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

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Q. Why cannot we see atoms from our naked eyes?

Ans. We cannot see atoms from our naked eyes because atoms are small in size that we cannot imagine. To see atom, specified tuned microscope is needed as size of atom is measured in nanometers (1nm = 10^–9 m). For example: an atom of hydrogen is having radius of 10 nanometers.

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

Ans. Molecular compounds are formed by the combination between two non-metal elements. For example: HCl and H₂S whereas ionic compounds are formed by combination of metals and non-metals. For example: CaCl₂and CaCO₃

Q. The visible universe is estimated to contain 1022 stars. How many moles of stars are present in the visible Universe?

Ans. 1 mole of stars equals 6.023 × 10²³ Number of moles of stars in the visible Universe

$$=\frac{\text{Number of estimated stars}}{\text{Avogadro’s number}}\\=\frac{10^{22}}{6.022×10^{23}}= 0.0166 moles$$

Q. (i) Define the term valency.
(ii)Identify the valency of the following elements: Magnesium, Aluminium, Chlorine and Copper.
(iii)Classify each of the following on the basis of their atomicity.


(a) F₂	(b) NO₂
(c) N₂O	(d) C₂H₆
(e) P₄	(f) H₂O₂
(g) P₄O₁₀	(h) O₃
(i) HCl	(j) CH₄
(k) He	(l) Ag

Ans. (i) The combining capacity of an element is known as valency of that element.
(ii)Valency of the following elements is as follows:

Element	Valency
Magnesium	2
Aluminium	3
Chlorine	1
Copper	2

(iii)


Monoatomic	He, Ag
Diatomic	F ₂, HCl
Polyatomic	NO ₂, N ₂O, C ₂H ₆, P ₄, H ₂O ₂, P ₄O ₁₀, O ₃, CH ₄

Q. Answer the following questions based on Dalton’s atomic theory:
(i)Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
(ii)Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Ans. (i) Atoms combine to form a compound in small whole numbers and relative number and kinds of atoms are constant in a given compound.
(ii)Atoms are indivisible particles, i.e., mass can neither be created nor be destroyed in a chemical reaction.

Q. (i) Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃. Given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u and, O= 16 u.
(ii)Calculate the molecular mass of the following:
(a) Ethanol
(b) Methanoic acid

Ans. (i) Formula unit mass of ZnO
= Atomic mass of Zn + atomic mass of O
= 65 + 16 u = 81 u
Formula unit mass of Na₂O
= 2 × atomic mass of Na + atomic mass of O
= 2 × 23 + 16 = 62 u
Formula unit mass of K₂CO₃
= 2 × At. mass of K + atomic mass of C + 3 × atomic mass of O
= 2 × 39 + 12 + 3 × 16
= 78 + 12 + 48 = 138 u
(ii) (a) Molecular mass of Ethanol (CH₃CH₂OH or C₂H₆O)
= (2 × atomic mass of carbon) + (6 × atomic mass of hydrogen) + (1 × atomic mass of oxygen)
= (2 × 12) + (6 × 1) + (1 × 16)u
= 24 + 6 + 16 u
= 46 u
(b) Molecular mass of methanoic acid (HCOOH)
= (2 × atomic mass of o xygen) + (2 × atomic mass of hydrogen) + (1 × atomic mass of carbon)
= (2 × 16) + (2 × 1) + (1 × 12) u
= (32 + 2 +12) u
= 46 u

Q.Raunak took 5 mole of carbon atoms in a container and Krish also took 5 mole of sodium atoms in another container of same weight.
(i)Whose container is heavier?
(ii)Whose container has more number of atoms?

Ans. (i) Mass of sodium atoms carried by Krish
= 5 × 23 = 115 g
While mass of carbon atom carried by Raunak
= 5 × 12 = 60 g
Thus,Krish’s container is heavier than Raunak’s container.
(ii)Both the containershave the same number of atoms as they have the same number of moles of atoms.

Q.Write the chemical formulae for the following compounds:
(i)Magnesium chloride
(ii)Calcium oxide
(iii)Copper nitrate
(iv) Aluminium chloride
(v)Calcium carbonate

Q.(i) Define ‘formula unit’ of an ionic compound. What is the formula unit of:
(a) Calcium chloride
(b) Potassium chloride
(ii)Calculate the formula masses of the following compounds:
(a) Calcium chloride
(b) Sodium carbonate (Given: Atomic Masses: Ca = 40 u; Cl = 35.5 u; Na = 23 u; C = 12 u; O = 16 u)
(iii)Calcium chloride when dissolved in water dissociates into its ions according to the following equation.
CaCl₂(aq) → Ca₂+ (aq) + 2 Cl^–
(aq)Calculate the number of ions obtained from CaCl₂when 222 g of it is dissolved in water.

(i) The simplest combination of ions that produces an electrically neutral unit is called ‘formula unit’ of the ionic compound.
(a)Formula unit of Calcium chloride is CaCl₂
(b) Formula unit of Potassium chloride is KCl
(ii)(a) Formula mass of Calcium chloride (CaCl₂)
= (1 × atomicmass of calcium) + (2 × atomic mass of Chlorine)
= (40 + [35.5 × 2]) = (40 + 71)
u = 111 u
(b) Formula mass of Sodium carbonate (Na₂CO₃)
= (2 × atomic mass of sodium) + (1 x atomic mass of Carbon) + (3 × atomic mass of oxygen)
= (2 × 23 + 1x 12 + 3 × 16) u = 106 u
(iii)Molar mass of CaCl₂ = 40 + 2 × 35.5
= 40 + 71 = 111 g mol–1
CaCl₂ ionises in water as follows:
CaCl₂(aq) → Ca₂+ (aq) + 2 Cl–(aq)
111 g of CaCl₂ produces ions = 3 moles
= 3 × 6.022 × 1023 ions
222 g of CaCl₂ produces ions = 3 × 6.022

$$=\frac{10^{23}}{111}×222\\ = 36.132 × 10^23 = 3.6132 × 10^24 \space ions$$

Q. (i) Calculate the number of calcium ions present in 0.032 g of calcium hydroxide.
(ii)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.
(iii)Calculate the molecular mass of chloroform.

$$(i)\space \text{Molar mass of calcium hydroxide}Ca(OH)_2\\ = 1 × 40 + 2 × (16 + 1) = 40 + 34 = 74 g\\ Therefore, 74 g of calcium hydroxide contains = 6.022 × 10^{23} calcium ions\\ So, 0.032 g of calcium hydroxide contains\\ \frac{6.022× 10^{23}}{74}× 0.032 = 2.60 × 10^{20} ions.\\ (ii)\space Given, molar mass of Mg = 24 g mol–1\\ 24 g \space of Mg = 1 mol\\ ⇒ 12 g \space of Mg =\frac{1×12}{24}=\frac{1}{2}= 0.5 mol\\ (iii)\space Molecular mass of chloroform (CHCl_3)\\= (1 × atomic mass of carbon) + (1 × atomic mass of hydrogen) + (3 × atomic mass of chlorine)\\= (1 × 12) + (1 × 1) + (3 × 35.5) u\\ = (12 + 1 + 106.5) u\\ = 119.5 u$$

Q. (i) Define gram molecular mass of a substance. How much is the gram molecular mass of oxygen? (ii)If sulphur exists as S8 molecule, calculate the number of moles in 100 g of sulphur. (S = 32 u) (iii)A sample of ethane (C₂H₆) gas has the same mass as 1.5 × 1020 molecules of methane (CH₄). How many C₂H₆ molecules does the sample of gas contain?

(i) The amount of substance whose mass in grams is numerically equal to its molecular mass is called gram molecular mass of that substance. Gram molecular mass of the oxygen is 32 g. (ii) Given mass of sulphur = 100 g Molar mass of S8 molecule = 32 × 8 g = 256 g Number of moles = Given mass / Molar mass = 100/ 256 = 0.39 moles. (iii)It is given that ethane (C₂H₆) gas has the same mass as 1.5 × 1020 molecules of methane (CH₄). Now calculating the mass of ethane. We know, 16g of methane contains 6.022 × 10²³molecules. Mass of 1.5 × 10²⁰

$$\text{molecules of methane} =\frac{16×1.5×10^{20}}{6.022×10^{20}}g$$

Q. (i) Define valency of an element.
(ii)Which has more atoms, 50 g of Aluminium or 50 g
of iron? Illustrate your answer with the help of calculations.
(Atomic masses: Al = 27 u; Fe = 56 u)
(iii)Write down the formulae of:
(a) Sodium oxide
(b) Aluminium chloride
(c) Sodium sulphide
(d) Magnesium hydroxide

Ans. (i) The valency of an element is defined as its capacity to combine with other elements. It can be used to find out how the atoms of an element will combine with the atoms of another element to form a chemical compound. Elements may have valency 1 (monovalent), 2 (divalent), 3 (trivalent), 4 (tetravalent) respectively. Some elements show variable valency in different compounds. For example: In red oxide of copper called Copper(I) oxide(Cu₂O), Cu has valency = 1. while in black oxide of Copper(II) oxide(CuO), Cu has valency = 2.
(ii)1 mole of aluminium weighing 27g has
= 6.022 × 10²³ atoms of Al
So, 1 g of Al has = 0.22 × 10²³ atoms of Al
Hence,
50 g of Al will have = 50 × 0.22 × 10²³ atoms
of Al = 11 × 10²³ atoms of Al
1 mole of iron weighing 56 g has
= 6.022 × 10²³ atoms of Fe
So, 1 g of Fe has = 0.10 × 10²³ atoms of Fe
Hence, 50 g of Fe will have = 50 × 0.10 × 10²³ atoms of Fe = 5 × 10²³ atoms of Fe
Thus, 50g of Al has more number of atoms as compared to 50 g of Fe.

Q. (i) What is the chemical formula of water molecule?
(ii)Which of the following symbols of elements are incorrect? Give their correct symbols.

(a) Cobalt, CO
(b) Carbon, c
(c) Aluminium, AL
(d) Helium, He
(e) Sodium, So

(iii)What is the ratio of number of atoms present in water molecule?
(iv) Calculate the ratio of masses of atoms of elements present in water molecule?

Ans. (i) The chemical formal of water molecule is H₂O (ii)(a) For cobalt CO is incorrect symbol. Its correct symbol is Co. (b) For carbon c is incorrect symbol. Its correct symbol is C. (c) For aluminium AL is incorrect symbol. Its correct symbol is Al. (d) Helium, He is the correct symbol. (e) For sodium So is incorrect symbol. Its correct symbol is Na. (It is derived from Latin name ‘Natrium’). (iii)The ratio of number of atoms present in water molecule is as H:O :: 2:1 (iv)Mass of hydrogen = 1 u Mass of oxygen = 16 u So, mass ratio of H:O = 2u : 16u = 1:8

Q. (i) One of the forms of a naturally occurring solid compound P is usually used for making the floors of houses. On adding a few drops of dilute hydrochloric acid to P, brisk effervescence are produced. When 50 g of reactant P was heated strongly, then 22 g of a gas Q and 28 g of a solid R were produced as products. Gas Q is the same which produced brisk effervescence on adding dilute HCl to P. Gas Q is said to cause global warming whereas solid R is used for white-washing.
(a) What is (1) solid P; (2) gas Q, and (3) solid R?
(b) What is the total mass of Q and R obtained from 50 g of P?
(c) How does the total mass of Q and R formed compare with the mass of P taken?
(d) What conclusion do you get from the comparison of masses of products and reactants?
(e) Which law of chemical combination is illustrated by the example given in this problem?
(ii)Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Ans. (i) According to question:
(a) Solid P - Calcium Carbonate (CaCO₃)
Gas Q - Carbon dioxide (CO₂)
Solid R - Calcium oxide (CaO)
(b) Total mass of Q and R = 22g + 28g = 50g
(c) Total mass of Q and R (50g) is equal to mass of reactant (50g).
(d) The law of conservation of mass is followed, i.e., total mass of product is equal to mass of reactant.
(e) Law of conservation of mass is illustrated by the example. (ii)1 g of hydrogen reacts with = 8 g of oxygen
3 g of hydrogen reacts with = 8 × 3 = 24 g of oxygen Thus,24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

Q. (i) Distinguish between cation and anion.
(ii)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

Ans.

Cations	Anions
Positively charged ions are known as cations.	Negatively charged ions are known as anions.
Metals loose electrons from their outermost shell to gain nearest noble gas configuration. For example: Na ⁺, Ca ⁺².	Non-metals gain electrons to complete their octet. For example: Cl ^–, O2 ^–.

Q. A silver ornament of mass “m” in grams 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.

Ans. Given, mass of silver = 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
Thus, ratio of number of atoms of gold to silver
= Au: Ag
= (m/100 × 197) × N_A : (m/108) × N_A = 108 : 100 × 197
= 108 : 19700
= 1 : 182.41

Q. What is the mass of:

Ans. (i) 0.2 mole of oxygen atoms?
(ii)0.5 mole of water molecules?

Q. (i) If 1 mole of carbon atoms weighs 12 g, what is the mass of 1 atom of carbon?
(ii) How many moles of O2 are there in 1.0 × 1022 oxygen molecules?
(iii) How many moles are there in 34.5 g of sodium? (Atomic mass of Na = 23 u)

Q. 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