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

NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules (MCQ, SAQ and LAQ)

Chapter Name	NCERT Exemplar Solutions for Class 9 Science Ch 3 Atoms and Molecules
Topics Covered	Objective Type Questions (MCQ's) Short Answer Type Questions Long Answer Type Questions
Related Study	NCERT Solutions for Class 9 Science NCERT Revision Notes for Class 9 Science Important Questions for Class 9 Science MCQ for Class 9 Science NCERT Exemplar Questions For Class 9 Science

Objective Type Questions for Atoms and Molecules

1. Which of the following correctly represents 360 g of water?
(i) 2 moles of H₂O
(ii) 20 moles of water
(iii) 6.022 × 10²³ molecules of water

(iv) 1.2044 × 10²⁵ molecules of water

(a) (i)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)

Solution

(d) (ii) and (iv)

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.

Solution

(a) Atoms are not able to exist independently.

3. The chemical symbol for nitrogen gas is :
(a) Ni
(b) N₂
(c) N⁺
(d) N

Solution

(b) N₂

4. The chemical symbol for sodium is :
(a) So
(b) Sd
(c) NA
(d) Na

Solution

(d) Na

5. Which of the following would weigh the highest?
(a) 0.2 mole of sucrose (C₁₂ H₂₂ O₁₁)
(b) 2 moles of CO₂
(c) 2 moles of CaCO₃
(d) 10 moles of H₂O

Solution

(c) 2 moles of CaCO₃

6. Which of the following has maximum number of atoms ?
(a) 18g of H₂O
(b) 18g of O₂
(c) 18g of CO₂
(d) 18 g of CH₄

Solution

(d) 18 g of CH₄

7. Which of the following contains maximum number of molecules ?
(a) 1 g CO₂
(b) 1 g N₂
(c) 1 g H₂
(d) 1 g CH₄

Solution

(c) 1 g H₂

8. Mass of one atom of oxygen is :
(a) 16/(6.023 × 10²³ g)
(b) 32/(6.023 × 10²³ g)
(c) 1/(6.023 × 10²³ g)
(d) 8u

Solution

(a) 16/(6.023 × 10²³ g)

9. 3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are :
(a) 6.68 × 10²³
(b) 6.09 × 10²²
(c) 6.022 × 10²³
(d) 6.022 × 10²¹

Solution

(a) 6.68 × 10²³

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.

Solution

(c) When energy is either given to, or taken out from the system.

Short Answer Questions Atoms and Molecules

11. Which of the following represents a correct chemical formula? Name it.
(a) CaCl
(b) BiPO₄
(c) NaSO₄
(d) NaS

Solution

(a) CaCl = Wrong (valency of Ca = 2, Cl = 1)
(b) BiPO₄ = Correct (valency of Bi = 3, PO₄ = 3)
(c) NaSO₄ = Wrong (valency of Na = 1, SO₄ = 2)
(d) NaS = Wrong (valency of Na = 1, Sulphide = 2)

12. Write the molecular formulae for the following compounds.

Solution

13. Write the molecular formulae of all the compounds that can be formed by the combination of given ions : Cu³⁺, Na⁺, Fe³⁺, Cl, SO³₄ PO₄³ .

Solution

CuCl₂, CuSO₄, Cu₃(PO₄), NaCl, Na₂SO₄, Na₃PO₄, FeCl₃ , Fe(SO₄)₃, FePO₄ .

14. Write the cations and anions present (if any) in the following compounds :
(a) CH₃COONa
(b) NaCl
(c) H₂
(d) NH₄NO₃

Solution

Compounds	Cations	Anions
CH3COONa	Na+	CH3COO
NaCl	Na+	Cl
H2	Nil	Nil
NH4NO3	NH4+	NO3

15. Give the formulae of the compounds formed from the following sets of elements.

Solution 

16. Write the symbols of given elements.

Solution

(a) Cobalt	Co	(b) Carbon	C
(c) Aluminium	Al	(d) Helium	He
(e) Sodium	Na

17. Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them.
(a) Ammonia
(b) Carbon monoxide
(c) Hydrogen chloride
(d) Aluminium fluoride
(e) Magnesium sulphide

Solution

(a) NH3	(b) CO	(c) HCl	(d) AlF3	(e) MgS
N : H × 3	C : O	H : Cl	Al : F × 3	Mg : S
14 : 1 × 3	12 : 16	1 : 35.5	27 : 19 × 3	24 : 32
14 : 3	3 : 4	1 : 35.5	9 : 19	3 : 4

18. State the number of atoms present in each of the following chemical species.
(a) CO^2-₃
(b) PO^3-₄
(c) P₂O₅
(d) CO

Solution

(a)	CO2-3	1 + 3 = 4
(b)	PO3-4	1 + 4 = 5
(c)	P2O5	2 + 5 = 7
(d)	CO	1 + 1 = 2

19. What is the fraction of the mass of water due to neutrons?

Solution

1. Mass of 1 molecule of water = 18 amu
2. No. of proton in 2 atoms of H = 2 and no. of neutron = 0
3. No. of proton in 1 atom of O = 8 and no. of neutron = 8
4. Fraction of mass due to neutron in water 8/18 = 4/9

20. Does the solubility of a substance change with temperature? Explain with the help of an example.

Solution

Yes, it is a temperature dependent property. The solubility generally increases with increase in temperature. For example : We can dissolve more sugar in hot water than in cold water.

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

Solution

On heating the powder, it will char if it is a sugar. Alternatively, the powder may be dissolved in water and checked for its conduction of electricity. If it conducts, it is a salt.

Long Answer Questions Atoms and Molecules

22. Verify by calculating that :
(a) 5 moles of CO₂ and 5 moles of H₂O do not have the same mass.
(b) 240 g of calcium and 240 g of magnesium elements have a mole ratio of 3 : 5.

Solution

(a) CO₂ has molar mass = 44 g mol ^{– 1}
5 moles of CO₂ have molar mass = 44 × 5 = 220 g
H₂O has molar mass = 18 g mol^-1
5 moles of H₂O have mass = 18 × 5g= 90 g
(b) Number of moles in 240 g Ca metal 240/40 = 6
Number of moles in 240 g of Mg metal 24 = 240/24 = 10
Ratio, 6 : 10 = 3 : 5

23. Find the ratio by mass of the combining elements in the following compounds :
(a) CaCO₃
(b) MgCl₂
(c) H₂SO₄
(d) C₂H₃OH
(e) NH₃
(f) Ca(OH)₂

Solution

(a) CaCO₃
Ca : C : O × 3
40 : 12 : 16 × 3
40 : 12 : 48
10 : 3 : 12
(b) MgCl₂
Mg : Cl × 2
24 : 35.5 × 2
24 : 71
(c) H₂SO₄
H × 2 : S : O × 4
1 × 2 : 32 : 16 × 4
2 : 32 : 64
1 : 16 : 32
(d) C₂H₅OH
C × 2 : H × 6 : O
12 × 2 : 1 × 6 : 16
24 : 6 : 16
12 : 3 : 8
(e) NH₃
N : H × 3
14 : 1 × 3
14 : 3
(f) Ca(OH)₂
Ca : O × 2 : H × 2
40: 16 × 2 : 1 × 2
40 : 32 : 2

24. Calcium chloride when dissolved in according to the following equation :
CaCl_2(aq) → Ca²⁺_(aq) + 2Cl_(aq)
Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water.

Solution

1 mole of calcium chloride = 111 g
∴ 222 g of CaCl₂ is equivalent to 2 moles of CaCl₂ since 1 formula unit CaCl₃ gives 3 ions, therefore, 1 mol of CaCl₂ will give 3 moles of ions.
2 moles of CaCl₂ would give 3 × 2 = 6 moles of ions
No. of ions = No. of moles of ions × Avogadro number
= 6 × 6.022 × 10²³
= 36.132 × 10²³ = 3.6132 × 10²⁴ ions

25. 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.

Solution

Molar mass of HgS = 200.6 + 32 = 323.6 g mol^-1
Mass of Hg in 232.6 g of HgS 200.6 g
Mass of Hg in 225 g of HgS = 200.6/232.6 ×225 = 194.04 g

26. A sample of vitamin C is known to contain 2.58 × 1024 oxygen atoms. How many moles of oxygen atoms are present in the sample?

Solution

1 mole of oxygen atoms = 6.022 × 1023 atoms Number of moles of oxygen atoms
= (2.58 × 10²⁴)/(6.022 × 10²³) = 4.28 mol
4.28 moles of oxygen atoms.

27. Raunak took 5 moles of carbon atoms in a container arid Krish also took 5 moles of a sodium atoms in another container of same weight.
(a) Whose container is heavier ?
(b) Whose countainer has more number of atoms ?

Solution

(a) Mass of sodium atoms carried by Krish
= (5 × 23) g = 115 g
While mass of carbon atom carried by Raunak
= (5 × 12) g = 60 g
Thus, Krish’s container is heavy.
(b) Both the bags have same number of atoms as they have same number of moles of atoms.

28. Fill in the missing data in the table :

Species property	Water	CO2	Na – Atom	MgCl2
No. of moles	2	………	………	0.5
No. of particles	…….	3.011 × 1023	……..	……..
Mass	36 g	……..	115 g	……

Solution

Species property	Water	CO2	Na – Atom	MgCl2
No. of moles	2	0.5 mole	5 moles	0.5
No. of particles	2×6.022×1023	3.011 × 1023	5×6.022×1023	0.5 × 6.022 × 1023 ×3
Mass	36 g	22 g	115 g	47.5 g

29. What is the SI prefix for each of the following multiples and submultiples of a unit?

(a)	103	(b)	10-3
(c)	10-2	(d)	10-6
(e)	10-9	(f)	10-19

Solution

(a)	kilo	(b)	deci
(c)	centi	(d)	micro
(e)	nano	(f)	pico

30. Express each of the following in kilograms :
(a) 5.84 × 10^–3 mg
(b) 58.34 g
(c) 0.584 g
(d) 5.873 × 10^–21 g

Solution

(a) 5.84 × 10^–9 kg
(b) 5.834 × 10^–2 kg
(c) 5.84 × 10^–4 kg
(d) 5.873 × 10^–24 kg

31. Which has more number of atoms?
100 g of N₂ or 100 g of NH₃

Solution
(i) 100 g of N₂ = 100/28 moles
Number of molecules = 100/28 × 6.022 × 10²³
Number of atoms = 2 × 100/28 × 6.022 × 10²³
= 43.01 × 10²³
(ii) 100 g of NH₃ = 100/17 moles
= 100/17 × 6.022 × 10²³ molecules
= 100/17 × 6.022 × 10²³ × 4 atoms
= 141.69 × 10²³
∴ NH₃ would have more atoms.

32. Compute the number of ions present in 5.85 g of sodium chloride.

Solution

5.85 g of NaCl = 5.85/58.5 = 0.1 mole
or 0.1 mole of NaCl particle.
Each NaCl particle is equivalent to one Na⁺ one Cl^- = 2 ions
Total moles of ions = 0.1 × 2 = 0.2 mole
No. of ions = 0.2 × 6.022 × 10²³
= 1.2042 × 10²³ ions

33. 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?

Solution

One gram of gold sample will contain = 90/100 = 0.9 g of gold
Number of moles of gold = Mass of gold/Atomic mass of gld
= 0.9/197 = 0.0046
One mole of gold contians N_a atoms = 6.022 × 10²³
∴ 0.0046 mole of gold will contain = 0.0046 × 6.022 × 10²³
= 2.77 × 10²¹ 

34. What are ionic and molecular compounds? Give examples.

Solution

Atoms of different elements join together in definite proportions to form molecules of compounds. Examples : Water, ammonia, carbon dioxide. Compounds composed of metals and non-metals contain charged species. The charged species are known as ions. An ion is a charged particle and can be negatively or positively charged. A negatively charged ion is called an anion and the positively charged ion is called cation. Examples : Sodium chloride, calcium oxide.

35. Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions (mass of an electron is 9.1 × 10^–28 g). Which one is heavier?

Solution

Mass of 1 mole of aluminium atom = the molar mass of aluminium = 27 g mol–1
An aluminium atom needs to lose three electrons to become an ion, Al³⁺
For one mole of Al3+ ion, three moles of electrons are to be lost.
The mass of three moles of electrons
= 3 × (9.1 × 10^-28) × 6.022 × 10²³ g
= 27.3 × 6.022 × 10^-5 g
= 164.400 × 10^-5 g = 0.00164 g
Molar mass of Al³⁺ = (27 – 0.00164) g mol^-1
= 26.9984 g mol^-1
Difference = 27 – 26.9984 = 0.0016 g

36. 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.

Solution

Mass of silver = mg
Mass of gold = m/100 g
Number of atoms of silver = [Mass/(Atomic mass)] × N_A
= m/108 × N_A
Number of atoms of gold = (m/100) × 197 × N_A
Ratio of number of atoms of gold to silver = An : Ag
= (m/100) × 197 × N_A : m/100 × N_A
= 108 : 100 × 197
= 108 : 19700 = 1 : 182.41

37. 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 Ca₃(PO₄)₂ is _____.
(d) Formula of sodium carbonate is _____ and that of ammonium sulphate is _____.

Solution

(a) Law of conservation of mass
(b) Polyatomic ion
(c) (3 × atomic mass of Ca) + (2 × atomic mass of phosphorus) + (8 × atomic mass of oxygen) = 310
(d) Na₂CO₃: (NH₄)₂SO₄

38. Write the formulae for the following and calculate the molecular mass for each one of them :
(a) Caustic potash
(b) Baking powder
(c) Limestone
(d) Caustic soda
(e) Ethanol
(f) Common salt

Solution

(a) KOH = (39 + 16 + 1) = 56 g mol^–1
(b) NaHCO₃ = 23 + 1 + 12 + (3 × 16) = 84 g mol^–1
(c) CaCO₃ = 40 + 12 + (3 × 16) = 100 g mol^–1
(d) NaOH = 23 + 16 + 1 = 40 g mol^–1
(e) C₂H₅OH = C₂H₆O = 2 × 12 + (6 × 1) + 16 = 46 g mol^–1
(f) NaCl = 23 + 35.5 = 58.5 g mol^–1 

39. 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 C₆H₁₂O₆. 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.

Solution

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
1 mole of glucose needs 6 moles of water.
180 g of glucose needs (6×18) g of water.
1 g of glucose will need 108/180 g of water.
18 g of glucose would need 108/180 × 18 g of water = 10.8 g
Volume of water used = Mass/Density
= 10.8 g/1 g cm^-3
= 10.8 cm³