NCERT Exemplar Class 9 Science Solutions Chapter 3 Atoms and Molecules 2021

NCERT Exemplar Class 9 Science Solutions Chapter 3 Atoms and Molecules

Edited By Sumit Saini | Updated on Sep 01, 2022 01:08 PM IST

NCERT exemplar Class 9 Science solutions chapter 3 deals with basic constituents of any substance or compound which is an atom. The combination of these atoms makes a molecule which in turn combines to make a substance. The NCERT exemplar Class 9 Science chapter 3 solutions are prepared by a skilled team of subject matter experts to ensure accurate and comprehensive learning of NCERT Class 9 Science. These Class 9 Science NCERT exemplar chapter 3 solutions clarify any doubts or queries as they provide a step-by-step approach to the basics of atoms and molecules. The NCERT exemplar Class 9 Science solutions chapter 3 are designed to cover all the topics of the CBSE Syllabus for Class 9 Chapter 3.

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NCERT Exemplar Class 9 Science Solutions Chapter 3-MCQ
NCERT Exemplar Class 9 Science Solutions Chapter 3-Short Answer
NCERT Exemplar Class 9 Science Solutions Chapter 3-Long Answer
NCERT Exemplar Solutions Class 9 Science Chapter 3 Important Topics:
NCERT Class 9 Science Exemplar Solutions for Other Chapters:
Features of NCERT Exemplar Class 9 Science Solutions Chapter 3:

NCERT Exemplar Class 9 Science Solutions Chapter 3-MCQ

Question:1

Which of the following correctly represents 360 g of water?
(i) 2 moles of $H_{2}O$
(ii) 20 moles of water
(iii) $6.022 \times 10^{23}$ molecules of water
(iv) $1.2044\times10^{25}$ molecules of water
(a) (i)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer: D
The average mass of one mole of $H_{2}O$ is 18.02 grams.
The molecular weight of water is 18 g/mol. Hence, one mole of water weighs 18 g.
One mole of water contains $6.023 \times 10^{23}$ water molecules.
2 moles of $H_{2}O$ is $36 g (2 \times 18 g)$ and 20 moles of water is $360g (20 \times 18 g)$ . The mass of $6.022 \times 10^{23}$ molecules of water is 18g.
Therefore, the mass of $1.2044\times10^{25}$ molecules ( $= 20 \times 6.022 \times10^{23}$ molecules) of water is $20 \times 18g = 360g$ .

Question:2

Which of the following statements is not true about an atom?
(a) Atoms are not able to exist independently
(b) Atoms are the basic units from which molecules and ions are formed
(c) Atoms are always neutral in nature
(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch
Answer: A
Atoms are not able to exist independently is not true, Atoms of most elements are not able to exist independently but inert gases can exists as atoms.

Question:3

The chemical symbol for nitrogen gas is
(a) Ni (b) $N_{2}$ (c) $N^{+}$ (d) N
Answer: B
Chemical formula of Nitrogen is N but nitrogen exist as a molecule of two atoms. Therefore, the chemical symbol of nitrogen gas is $N_{2}$

Question:4

The chemical symbol for sodium is
(a) So (b) Sd (c) NA (d) Na
Answer: D
Sodium word is derived from Latin word Natrium hence the chemical name of sodium is Na.

Question:5

Which of the following would weigh the highest?
(a) 0.2 mole of sucrose $(C_{12} H_{22} O_{11})$
(b) 2 moles of $CO_{2}$
(c) 2 moles of $CaCO_{3}$
(d) 10 moles of $H_{2}O$
Answer: C
From mole concept,
1 mole = molecular weight in gram.
Weight of a sample in gram = Number of moles × Molar mass
Mass of 1 mole of sucrose $(12 \times 12) + (1 \times 22) + (16 \times 11) = 342 g$
Mass of 1 mole of $CO_{2} = 12 + (16 \times 2) = 44g$
Mass of 1 mole of $CaCO_{3} = 40 + 12 + (16 \times 3) = 100g$
Mass of 1 mole of $H_{2}O = 2 + 16 = 18g$
(a) 0.2 moles of $C_{12} H_{22} O_{11} = 0.2 \times 342 = 68.4 g$
(b) 2 moles of $CO_{2}$ is $2 \times 44 = 88 g$
(c) 2 moles of $CaCO_{3} = 2 \times 100 = 200 g$
(d) 10 moles of $H_{2}O = 10 \times 18 = 180 g$

Question:6

Which of the following has maximum number of atoms?
(a) 18g of $H_{2}O$
(b) 18g of $O_{2}$
(c) 18g of $CO_{2}$
(d) 18g of $CH_{4}$
Answer: D
$N_{A} = 6.023 \times 10^{23}$
(a) 18 g of water $=18 \times \frac{3}{18}\times N_{A} = 3 N_{A}$
(b) 18 g of oxygen = $18 \times \frac{3}{32} \times N_{A} = 1.12 N _{A}$
(c) 18 g of $CO_{2} = 18 \times \frac{3}{44} \times N_{A} = 1.23 N_{A}$
(d) 18 g of $CH_{4} =18\times \frac{5}{16}\times N_{A} = 5.63 N_{A}$

Question:7

Which of the following contains maximum number of molecules?
(a) 1g $CO_{2}$
(b) 1g $N_{2}$
(c) 1g $H_{2}$
(d) 1g $CH_{4}$
Answer: C
Molar mass of other molecules are much higher than given mass, so number of molecules in them will be less than that in hydrogen.
1 g of $H_{2} = \frac{1}{2} \times N_{A}$
$= 0.5 N_{A}$
$= 0.5 \times 6.022 \times 10^{23}$
$= 3.011 \times 10^{23}$

Question:8

Mass of one atom of oxygen is
(a) $\frac{16}{6.023\times10^{23}}g$
(b) $\frac{32}{6.023\times10^{23}}g$
(c) $\frac{1}{6.023\times10^{23}}g$
(d) 8u
Answer:
Mass of one atom of oxygen $=\frac{Atomic \;mass}{N_{A}}$
$=\frac{16}{6.023\times10^{23}}g$

Question:9

3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
(a) $6.68\times 10^{23}$
(b) $6.09\times 10^{22}$
(c) $6.022\times 10^{23}$
(d) $6.022\times 10^{21}$
Answer: A
1 mol of sucrose $( C_{12}H_{22}O_{11})$ contains $= 11\times N_{A}$ atoms of oxygen,
Here, $N_{A} = 6.023\times10^{23}$

1. mol of sucrose $( C_{12}H_{22}O_{11})$ contains $= 0.01 \times 11 \times N_{A}$ atoms of oxygen
  $= 0.11\times N_{A}$ $=0.11\times N_{A}$ atoms of oxygen
  $=18 g/(1\times2+16)$ gmol-1
  = 1mol
  1mol of water $(H_{2}O)$ contains $1 \times N_{A}$ atom of oxygen
  Total number of oxygen atoms = Number of oxygen atoms from sucrose + Number of oxygen atoms from water
  $= 0.11 N_{A} + 1.0 N_{A}$
  $?????=1.11 N_{A}$ $=1.11 N_{A}$
  Number of oxygen atoms in solution
  $=1.11 \times$ Avogadro’s number
  $=1.11 \times 6.022 \times 10^{23}=6.68\times 10^{23}$

Question:10

A change in the physical state can be brought about
(a) only when energy is given to the system
(b) only when energy is taken out from the system
(c) when energy is either given to, or taken out from the system
(d) without any energy change
Answer: C
A change in the physical state can be brought about when energy is either given to, or taken out from the system. When a solid change into liquid, it absorbs energy. It is because, energy change helps in changing the magnitude of attraction forces between the particles, thus helps in changing the physical states (i.e., solid, liquid, gas) of matter. When a liquid change into solid, it releases energy.

NCERT Exemplar Class 9 Science Solutions Chapter 3-Short Answer

Question:11

Which of the following represents a correct chemical formula? Name it.
(a) CaCl
(b) $BiPO_{4}$
(c) $NaSO_{4}$
(d) NaS
Answer: B
The correct formula is CaCl₂ (valency of Ca = 2, valency of Cl = 1).
The correct formula is $Na_{2}SO_{4}$ (valency of Na = 1, Valency of $SO_{4} = 2$ ).
The correct formula is $Na_{2}S$ (valency of Na = 1, valency of sulphide = 2).
$BiPO_{4}$ , is the correct formula, its name is bismuth phosphate. Bismuth phosphate is right because both ions are trivalent.

Question:12

Write the molecular formulae for the following compounds
(a) Copper (II) bromide
(b) Aluminium (III) nitrate
(c) Calcium (II) phosphate
(d) Iron (III) sulphide
(e) Mercury (II) chloride
(f) Magnesium (II) acetate
Solution:
(a) Copper(II) Bromide – $CuBr_{2}$
(b) Aluminium(III) nitrate – $Al (NO_{3})_{3}$
(c) Calcium (II) phosphate – $Ca_{3}(PO_{4})_{2}$
(d) Iron(III) sulphide – $Fe_{2}S_{3}$
(e) Mercury(II) chloride – $HgCl_{2}$
(f) Magnesium(II) acetate – $Mg(CH_{3}COO)_{2}$

Question:13

Write the molecular formulae of all the compounds that can be formed by the combination of following ions
$Cu^{2+}, Na^{+}, Fe^{3+}, Cl^{-}, SO_{4}^{2-} , PO_{4}^{3-}$
Solution.
$CuCl_{2}/ CuSO_{4} / Cu_{3} (PO_{4})_{2}$
$NaCl/ Na_{2}SO_{4}/ Na_{3} PO_{4}$
$FeCl_{3}/ Fe_{2} (SO_{4})_{3} / FePO_{4}$

Question:14

Write the cations and anions present (if any) in the following compounds
(a) $CH_{3}COONa$
(b) $NaCl$
(c) $H_{2}$
(d) $NH_{4}NO_{3}$
Solution.
Anions Cations
(a) $CH_{3} COO^{-}$ $Na^{+}$
(b) $Cl^{-}$ $Na^{+}$
(c) It is a covalent compound
(d) $NO_{3}^{-}$ $NH_{4}^{+}$

Question:15

Give the formulae of the compounds formed from the following sets of elements
(a) Calcium and fluorine
(b) Hydrogen and Sulphur
(c) Nitrogen and hydrogen
(d) Carbon and chlorine
(e) Sodium and oxygen
(f) Carbon and oxygen
Answer:
(a) Calcium and fluoride - Calcium Fluoride $(CaF_{2})$
(b) Hydrogen and sulphur - Hydrogen Sulphide $(H_{2}S)$
(c) Nitrogen and hydrogen - Ammonia $(NH_{3})$
(d) Carbon and chlorine - Carbon Tetrachloride $(CCl_{4})$
(e) Sodium and oxygen - Sodium Oxide $(Na_{2}O)$
(f) Carbon and oxygen - Carbon dioxide $(CO_{2})$ ; Carbon Monoxide (CO)

Question:16

Which of the following symbols of elements are incorrect? Give their correct symbols
(a) Cobalt CO
(b) Carbon c
(c) Aluminum AL
(d) Helium He
(e) Sodium So
Answer:
(a) Incorrect, the correct symbol of cobalt is Co
(b) Incorrect, the correct symbol of carbon is C
(c) Incorrect, the correct symbol of aluminum is Al
(d) Correct (He)
(e) Incorrect, the correct symbol of sodium is Na

Question:17

Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them. (You mayuse appendix-III).
(a) Ammonia
(b) Carbon monoxide
(c) Hydrogen chloride
(d) Aluminum fluoride
(e) Magnesium sulphide
Answer:

(a) $NH_{3}$
$N:H\times3$
$14:1\times3$
(b) $CO$
$C:O$
$12:16$
$3:4$
(c) $HCl$
$H:Cl$
$1:35.5$
$2:71$
(d) $AlF_{3}$
$Al:F\times3$
$27:19\times3$
$9:19$
(e) $MgS$
24:32 $3:4$

Question:18

State the number of atoms present in each of the following chemical species
(a) $CO_{3}^{2-}$
(b) $PO_{4}^{3-}$
(c) $P_2O_5$
(d) $CO$
Solution.
(a) $CO_{3}^{2-} = 1C + 3(O) = 4$
(b) $PO_{4}^{3-} = 1P + 4(O) = 5$
(c) $P_{2}O_{5} = 2P + 5(O) = 7$
(d) CO = 1C + 1(O) = 2

Question:19

What is the fraction of the mass of water due to neutrons?
Solution.
In water molecule $(H_{2}O)$ , number of neutrons = [(number of neutrons in H) $\times2+$ (number of neutrons in O)] $= 0\times 2 + 8 = 8$ (as number of
neutrons in H = 0)
Mass of 8 neutrons $= 8\times 1.00893 = 8.07$ (mass of one neutron = 1.008934)
Molar mass of water $= 1.008\times 2 + 16.0 = 18.016 u$
There are 8 neutrons in one atom of oxygen
Mass of 8 neutrons $=\frac{8}{N_{A}} g$
Fraction of mass of water due to neutrons = $\frac{8}{18}$

Question:20

Does the solubility of a substance change with temperature? Explain with the help of an example.
Solution.
Yes, the solubility of a substance depends on its temperature. The solubility of solids in liquids usually increases on increasing the temperature and decreases on decreasing the temperature. The solubility of gases in liquids usually decreases on increasing the temperature and increases on decreasing the temperature.
For example, one can dissolve more sugar in hot water than in cold water.

Question:21

Classify each of the following on the basis of their atomicity.
(a) $F_{2}$
(b) $NO_{2}$
(c) $N_{2}O$
(d) $C_{2}H_{6}$
(e) $P_{4}$
(f) $H_{2}O_{2}$
(g) $P_{4}O_{10}$
(h) $O_{3}$
(i) HCl
(j) $CH_{4}$
(k) He
(l) Ag
Answer:
(a) Diatomic (2 atoms)
(b) Triatomic (3 atoms)
(c) Triatomic (3 atoms)
(d) Polyatomic (8 atoms)
(e) Polyatomic (4 atoms)
(f) Polyatomic (4 atoms)
(g) Polyatomic (14 atoms)
(h) Triatomic (3 atoms)
(i) Diatomic (2 atoms)
(j) Polyatomic (5 atoms)
(k) Monoatomic (1 atom)
(l) Monoatomic (1 atom)

Question:22

You are provided with a fine white colored powder which is either sugar or salt. How would you identify it without tasting?

Answer.
There are many ways by which we can differentiate between sugar and salt without testing. We can identify the powder by heating. The powder will char if it is a sugar. Dissolve salt and sugar separately in alcohol, sugar will be dissolved in it while salt will not be dissolved. the powder can be dissolved in water and checked for electrical conductivity. If it conducts, then the powder is salt.

Question:23

Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of
magnesium is $24\;g \;mol^{-1}$ .
Answer:
Weight of a sample in grant = Number of moles × Molar mass
Atomic mass of Mg $= 24 g mol^{-1}$
24 g of Mg = 1 mol
12 g of Mg = 12/24 = 0.5 mol

NCERT Exemplar Class 9 Science Solutions Chapter 3-Long Answer

Question:24

Verify by calculating that
(a) 5 moles of $CO_{{2}}$ and 5 moles of $H_{2}O$ do not have the same mass.
(b) 240 g of calcium and 240 g magnesium elements have a mole ratio of $3:5$ .
Answer:
(a) $CO_{2}$ has molar mass $= 44g mol^{-1}$
5 moles of $CO_{2}$ have molar mass $= 44 \times 5 = 220 g$
$H_{2}O$ has molar mass $= 18 g mol^{-1}$
5 moles of $H_{2}O$ have mass $= 18 \times 5 g = 90 g$
Hence, they do not have same mass.
(b) Number of moles in 240g Ca metal $=\frac{240}{40} = 6$
Number of moles in 240g of Mg metal $=\frac{240}{24} = 10$
Ratio $6:10 = 3: 5$

Question:25

Find the ratio by mass of the combining elements in the following compounds. (You may use Appendix-III)
(a) $CaCO_{3}$ (d) $C_{2}H_{5}OH$
(b) $MgCl_{2}$ (e) $NH_{3}$
(c) $H_{2}SO_{4}$ (f) $Ca(OH)_{2}$
Answer:
(a) $CaCO3 \rightarrow Ca: C : O = 40 : 12 : 48 = 10 : 3 : 12$
(b) $MgCl_{2} \rightarrow Mg : Cl = 24 : 2 \times 35.5 = 24 : 71$
(c) $H_{2}SO_{4} \rightarrow H : S : O = 2 \times 1 : 32 : 4 \times 16 = 2 : 32 : 64 = 1 : 16 : 32$
(d) $C_{2} H_{5} OH \rightarrow C : H : O = 2 \times 12 : 6 \times 1 : 16 = 24 : 6 : 16 = 12 : 3: 8$
(e) $NH_{3}\rightarrow N : H = 14 : 3\times 1 = 14 : 3$
(f) $Ca (OH)_{2} \rightarrow Ca : O : H = 40 : 2 \times 16 : 2 \times 1 = 40 : 32 : 2 = 20 : 16 : 1$

Question:26

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.
$CaCl_{2} (aq) \rightarrow Ca^2+ (aq) + 2Cl^{-} (aq)$
Calculate the number of ions obtained from $CaCl_{2}$ when 222 g of it is dissolved in water.
Answer:
$111g(1mole)\rightarrow 1mole+2moles$
$222g(2mole)\rightarrow 2moles+2moles$
Total number of ions by 2 moles of $CaCl_{2} = 6$
Number of ions = Number of moles of ions $\times N_{A}$
$=6\times 6.023\times 10^{23}$
$=3.6132\times 10^{24}$ ions

Question:27

The difference in the mass of 100 moles each of sodium atoms and sodium ions is 5.48002 g. Compute the mass of an electron.
Answer:
Number of electrons in Na atom = 11
Number of electrons in $Na^{+} = 10$
For 1 mole of Na atom and Na⁺ the difference in electrons = 1 mole
For 100 moles of Na atoms and Na⁺ ions the difference = 100 moles of electrons
Mass of 100 moles of electrons = 5.48002 g
Mass of 1 mole of electrons $=\frac{5.48002}{100}g$
Mass of one electron $=\frac{5.48002}{100\times6.022\times10^{23}}g$
$=9.1\times10^{-28}g$
$=9.1\times 10^{-31}kg$

Question:28

Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? Molar mass of Hg and S are 200.6 g mol^–1 and 32 g mol^–1 respectively.

Answer:
Molar mass of HgS $=200.6+32=232.6gmol^{-1}$
Mass of Hg in 232.6g of HgS =200.6g
Mass of Hg in 225g of HgS $=\frac{200.6}{232.6}\times225$
$=194.04g$

Question:29

The mass of one steel screw is 4.11g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth $(5.98 \times 10^{24}kg)$ . Which one of the two is heavier and by how many times?
Answer:

One way of srews weigh $=2.475\times 10^{24}g$
$=2.475\times 10^{21}kg$
$\frac{Mass \;of\; the\; earth}{Mass\; of\; one\; mole\; of\; screws}=\frac{5.98\times 10^{24}}{2.475\times 10^{21}}=2.4\times 10^{3}$
Mass of earth is $2.4 \times 10^{3}$ times the mass of screws. The earth is 2400 times heavier than one mole of screws

Question:30

A sample of vitamic C is known to contain $2.58 \times10^{24}$ oxygen atoms. How many moles of oxygen atoms are present in the sample?
Answer:

1 mole of oxygen atoms $= 6.023 \times 10^{23} atoms$
∴ Number of moles of oxygen atoms $=\frac{2.58\times 10^{24}}{6.023\times 10^{23}}$
$=4.28moles$

Question:31

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of same weight. (a) Whose container is heavier? (b) Whose container has more number of atoms?
Solution.
(a) Mass of sodium atoms carried by Krish $= (5 \times 23) g = 115 g$
While mass of carbon atom carried by Raunak $= (5 \times12) g = 60g$
Thus, Krish’s container is heavy.
(b) As the two bags have the same number of moles of atoms, both the bags will have the same number of atoms.

Question:32

Fill in the missing data in the Table

Answer.

Question:33

The visible universe is estimated to contain $10^{22}$ stars. How many moles of stars are present in the visible universe?
Answer:
1 mole of stars equals $6.023\times10^{23}$
Number of moles of stars $=\frac{10^{22}}{6.022\times 10^{23}}$
Number of moles of stars = 0.0166

Question:34

What is the SI prefix for each of the following multiples and submultiples of a unit?
(a) $10^{3}$ (b) $10^{-1}$ (c) $10^{-2}$ (d) $10^{-6}$
(e) $10^{-9}$ (f) $10^{-12}$
Answer:
(a) kilo $10^{3}$
(b) deci $10^{-1}$
(c) centri $10^{-2}$
(d) micro $10^{-6}$
(e) nano $10^{-9}$
(f) pico $10^{-12}$

Question:35

Express each of the following in kilograms
$(a) 5.84 \times 10^{-3} mg$
$(b) 58.34 g$
$(c) 0.584g$
$(d) 5.873 \times 10^{-21}g$
Answer:

$(a) 1 mg = 10^{-6} kg\: \: \:$
$5.84 \times 10^{-3} mg = 5.84 \times 10^{-3} \times10^{-6} kg = 5.84 \times10^{-9} kg$
$(b) 1 g = 10^{-3}kg$
$58.34 g = 58.34 \times 10^{-3} kg = 5.834 \times10^{-2} kg$
$(c) 1 g = 10^{-3} kg$
$0.584g = 0.584g \times 10^{-3} kg = 5.84 \times10^{-4}kg$
$(d) 1 g = 10^{-3} kg$
$5.873 \times 10^{-21} g = 5.873 \times 10^{-24} \times 10^{-3} kg = 5.873 \times 10^{-24} kg$

Question:36

Compute the difference in masses of $10^3$ moles each of magnesium atoms and magnesium ions. $(Mass \: of \: an\: electron = 9.1\times 10^{-31} kg)$
Answer:

$Mg^{2+}$ ion and Mg atom differ by two electrons.
$10^3$ moles of $Mg^{2+}$ and Mg atoms would differ by $2 \times 10^3$ moles of electrons.
Mass of $2 \times 10^3$ moles of electrons.
$= 2 \times 10 ^3 \times 6.023 \times 10^{23} \times 9.1 \times 10^{-31} kg$
$\Rightarrow 2 \times 6.022 \times 9.1 \times 10^{-5} kg$
$\Rightarrow 109.6004 \times 10^{-5} kg$
$\Rightarrow 1.096 \times 10^{-3} kg$

Question:37

Which has more number of atoms?
$100g \: \: of \: \: N_2 \: \: or\: \: 100 g\: \: of \: \: NH_3$
Answer:

$(i)100g\: \: of\: \: N_2 = \frac{100}{28}moles$
$Number \: of \: molecules = \frac{100}{28} \times 6.022 \times 10 ^{23}$
$Number \: of \: atoms = \frac{2 \times 100}{28} \times 6.022 \times 10 ^{23}$
$=43.01 \times 10^{23}$
$(ii)100g\: \: of\: \: NH_3 = \frac{100}{17}moles$
$Number \: of \: molecules = \frac{100}{17} \times 6.022 \times 10 ^{23}$ $Number \: of \: atoms = \frac{4 \times100}{28} \times 6.022 \times 10 ^{23}$
$=141.69 \times 10^{23}$
$NH_3$ will have more atoms.

Question:38

Compute the number of ions present in 5.85 g of sodium chloride.
Answer:
$5.85g$ of $NaCl=\frac{5.85}{58.5}=0.1moles$
Or, 0.1 moles of NaCl particles.
Each NaCl particle is equivalent to one $Na^{+}$ one $Cl^{-}$ i.e. 2 ions.
Total moles of ions $= 0.1 \times 2 = 0.2 moles$
No. of ions $= 0.2 \times 6.022 \times 10^{23} = 1.2042 \times 10^{23} ions$

Question:39

A gold sample contains $90\%$ of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?

Answer:
One gram of gold sample will contain $\frac{90}{100}=0.9g$ of gold
Number of moles of gold $=\frac{Mass\; of \;gold}{Atomic\; mass\; of\; gold}$
$=\frac{0.9}{197}$
$=0.0046$
One mole of gold contains $N_{A}$ atoms $=6.022\times 10^{23}$
$\therefore0.0046mole$ of gold will contain $=0.0046\times 6.022\times10^{23}atoms$
$=2.77\times10^{21}atoms$

Question:40

What are ionic and molecular compounds? Give examples.
Answer:
While forming some compounds, atoms may gain or lose electrons, thereby forming electrically charged particles called ions. Compounds that are formed by the attraction of cations and anions are known as ionic compounds.
Ex : $2Na + Cl_2 \rightarrow 2Na + Cl^{-} \rightarrow 2NaCl$ (sodium chloride- common salt.)
Compounds formed by the bonding of uncharged species are known as molecular compounds. The bonding is called covalent bonding. Molecular compounds are formed by sharing of electrons between the two atoms. Ex: $2C + O_{2} \rightarrow 2CO$ (Carbon monoxide)

Question:41

Compute the difference in masses of one mole each of aluminum atoms and one mole of its ions.
$\left ( Mass \: \: of\: \: an\: \: electron\: \: is\: \: 9.1 \times 10^{-28} g \right )$ Which one is heavier?
Answer:

Ionization of Al atom occurs as
$Al\rightarrow Al^{3+}+3e^-$
Therefore, $Al^{3+}$ ion is formed from Al atom by loss of 3 electrons. Difference in mass of 1 mole of Al atoms and 1 mole of $Al^{3+}$ ions
$= mass\: \: of \: \: 3 \times 6.022 \times 10^{23} electrons$
$= (3 \times 6.022 \times 10^{23}) \times (9.1 \times 10^{-28} g)$
$(as \: \: mass\: \: of\: \: an \: \: electron = 9.1 \times 10^{-28} g)$
$=164.4 \times 10^{-5}g=1.644 \times 10^{-3}g$
$1$ mole of Al atoms is heavier than 1 mole of $Al^{3+}$ ions.
It is because 1 mole of Al atom have $e^-= 13\times 6.022 \times 10^{23 }$ electrons and $1$ mole of Al ion have
$e=10\times 6.022 \times 10^{23 }$ electrons

Question:42

A silver ornament of mass’m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
Answer:
Mass of silver =mg
Mass of gold deposited $=\frac{m\times1}{100}=\frac{m}{100}g$
No. of atoms of silver $=\frac{Mass}{Atomic mass}\times N_{A}$ $=\frac{m}{108}\times N_{A}$
No. of atoms of gold $=\frac{Mass}{Atomic mass}\times N_{A}=\frac{m}{100\times 197}\times N_{A}$
Ratio of the number of atoms of gold to silver $=\frac{m}{100\times 197}\times N_{A}:\frac{m}{108}\times N_{A}$
$108:100\times 197=108:19700=1.182.41$

Question:43

A sample of ethane $(C_2H_6)$ gas has the same mass as $1.5 \times 10^{20}$ molecules of methane $(CH_4)$ . How many $(C_2H_6)$ molecules does the sample of gas contain?
Answer:

$Mass \: of\: 1\: molecule\: of\: CH_4=\frac{16}{N_A}g$
$Mass \: of\: 1.5 \times 10^{20}\: molecule\: of\: CH_4=\frac{1.5\times 10^{20}\times 16}{N_A}g$
$Mass \: of\: 1\: molecule\: of\: C_2H_6=\frac{30}{N_A}g$
$Mass \: of\: C_2H_6\: molecules\: =\frac{1.5 \times 10^{20} \times 16}{N_A}g$
Therefore, number of molecules of $C_2H_6 = \frac{1.5 \times 10^{20} \times 16}{N_A}g \times \frac{N_A}{16}$
$=0.8 \times 10^{20}$

Question:44

Fill in the blanks
(a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called ————.
(b) A group of atoms carrying a fixed charge on them is called ————.
(c) The formula unit mass of $Ca_3 (PO_4)_2$ is ————.
(d) Formula of sodium carbonate is ———— and that of ammonium sulphate is ————.
Answer:

(a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called Law of conservation of mass.
It states that mass is neither created nor destroyed in chemical reactions.
(b) A group of atoms carrying a fixed charge on them is called ions.
An ion is a charged atom or molecule. It is charged because the number of electrons are not equal to the number of protons in the atom or molecule.

(c) The formula unit mass of $Ca_3(PO_4)_2$ is $310$ .
$Ca_3 (PO_4)_2 = (40) 3 + [31 + (16) 4] 2 = 120 + 190 = 310$
(d) Formula of sodium carbonate is $Na_2CO_3$ and that of ammonium sulphate is $(NH_4)_2SO_4$

Question:45

Complete the following crossword puzzle (Figure) by using the name of the chemical elements. Use the data given in the table following.

Answer:

Question:46

(a) In this crossword puzzle (Fig 3.2), names of 11 elements are hidden. Symbols of these are given below. Complete the puzzle.
1. Cl 7. He
2. H 8. F
3. Ar 9. Kr
4. O 10. Rn
5. Xe 11. Ne
6. N

(b) Identify the total number of inert gases, their names and symbols from this cross word puzzle.
Answer:
(a)

(b) Six inert gases:
Helium (He); Neon ( Ne); Argon (Ar); Krypton (Kr); Xenon (Xe); Radon (Rn).

Question:47

Write the formulae for the following and calculate the molecular mass for each one of them.
(a) Caustic potash
(b) Baking powder
(c) Lime stone
(d) Caustic soda
(e) Ethanol
(f) Common salt
Answer:
(a) $KOH = 39 + 16 + 1 = 56 u$
(b) $NaHCO_{3} = 23 + 1 + 12 + 3 \times 16 = 84 u$
(c) $CaCO_3= 40 +12+ 3\times 16 = 100u$
(d) $NaOH = 23 + 16 + 1 = 40 u$
(e) $C_{2}H_{5}OH = 2 \times 12 + 6 \times 1 + 16 = 46 u$
(f) $NaCl = 23 + 35.5 = 58.5 u$

Question:48

In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula $C_6 H_{12} O_6$ . How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed assuming the density of water to be $1 \: g\: cm^{-3}$ .
Answer:

The chemical reaction involved is :
$6CO_2 + 6H_2O \xrightarrow[Sunlight (6 \times 10 g)]{Chlorophyll}C_6H_{12}O_6 \: \: (180g) + 6O_2$
To produce 18g of glucose, mass of water consumed = $108g$
To produce 18g of glucose, mass of water consumed $= \frac{(108g)}{(180g)} \times (18g) = 10.8g$
$Density \: of\: water=1\: g\: cm^{-3}$
$Volume \: of\: water = \frac{Mass\: of\: water}{Density\: of\: water}=\frac{(10.8g)}{(1gcm^{-3})}=10.8\: g\: cm^{3}$

NCERT Exemplar Solutions Class 9 Science Chapter 3 Important Topics:

NCERT exemplar Class 9 Science solutions chapter 3 includes the following topics :

Definitions of atoms and molecules.
The fundamental quantity which defines the amount of substance is called a mole.
Molecular weight and atomic weight of any substance.
How to find out the molecular weight of any compound if we know the chemical formula of the compound.
NCERT exemplar Class 9 Science solutions chapter 3 discusses the Avogadro number which will tell a number of atoms or molecules in one mall of the substance.
Study of the molecules and atoms and their molecular weight.

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Science Exemplar Solutions for Other Chapters:

Chapter wise solutions

Chapter 1 Matter in our Surroundings

Chapter 2 Is Matter Around Us Pure?

Chapter 4 Structure of the Atom

Chapter 5 The Fundamental Unit of Life

Chapter 6 Tissues

Chapter 7 Diversity in Living Organisms

Chapter 8 Motion

Chapter 9 Forces and Laws of Motion

Chapter 10 Gravitation

Chapter 11 Work and Energy

Chapter 12 Sound

Chapter 13 Why do We Fall ill?

Chapter 14 Natural Resources

Chapter 15 Improvement in Food Resources

Features of NCERT Exemplar Class 9 Science Solutions Chapter 3:

These Class 9 Science NCERT exemplar chapter 3 solutions provide a basic understanding of atoms and molecules. This chapter discusses different chemical formulas for different type of compounds and enables a student to judge if the chemical formula given is correct or not. Atoms and Molecules can be practiced by the students of Class 9 through these NCERT exemplar Class 9 Science solutions chapter 3 and can be leveraged to solve other reference books such as NCERT Class 9 Science TextBook, Lakhmir Singh and Manjit Kaur by S. Chand et cetera.

NCERT exemplar Class 9 Science solutions chapter 3 pdf download enables the students to download/view these solutions in an offline environment while studying NCERT exemplar Class 9 Science chapter 3.

Check the Solutions of Questions Given in the Book

Chapter No.	Chapter Name
Chapter 1	Matter in Our Surroundings
Chapter 2	Is Matter Around Us Pure
Chapter 3	Atoms and Molecules
Chapter 4	Structure of The Atom
Chapter 5	The Fundamental Unit of Life
Chapter 6	Tissues
Chapter 7	Diversity in Living Organisms
Chapter 8	Motion
Chapter 9	Force and Laws of Motion
Chapter 10	Gravitation
Chapter 11	Work and Energy
Chapter 12	Sound
Chapter 13	Why Do We Fall ill?
Chapter 14	Natural Resources
Chapter 15	Improvement in Food Resources

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Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Q1. What is Avogadro’s number?

A1. Avogadro number tells us a number of molecules or atoms in one mole of substance

Its value is 6.023×1023. For example, molecular weight of water is 18 therefore one mole of water will have 18 g and it will have number of water molecules equal to Avogadro number.

2. Q2. Does oxygen exist in atomic form?

A2. No, oxygen does not exist in the atomic form in nature. Oxygen exists either in the form of a diatomic molecule or triatomic molecules. They are called oxygen gas and ozone respectively.

3. Q3. 2 g of oxygen and 2 g of helium gas are kept in a chamber. What is the ratio of their moles?

A3. The molecular weight of oxygen is 32 therefore 2 g of oxygen will have 1 /16 moles.

The molecular weight of helium is four therefore 2 g of helium will have 1 /2 moles.

Hence, in the mixture, the ratio of oxygen and helium will be 1:8.

4. Q4. Does the atomic weight of any atom change when it gets ionised?

A4. The atomic weight of any atom does not change when it gets ionised. In ionisation, only electrons are transferred which have negligible weight. Thus, they do not affect the atomic weight of the atom.

5. Q5. How many questions are expected from the chapter on Atoms and Molecules?

A4. Generally, the CBSE pattern suggests that the final paper for Class 9 should consist of a mixture of objective type (1 question) and short answer type(2 questions). A student going with the NCERT exemplar Class 9 Science solutions chapter 3 thoroughly can easily score well in the final paper

Also Read

NCERT Solutions for Class 9 - Subject-wise PDFs Updated for 2024-25

NCERT Solutions for Class 9 Science

NCERT Solutions for Class 9 Maths

NCERT Syllabus for Class 9 Maths 2023 - Download Syllabus Pdf

NCERT Solutions For Class 9 Science Chapter 1 Matter in Our Surroundings

NCERT Solutions for Exercise 2.5 Class 9 Maths Chapter 2 Polynomials

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

NCERT Exemplar Class 9 Maths Solutions - Chapter Wise Problems with Solutions

NCERT Exemplar Class 9 Solutions - Subject Wise Problems with Solutions

NCERT Exemplar Class 9 Science Solutions Chapter 15 Improvement in Food Resources

NCERT Exemplar Class 9 Science Solutions Chapter 14 Natural Resources

NCERT Exemplar Class 9 Science Solutions Chapter 13 Why do we fall ill?

NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 9 Science Solutions Chapter 3 Atoms and Molecules

NCERT Exemplar Class 9 Science Solutions Chapter 3-MCQ

NCERT Exemplar Class 9 Science Solutions Chapter 3-Short Answer

NCERT Exemplar Class 9 Science Solutions Chapter 3-Long Answer

NCERT Exemplar Solutions Class 9 Science Chapter 3 Important Topics:

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