −10 23

62. Analysis of chlorophyll shows that it contains2.40 per cent magnesium. Thus, number of atoms in1 g chlorophyll is(a) 0.001 (b) 1.00(c) 6.02 × 1020 (d) 1.445 × 1019

63. Radius of water molecule is (assuming it spherical)(a) 19.27 nm (b) 19.27 Å(c) 192.7 pm (d) 19.27 µm

64. Atoms of the element X are spherical. Each atom of theelement (atomic mass 23) is at the corner of the cubeand is in contact along the edge length, then edgelength is (density = −6.2 g cm 3 )

(a) 2.274 Å (b) 1.137 Å(c) 4.548 Å (d) 1.183 Å

65. Number of moles in 1.8 gH O2 is equal to the number of

moles inI : 1.8 g glucoseII : 6 g ureaIII : 34.2 g sucrose

Select the correct group.

(a) I, II, III (b) I, II(c) I, III (d) II, III

66. A sample of ammonium dihydrogen phosphate (I)contains 3.18 moles of hydrogen atom. The number ofmoles of oxygen atoms in the sample is(a) 0.265 (b) 0.795(c) 1.06 (d) 3.18

67. A sample of copper sulphate pentahydrate contains3.782 g of Cu. How many grams of oxygen are in thesample?(a) 0.952 g (b) 3.809 g(c) 4.761 g (d) 8.576 g

68. An unknown amino acid has 0.032% sulphur. If eachmolecule has one S-atom, then 1 g of this amino acidhas molecules(a) 6.02 × 1018 (b) 6.02 × 1019

(c) 6.02 × 1021 (d) 6.02 × 1023

69. Consider the following casesI : 60 g CH COOH3 II : 30 g HCHOIII : 60 g NH CONH2 2 IV : 180 g C H O6 12 6

Percentage of carbon is identical in

(a) I, II (b) I, III(c) I, II, III (d) I, II, IV

70. Ethanol-water mixture has 46 g ethanol in 100 gmixture. By a suitable technique volatile componentgoes off. Thus(a) 3 moles of non-volatile component are left(b) 9 N0 atoms of non-volatile component are left(c) 9 0N atoms of volatile component are separated(d) All the above statements are correct

1. Which is/are correct about 4.25 g NH3?

(a) It contains 0.25 mole of NH3

(b) It contains 0.75 mole of H-atoms(c) It contains total of 1.0 mole of N and H atoms(d) It contains 1.5 1023× molecules of NH3

2. Which of the following are isoelectronic of O2 −?(a) N3− (b) F− (c) Ti+ (d) Na+

3. Among the following groupings which represents thecollection of isoelectronic species?(a) NO, CN ,− N ,2 O2

− (b) NO ,+ C ,22− O ,2

− CO

(c) N2,C22−, CO, NO (d) CO, NO ,+ CN ,− C2

2−

4. The atomic number of an element is always equal to(a) number of neutrons in the nucleus(b) half of the atomic weight(c) electrical charge of the nucleus(d) number of protons

5. A bivalent metal ion has equivalent mass of 12. Then(a) equivalent mass of its oxide is 28(b) molar mass of its oxide is 40(c) equivalent mass of its hydride is 13(d) molar mass of its hydride is 14

6. Volume of a gas at NTP is 1.12 10 7× − cc. The numberof molecules is thus equal to(a) 3.01 1012× (b) 3.01 1018×(c) 3.01 1024× (d) 3.01 1030×

7. Which of the following may contain one proton andone neutron?(a) H2

+ (b) 24 He (c) 1

2 D (d) 13 T

8. 1735 Cl and 17

37 Cl differ in

(a) atomic number (b) number of neutrons(c) number of electrons (d) atomic mass

9. Isoelectronic species are represented by pairs

(a) N ,3− O2− (b) CO, CN−

(c) O ,22− F2 (d) O ,2

− CN−

10. X − is isoelectronic of CO and has ( )Z + 2 neutrons

(Z = atomic number of X −). Thus,

(a) ionic mass of X − is 28(b) ionic mass of X − is 30(c) atomic number of X − is 13(d) atomic number of X − is 14


Format II MCQs with One or More than Correct OptionONE

1. 1021 molecules are removed from 200 mg of CO .2 Thus,

number of moles of CO2 left is

2. Al (SO ) H O2 4 3 2⋅x has 8.20 per cent Al by mass.Calculate the value of x.

3. In addition to carbon monoxide (CO) and carbondioxide (CO ),2 there is a third compound of carboncalled carbon suboxide. If a 2.500 g sample of carbonsuboxide contains 1.32 g of C and 1.18 g of O, showthat the law of multiple proportions is followed. Whatis the possible formula of carbon suboxide?

4. XHCO3 and YCO3 are two pure substances of equalmolar masses decomposing as shown

2X XHCO H O + CO + CO3 2 2 2 3→∆

Y YCO O + CO3 2→∆

16.8 g of XHCO3 gave 6.2 g of mixture of H O2 andCO2 . Identify the substances.

5. Cesium atoms are the largest naturally occurringatoms. The radius of cesium atom is 2.62 Å. How manycesium atoms would have to be laid side by side to givea row of cesium atoms 2.50 cm long? Assume that theatoms are spherical.

Example 1 Study the following observations and answerthe questions at the end of it.

The following is a crude but effective method for estimating theorder of magnitude of Avogadro’s number using stearic acid(C H O ).18 36 2 When stearic acid is added to water, its moleculescollect at the surface and form a monolayer; that is, the layer isonly one molecule thick. The cross-sectional area of eachstearic acid molecule has been measured to be 0.21nm2. In oneexperiment it is found that1.4 10 g4× − of stearic acid is needed

to form a monolayer over water in a dish of diameter 20 cm.(the area of a circle of radius r is πr2.)

1. Based on these measurements value of Avogadro’snumber is(a) 3 1023× (b) 6 1023× (c) 4 1023× (d) 1 1023×

2. What is the equivalent of 1g H in amu for this value ofAvogadro’s number?(a) 1.66 10 g24× − (b) 3.33 10 g24× −

(c) 2.5 10 g24× − (d) 1 10 23× − g

Example 2 Read the following passage regarding fertiliserK O2 and answer the questions at the end of it.

Potash is only potassium mineral that is used for its potassiumcontent. Most of the potash produced in the United States goesinto fertilizer. The major sources of potash are potassiumchloride (KCl) and potassium sulphate (K SO ).2 4 Potashproduction is often reported as the potassium oxide (K O)2

equivalent or the amount of K O2 that could be made from agiven mineral. KCl costs ` 50 per kg.

1. What is the cost of K per mole of the KCl sample?(a) ` 13.42 mol 1− (b) ` 3.73 mol 1−

(c) ` 1.00 mol 1− (d) ` 2.00 mol 1−

2. For what price must K SO2 4 be sold in order to supplythe same amount of potassium as in KCl?(a) ` 58.40 kg−1 (b) ` 50.00 kg 1−

(c) ` 42.82 kg 1− (d) ` 25.00 kg 1−

3. What mass (in kg) of K O2 contains the same number of

moles of K atoms as 1.00 kg KCl?(a) 0.158 kg (b) 0.315 kg(c) 1.262 kg (d) 0.631 kg

Example 3 The mass spectrum (given below) of magnesiumhas three peaks, which indicates that magnesium has threeisotopes. Questions given below are based on this mass spectrum.

1. Which isotope has maximum number of atoms pergram of each?(a) Mg-24 (b) Mg-25(c) Mg-26 (d) Equal

2. Number of atoms in one mole of each isotope is placedin increasing order(a) 24 25 26Mg < Mg < Mg (b) 26 25 24Mg < Mg < Mg

(c) 26 25 24Mg = Mg = Mg (d) given data is insufficient

3. Average atomic mass of Mg is approximately(a) 25.0 (b) 24.3(c) 25.2 (d) 25.8


Format IV Comprehension Based MCQs

Format III Testing of Numerical Skill

20

40

60

80

100

Re

lative

abu

nd

an

ce

24 25 26Atomic mass (amu)

Example 4 The equivalent volume is defined as the volumeoccupied by one equivalent of a substance in the givencondition. The equivalent volume of a gaseous substance canbe derived from the molar volume.

1. Equivalent volume of O2 gas at STP is

(a) 5.6 L (b) 11.2 L(c) 16.8 L (d) 22.4 L

2. Number of hydrogen atoms in H2 gas in terms of

Avogadro’s number at STP is

(a) N0 (b)N0

2

(c)N0

4(d)

N0

8

3. A certain amount of a metal whose equivalent mass is28 g mol−1 displaces 0.7 L of H2 gas in standard

conditions. Thus, mass of the metal is(a) 0.07 g (b) 0.70 g(c) 1.75 g (d) 0.875 g

1. Match the properties in Column I with the compoundsin Column II.

Column I Column II

A. Identical % of C 1. HCHO

B. Identical % of H 2. C H O6 12 6

C. Identical % of O 3. CH COOH3

D. Same molar mass 4. NH CONH2 2

2. Match the gases in Column I with their correspondingproperties in Column II.

Column I Column II

A. H2 gas 1. 2 0N atoms per mole

B. O2 gas 2. Specific heat ratio : 1.40

C. N2 gas 3. 0.8 L g−1 at STP

D. CO gas 4. Lightest gas

E. HCl gas 5. Heaviest gas

3. Match percentage of carbon (in Column I) with thecompound (in Column II).

Column I Column II

A. 20% 1. CH4

B. 52.2% 2. CO2

C. 75% 3. C H O2 6

D. 27.3% 4. CN OH2 4

E. 12% 5. CaCO3

4. Match the compounds in Column I (1 mole of each)with corresponding properties in Column II.

Column I(1 mole each)

Column II

A. H SO2 4 1. 60 g

B. H PO3 4 2. 98 g

C. H PO3 3 3. N0 ionisable H

D. H PO3 2 4. 2 0N ionisable H

E. CH COOH3 5. 3 0N ionisable H

F. N H CO2 4 6. 4 0N H

1. Disilane (Si H )2 x is analysed and found to contain

90.28% silicon by weight. What is the value of x ?(Si = 28)

2. The aluminium sulphate hydrate Al (SO ) H O2 4 3 2⋅ y

contains 8.20 per cent Al by mass. What is the value ofy? ( , , )Al S O= = =27 32 16

3. Na SO H O2 3 2⋅ z has 50% H O2 by mass. What is thevalue of z? (Na = 23)

4. In Japan, during earth-quake on April 4, 2011, 0.060 kgof radioactive water leaked into sea from nuclearreactors. How many moles of radioactive water leakedinto sea?

5. When 2.495 g of CuSO H O24 ⋅ x is heated, 0.05 mole of

H O2 is lost forming CuSO4. What is the value of x ?

6. Mass of one atom of an element is 6.66 g.× −10 23 How

many moles of element are there in 0.320 kg?

7. X2 − is isoelectronic of 10Ne. 1.6 g of X2 − has 0.1 mole.

Calculate the number of neutrons in X2 −.

8. Haemoglobin (molar mass 67200) contains 0.33% ironby weight. How many iron atoms are present in onemolecule of haemoglobin?


Format V Matrix Matching

Format VI Integer Answer Type

1. (a) CaCO CaO CO3 2( ) ( ) ( )s s g→ ↓ + ↑∆

1 mol = 100 g 1 mol = 56 g 22.4 L at STP0.05 mol = 5 g pure 0.05 mol = 2.8 g 1.12 L at STP

Impure CaCO3 taken 10 g (5 g= pure CaCO3 + 5 gimpurity)

CaO(s) left = 2.8 gI mpurity 5.0 g=

Total residue 7.8 g=

2. (d) MgCO MgO + CO3 2( ) ( )s s g→∆ ( )

1 mol = 84 g 40 g 1 mol

8.4 g (pure) 4.0 g 0.1 mol

in 10 g sample

Thus, % of pure MgCO = 84%3

3. (d) I :C : H mass ratio in CHC : H mass ratio in C H

4

2 6

= 3 gc / gH4 gc / g H

3 : 4=

II :C : O mass ratio in COC : O mass ratio in CO

6

2

= g C / 16g O6 g C / 32g O

2 : 1=

III :N : O mass ratio in NON : O mass ratio in NO

14

2

= g N / 10g O14 gN/ 32g O

2 : 1=

IV :H : O mass ratio in H OH : O mass ratio in H O

2

2 2

= 2 gH / 10g O2gH / 32g O

2 : 1=

Thus, in all cases law of multiple proportion is followed.

4. (a) Substance I C O

27.27 72.73

mole : 2.2725 4.5456

ratio : 1 : 2

Substance II 52.94 47.06

mole : 4.411 2.94

ratio : 1.5 1.0

3 : 1

5. (b) (a) CaCO CaO + CO3 21 mol 22.4 L

→∆

actual 11.2 L=Thus, 50%

(b) MgCO MgO + CO3 21 mol 40 g

→∆

Thus, 100%

(c) 2NaHCO Na CO +H O + CO3 2 3 2 22 mol1 mol

18 g9 g

→∆

actual = 4 g

Thus, 44.4%

(d) Ca(HCO ) CaO H O +2CO3 2 2 21 mol 2 mol

→ +∆

actual = 1 mol

Thus, 50%

6. (a) I : N O2 : NO : NO2

28 gN/16g O : 14 gN/16g O : 14 gN/32g O

14 gN/8g O : 14 gN/16g O : 14 gN/32g O

4 2 1: :

Ratio of H and S in H S2 : 2 g H/32 g S

Ratio of H and O in H O2 : 2 g H/16 g O

Ratio of S and O in SO2 : 32 g S/32 g O

Thus, law of reciprocal proportion is followed.

Thus, I and II both.

7. (b) H S2 5.89 g H combines with = 94.11 gS

hence, 1 g H combines with = 16 g SSO2 50 g O combine with = 50 g S

hence, 1 g O combines with = 1g S

H O2 11.11 g H combines with 88.89 g O=1 g H combine with 8 g O=Thus, law of reciprocal proportion is followed.

8. (d) There are two types of NaCl formed. They differ in molar

masses due to different isotopes of Cl.

9. (a) By Dalton’s theory, atom is the smallest particle.

10. (d) Nucleus consists of proton and neutron and molar mass

= neutron + proton

11. (c) Average atomic mass AA X A X

X X= +

+1 1 2 2

1 2

35.5 = ++

35 371 2

1 2

X X

X X

On solvingX

X

1

2

31

=

12. (a) X3− has 10 electrons.

Thus, protons (= atomic number) in X3 7− =neutron proton = ionic mass = 17+

Thus, neutrons = 17 – 7 = 10

13. (a) M2+ has electrons = 32 (isoelectronic of SO )2

protons = =34 Z

neutrons = 36Thus, ionic mass = 70

Format I MCQs with only Correct OptionONE

S O

H

H S2 H O2

SO2

II.

Check your Solutions

14. (c) Number of electrons in CO2 = + =6 16 22

Electrons Electrons in neutral species = proton

X+ 22 23

Y2+ 22 24

Z− 22 21

Thus, increasing order of proton is

Z X Y− + +< < 2

15. (b) Number of neutrons = number of electrons in X−, Y2−

and Z3−

Electrons Z (atomic number)

X−E ( )E −1

Y2−E ( )E −2

Z3−E ( )E − 3

Thus, increasing order of atom number is

Z Y X< <

16. (b) Number of moles in Molecules Atoms

CH4 0.1 0.1 N0 0.5 N0

NH3 0.1 0.1 N0 0.4 N0

H O2 0.1 0.1 N0 0.3 N0

17. (a) H-atoms per gram

(a) CH4416 4

0 00

N NN= = 0.25

(b) CuSO 5H O4 2⋅ 10 00

NN

249.50.04=

(c) H O2 2234

00

NN= 0.0588

(d) H O2217

00

NN= 0.1176

18. (c) 1 mole O2 = 2 O-atoms = 4 equivalents oxygen

Volume of 1 mole 22.4 L= at STP

Volume of 1 equivalent = 5.6 L at STP

19. (b) 0.018 mL = 0.018 g (density of water = 1g/mL)

= 0.01818

= 0.001 mol

= ×6.02 10 molecules20

20. (d)116

44

gCH

g= X

m

m = −64 1g mol

21. (b) Mass of one atom 6.64 10 g23= × −

Thus, atomic mass = 6.64 10 6.02 10–23 23× × × = 40

22. (a) A new Avogadro’s number = X

Mass of one H-atom = =11

xamu

23. (b) Mass of one atom 2.66 10 g–23= ×

Mass of N0 atoms 2.66 10 6.02 10 g mol23 23 1= × × ×− −

= atomic mass

Thus, number of moles in 32 g

=32

2.66 10 6.02 10–23 23× × ×

24. (d) 1 mol 612C 12 g=

1 amu = 1

0N

Number of moles in amu = 112 0N

25. (a) N0 atoms = 14 g

1 atom = 14

0N= 14 amu

26. (c) Same empirical formula, it means ratio of atoms is

identical. Hence, they differ in molecular weight.

27. (d) Equivalent weight of element 32 g=

and that of oxygen 8 g=Thus, one equivalent of oxide 40 g=

Percentage of oxygen in oxide = ×840

100

= 20%

28. (a) Amount Moles Ratio

C 3 g3

12= 0.25 1

H 1 g11

= 1.0 4

Thus, simplest formula is CH .4

29. (c) Na SO : H O2 3 2

Moles 1 7:Mass 126 126

Per cent 50 50

30. (d) (a) Cl%35.5

35.5 12 135.548.5

=+ +

= = x

y

(b) =+ +

=35.572 5 35.5

35.5112.5

(c) =+

=35.515 35.5

35.550.5

(d) = ×+ ×

=+

=35.5 412 35.5 4

35.53 35.5

35.538.5

In all cases value of X = 35.5Smaller the value of Y, larger the percentage of Cl.

31. (a) 11H and 1

2H are isotopes. Thus, they resemble very

closely in their chemical properties.

32. (b) ZA

A NAX X X X= = =− −127 74

12753127

There are 54 electrons. Hence, ionic species is 53127 X−

33. (d) (a) NH3 in NH NO1780

0.21254 3 = =

(b) NH3 is NH CONH =3462

0.54842 2 =


(c) NH3 in NH Cl17

53.50.31784 = =

(d) NH3 in HNC(NH )51

59 40.80952 2 =

+=

(NH CONH 2NH – 2H)2 2 3≡∴ NH CONH + 2H 2NH2 2 3≡Similarly, NHC(NH ) 3NH – 4H2 2 3=∴ NHC(NH ) + 4H 3NH2 2 3≡

34. (b) (a) 5.0 g Cl2(b) 0.5 mol Cl 0.5 71 g 35.5 g Cl2 2= × =(c) 0.10 mol KCl 0.1= mol Cl = 3.55 g Cl

(d) 30.0 g MgCl30.095

mol 0.25262 = = mol = 17.49 g Cl2

35. (a) Mass of Mg 0.273 g=

Mass of magnesium and nitrogen compound 0.378 g=Thus, nitrogen combined 0.105 g=

Mass Mole Mole ratio

Mg 0.273 g 0.011375 1.51 = 3N 0.105 g 0.0075 1.00 = 2

Thus, Mg N3 2

36. (d) M MS2 = + ×32 2

= +M 64

% of sulphur =+

× =64

64100

M40.06

∴ M + =64640040.06

M + =64 160M = − =160 64 96

37. (c) 0.372 mol = 186 g

1 mol186

0.372500 g= =

38. (c) (a) 4.4 g CO4.444

mol CO 0.1 mol CO2 2 2= =

= 0.1 mol C

(b) 3.0 g C H3.030

mol C H = 0.1 mol C H2 6 2 6 2 6=

= 0.2 mol C

(c) 4.4 g C H =4.444

mol C H = 0.1 mol C H3 8 3 8 3 8

= 0.3 mol C

(d) 1.3 g C H1.378

mol C H = 0.017 mol C H6 6 6 6 6 6=

= 0.1 mol C

39. (a) Mg C H O Mg(OH) CH C CH2 3 2 2 3( )

1 mol( )

proX Y

+ → + ≡≡

pyne

8.4 g 1 mol

8.484

= 0.1 mol 0.1 mol

40. (b) 20 g Ca2040

mol 0.5 mol Ca= = = 0.5 atomsN0

(a) 20 g Mg2024

mol 0.833 mol Mg= = = 0.833 atomsN0

(b) 1.6 g CH1.616

mol = 0.1 mol CH4 4= = 0.5 atomsN0

(c) 1.8 g H O =1.818

mol = 0.1 mol H O2 2 = 0.3 atomsN0

(d) 1.7 g NH1.717

mol 0.1 mol NH3 3= = = 0.4 atomsN0

41. (d) Mass Moles Ratio

Na SO2 3 = 50 0.3968 1

H O2 = 50 2.7778 7

Thus, x = 7

42. (a) Mass of one atom of X = × −6.66 10 g23

If atomic mass = A

then mass of one atom = A

N0

∴ A

N0

= × −6.66 10 g23

A N= × ×−6.66 10 230

= × × ×−6.66 10 6.02 1023 23

= −40 g mol 1

Hence, 40 kg X gX= = =4000040000

401000 mol

43. (c) Molar mass 0.01 mol

(a) NaHCO = 84 g3 0.84 g

(b) Na CO = 106 g2 3 1.06 g

(c) Na SO = 142 g2 4 1.42 g

(d) Na C O = 134 g2 2 4 1.34 g

44. (d) 18 g glucose 0.10= mol glucose

withdrawn 0.08 mol=left 0.10 0.08 0.02 mol = 3.6 g= − =

45. (c) Rest mass of electron = × −9.11 10 kg31

mass of one mole of electrons 9.11 10 6.02 1031 23= × × ×−

= × −5.48 10 kg mol–7 1

46. (c) x mol (NH ) PO4 3 4 ≡ 12 x mol H atoms 3.18 mol=

∴ x = =3.1812

0.265 mol

Thus, O-atoms = 4x mol 0.265 4 1.06 mol= × =

47. (d) x mol CuSO 5H O4 2⋅

= ⋅249.5x g CuSO 5H O4 2

= 63.5 x g Cu= 3.782 g Cu

∴ 63.5 3.782x =x = 0.05956 mol

Every 1 mole of salt has 9 mol= es O-atoms


∴ O-atoms 0.05956 9 mol= ×= × ×0.05956 9 16 g= 8.576 g O -atoms

48. (d) If carbon content is 69.98 g then molar mass = 100 g

If carbon content is 21 12× g then molar mass is

= × ×10021 12

69.98= −360.1 g mol 1

49. (a)1 10

125 10

6

016× × = ×

−N atoms

50. (b) One equivalent magnesium oxide = 20 g

Since, equivalent mass of O = 8hence, equivalent mass of Mg = − =20 8 12Also, equivalent mass of chlorine 35.5=hence equivalent mass of magnesium chloride 35.5 12= +

= 47.5

51. (c)Mass

Volume= Density

∴ Mass of spherical ball = ×V d

= ×43

3πr d

= × × ×43

227

7 3( ) 1.4

= 2012.27 g

pure Fe content (56%) 2012.2756

100g= × = 1126.87 g

Thus, moles of Fe 20.12=52. (a) I. 0.1 mol CH = 0.4 mol H -atoms4

II. 3.01 1023× CH4 molecules 0.5 mol CH = 2.04= mol

H-atoms

III. 9.6 g CH = 0.6 mol CH = 2.4 mol H -atoms4 4

Thus, I + II – III = 0

53. (a)

54. (b) Co : CO2 : C O2 3 (CO )1.5

1 mol C 12 g C in CO = 16 g O== 12 g C in CO = 32 g O2

= 12 g C in C O = 24 g O2 3

Thus, ratio of O that combines with 12 g C = 2 4 3: :

55. (b) Mass Moles Ratio

S 50 g5032

= 1.5625 1

O 50 g5016

= 3.125 2

56. (a) 3N 1 (NH ) PO4 3 4≡

42 g N is in 1 mol (NH ) PO4 3 4=

14 g N is in =1442

mol (NH ) PO4 3 4

= 13

mol

= ×13

149 g (NH ) PO4 3 4

= 49.67 g ≈ 50.0 g

57. (c) 1 mol BaCO 1 mol C3 ≡

197 g BaCO3 contains = 12 g C

1.35 g BaCO3 contains12 1.35

197= × = 0.0822 g C

C H 3C3 8 ≡3 12× g carbon is in 44 g C H3 8=

∴ 0.0822 g carbon is in44 0.0822

360.100 g= × =

58. (c) Iron (III) Sulphate (IV) is

Fe3+ SO32− (oxidation of sulphur is = + 4 in SO3

2−)

Fe (SO )2 3 3

Fe 56 2 112 g= × =S 32 3 96 g= × =O 16 9 144 g= × =

Thus, percentage of O is maximum out of total molar mass

of Fe (SO ) = 352.2 3 2

59. (d) Oxygen content in 1 L at STP = × =1 21100

0.21 L

22.4 L at STP = 1mol

hence 0.21 L at STP1 0.21

22.4= × = 0.009375 mol

60. (c) R R—NH + CH —C

O

—Cl —NH—C

O

—CH2 3 3(–HCl)

→

Since each —COCH3 group displace one H atom in the

reaction of one mole of CH —C

O

—Cl3

with one —NH2

group, the molecular mass increases with 42 unit. Since the

mass increases by ( )390 180 210− = hence the number of

—NH2 group 4 is21042

= 5.

61. (a) 3.9854 10 g 1 atom23× =−

Thus, 1 g1

3.9854 10atoms23=

× −

= ×2.5092 10 atoms23

62. (c) Mg in 1 g chlorophyll1 2.4

100g= ×

= ××

1 2.4100 24

mol

= × × ××

1 2.4 6.02 10100 24

atoms23

= ×6.02 10 atoms20


63. (c) Density of water molecule = 1g/mL

1 g = 1 mL

Thus,1

18mol 1 mL=

118

6.02 10 molecules 1 mL23× × =

Thus, volume of 1 molecule18

6.02 1023=×

= × −3 10 mL23

Volume of one spherical molecule = 43

3πr

43

3 103 23πr = × − cm3

∴ r 3233 3 10

4= × × −

πcm3

r 3 23 310= × −7.162 cmr = × −1.927 10 cm8

= × −1.927 10 m10

= 1.927 Å

= × × −1.927 10 1010

m10

= 0.1927 nm

= × × −1.927 100 10100

m10

= × −192.7 m10 12 = 192.7 pm

64. (a) Edge length = =AB r2

Volume of the spherical atom = 43

3πr

MassVolume

= density

Volumemass

density=

Mass of one atom = =×

m

N0

236.02 1023

∴ Volume =× ×

2310236.023 6.20

43

3πr =× ×

236.02 10 6.223

r 323

3 234 10

= ×× × ×π 6.02 6.2

r 3 = × −1.47 10 cm24 3

r = × −1.137 10 cm8

2r = × −2.274 10 cm8 = 2.274 Å

65. (d) 1.8 g H O1.818

0.1 mol2 = =

I : 1.8 g glucose1.8180

0.01 mol= =

II : 6 g urea = =660

0.1 mol

III : 34.2 g sucrose34.2342

0.1 mol= =

Thus, II, III

66. (c) Oxidation number of P in phosphate (I) = +1

Thus, anion is : H PO2 2–

Thus, salt is NH H PO4 2 2

(Ammonium hypophosphite)

6 mol H ≡1 mol NH H PO 2 mol (O)4 2 2 =

Thus, 3.18 mol H =2 3.18

6mol (O)

×= 1.06 mol (O)

67. (d) CuSO 5H O4 2⋅

1 mol solute ≡ 1mol (Cu) ≡ 10 mol (H) ≡ 9 mol (O)

63.5 g (Cu) = ×9 16 g (O)

Thus, 3.782 g (Cu)9 16 3.782

63.5g (O)= × ×

= 8.5765 g

68. (a) If 0.032 g sulphur then molar mass = 100 g

If 32 g sulphur then molar mass = × =1032 105

0.032g

Thus, molar mass 10 g mol5 1= −

1 g amino acid = 1105 mol amino acid

= N0510

molecules

= ×6.02 1018 molecules

69. (d) Percentage is irrespective of amount given

I. CH COOH 2C3 ≡60 g 24 g

C% = × =24 10060

40

II. HCHO 1 C≡30 g 12 g

C% = × =12 10030

40

NH CONH 1C2 2 ≡III. 60 g 12 C

C% = × =12 10060

20

IV. C H O 6 C6 12 6 ≡180 g 72 g

C% = × =72 100180

40

Thus, I, II and IV


AB

70. (d) Volatile component is CH CH OH = 46 g3 2

= =4646

1mol

1 mole = N0 molecules

= 9 0N atoms thus, (c) is correct.

Non-volatile component is H O = 54 g =54182

= 3 molesThus, (a) is correct

= 3 0N atoms

Thus, (b) is correct

1. (a,b,c,d) 2. (a,b,d) 3. (d) 4. (c,d)

5. (b, c) 6. (a) 7. (c) 8. (b,d) 9. (a,b,c)

10. (a, c)

1. 2.88 10 mol CO–32× 2. 18

3. Thus, CO, CO2 and C O3 2 follow law of multiple proportion.

4. Thus, Y is Mg and YCO3 is MgCO3

5. 4.77 10 atoms7×

Ex. 1 1. (a) 2. (b) Ex. 2 1. (b) 2. (c) 3. (d)

Ex. 3 1. (a) 2. (b) 3. (b)

Ex. 4 1. (a) 2. (a) 3. (c)

1. A—(1,2,3,4); B—(1,2,3,4); C—(1,2,3); D — (3,4)

2. A—(2,4); B—(2,5); C—(4); D—(3); E—(1,3,6); F—(6)

3. A—(1,2,4); B—(1,2); C—(1,2,3); D—(1,2,3); E—(1,2,5)

4. A—(4); B—(3); C—(1); D—(2); E—(5)

Questions 1 2 3 4 5 6 7 8

Answers 6 18 7 3 5 8 8 4


Format II MCQs with One or More thanCorrect OptionONE

Format III Testing of Numerical Skill

Format IV Comprehension Based MCQs

Format V Matrix Matching

Format VI Integer Answer Type

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