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Practice Book of Chemistry for Jee Main
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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
Atoms, Molecules and Ions 7
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
8 Practice Book of Chemistry for JEE Main & Advanced
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
Atoms, Molecules and Ions 9
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
10 Practice Book of Chemistry for JEE Main & Advanced
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?
Atoms, Molecules and Ions 11
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 =
Atoms, Molecules and Ions 13
(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
14 Practice Book of Chemistry for JEE Main & Advanced
∴ 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
Atoms, Molecules and Ions 15
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
16 Practice Book of Chemistry for JEE Main & Advanced
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
Atoms, Molecules and Ions 17
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