 # ATOMS AND MOLECULES – NCERT TEXT BOOK Questions Solutions

Science – Text Book for Class IX
Page No. – 32

Q. In a reaction, 5.3 gm of sodium carbonate reacted with 6 gm of ethanoic acid. The products were 2.2 gm of carbon dioxide, 0.9 gm water and 8.2 gm sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.

You can hear this Question and its Answer –

Ans –Mass of reactants = Mass of sodium carbonate + Mass of ethanoic acid

= 5.3 + 6.0 = 11.3 gm

Mass of products = Mass of carbon dioxide + Mass of water + Mass of sodium ethanoate

= 2.2 + 0.9 + 8.2 = 11.3 gm

The mass of products is equal to the mass of reactants. Thus, mass is neither created nor lost during the given chemical change which is in agreement with the law of conservation of mass.

Q. 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 gm of hydrogen gas ?

Ans-
Since hydrogen and oxygen combine in the ratio of 1:8 by mass, it means that x gm of hydrogen and 8x gm of oxygen will be required to form water.
So, Oxygen required to react with 3 gm of hydrogen to form water = 3 x 8 = 24 gm.

Q. Which postulate of Dalton’s Atomic Theory is the result of the law of conservation of mass ?
Ans –

“Atoms are indivisible particles, which can not be created nor destroyed in a chemical reaction.” The above postulate of Dalton’s Atomic Theory is the result of the law of conservation of mass.

Q. Which postulate of Dalton’s Atomic Theory can explain the law of definite proportions ?
Ans-
“The relative number and kinds of atoms are constant in a given compound.” The above postulate of Dalton’s Atomic Theory can explain the law of definite proportions.

Law of Conservation of Mass

Problems based on Law of Conservation of Mass

Law of Definite Proportions / Constant Composition

Problems based on law of definite proportions or constant composition

Dalton’s Atomic Theory of Matter

Page No. – 35

Q. Define the Atomic Mass Unit.
Ans-
The atomic mass of an element is the relative mass of its atom as compared with the mass of a particular atom of carbon isotope taken as 12 units. Thus the atomic mass of an element indicates the number of times one atom of the element is heavier than 1/12 th of a 12C isotope. For example the atomic mass of oxygen is 16 which indicates that an atom of oxygen is 16 times heavier than 1/12thof a 12C isotope atom.
Thus, one atomic mass unit (u) = 1/12th  mass of 12C isotope atom.

Q. Why is it not possible to see an atom with naked eyes or even with the most powerful microscope?
Ans-
The size of an atom is very small and is measured in nanometers. Therefore, it is not possible to see an atom with naked eyes.

The Atom – its Size

The Atom – its Symbols

The Atom – its Atomic Mass

Page No. – 39

Q. Write down the formulae of –
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide

Ans –
(i) Na2O
(ii) AlCl3
(iii) Na2S
(iv) Mg(OH) 2

Q. Write down the names of compounds represented by the following formulae –
(i) Al2(SO4) 3,         (ii) CaCl2,            (iii) K2SO4,             (iv) KNO3,           (v) CaCO3

Ans –
(i) aluminium sulphate
(ii) calcium chloride
(iii) potassium sulphate
(iv) potassium nitrate
(v) calcium carbonate

Q. What is meant by the term chemical formula ?
Ans –
The chemical formula of a compound is the symbolic representation of its composition. A chemical formula of a compound shows its constituent elements and the number of atoms of each combining element. For example, the chemical formula of hydrogen sulphide is H2S. It indicates that in hydrogen sulphide two hydrogen atoms and one sulphur atom are chemically united.

Q. How many atoms are present in a
(i) H2S molecule              and                          (ii) PO43–  ion ?
Ans-
(i) Three atoms (2H and 1S)
(ii) Five atoms (1P and 4O)

Molecules of Elements and Compounds

Ions – Anions and Cations

Common Simple and polyatomic Ions

Valency – Combining Capacity or Combining Power

Chemical Formulae of Simple Compounds

Chemical Formulae of Ionic Compounds

Page No. – 40

Q.  Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
Ans –      Molecular mass of H2  = 1 + 1 = 2 u
Molecular mass of O2= 16 + 16 = 32 u
Molecular mass of Cl2 = 35.5 + 35.5 = 71 u
Molecular mass of CO2 = 12 + 32 = 44 u
Molecular mass of CH4= 12 + 4 = 16 u
Molecular mass of C2H6 = 24 + 6 = 30 u
Molecular mass of C2H4 = 24 + 4 = 28 u
Molecular mass of NH3 = 14 + 3 = 17 u
Molecular mass of CH3OH = 12 + 3 + 16 + 1 =32 u

Q.  Calculate the formula unit masses of ZnO, Na2O, K2CO3,
given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u and O = 16u.
Ans –
Formula unit mass of ZnO = 65 + 16 = 81u
Formula unit mass of Na2O = 23 x 2 + 16 = 62u
Formula unit mass of K2CO3 = 39 x 2 + 12 x 1 + 16 x 3 = 138u.

Formula Unit Mass or Formula Mass

Page No. – 42

Q. If one mole of carbon atoms weighs 12 gm, what is the mass (in gm) of 1 atom of carbon ?
Ans
– 1 mole of carbon atoms = 6.022 x 1023 atoms.
or  6.022 x 1023 atoms of carbon weighs = 12 gm
So, 1 atom of carbon weighs = 12 / 6.022 x 1023 = 1.99 x 10-23gm.

Q.  Which has more numbers of atoms, 100 g of sodium or 100 g of iron?
(given, atomic mass of Na = 23u, Fe = 56u)
Ans – Sodium.
Number of sodium atoms = (100 / 23) x 6.022 x 1023
Number of iron atoms      = (100 / 56) x 6.022 x 1023

Mole Concept – Introduction

Mole Concept – Problems / Numericals

Back Exercises

Page Number – 43

Q. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Ans-                Compound of oxygen and boron → Boron + Oxygen
0.24 g → 0.096 g + 0.144 g

Percentage composition of compound of oxygen and boron

(a) For Boron –

0.24g of compound of oxygen and boron gives 0.096g of Boron
So, 100g of compound of oxygen and boron will give – (100 x 0.096)/0.24 g  = 40% of boron by weight.

(b) For Oxygen –

24g of compound of oxygen and boron gives 0.144g of oxygen
So, 100g of compound of oxygen and boron will give – (100 x 0.144)/0.24 g  = 60% of oxygen by weight.

Q. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Answer: The reaction of burning of carbon in oxygen may be written as:

C         +          O2        →       CO2
1mole             1mole             1mole
12g                 32g                 44g
3g                    8g                    11g
According to the law of constant proportion, 12 g of carbon bums in 32 g oxygen to form 44 g of carbon dioxide or 3 g of carbon reacts with 8 g of oxygen to  form 11 g of carbon dioxide. It is given the same in the question. Further , it is clear that when 3.00g of carbon consumes only 8.00g of oxygen out of 50.00g of oxygen, leaving behind 50 – 8 = 42 g of O2. The answer governs the law of constant proportion.

Q. What are polyatomic ions ? Give examples.

Ans – When two or more atoms combine together and behave like one entity with a net charge, then it is called polyatomic ion. For example, oxygen atom and hydrogen atom combine to form hydroxide ion (OH). One carbon atom and three hydrogen atom combine to form carbonate ion (CO3–2).

Q. Write chemical formulae of the following –
(a) Magnesium chloride                      (b) Calcium oxide                   (c) Calcium nitrate
(d) Aluminium chloride                      (e) Calcium carbonate

Ans –
(a) MgCl2
(b) CaO
(c) Cu(NO3) 2
(d) AlCl3
(e) CaCO3

Q.  Give the names of the elements present in the following compounds –

(a) Quick lime                         (b) Hydrogen bromide                        (c) Baking powder      (d) Potassium sulphate

Ans –
(a) Calcium and oxygen
(b) Hydrogen and bromine
(c) Sodium, hydrogen, carbon and oxygen
(d) Potassium, sulphur and oxygen.

Q. Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4(Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3
Ans-
The molar mass of the following: [Unit is ‘g’]
(a) Ethyne, C2H2 = 2 x 12 + 2 x 1 = 24 + 2 = 26 g
(b) Sulphur molecule, S8 = 8 x 32 = 256 g
(c) Phosphorus molecule, P4=4 x 31 = 124g
(d) Hydrochloric acid, HCl = 1 x 1 + 1 x 35.5 = 1 + 35.5 = 36.5 g
(e) Nitric acid, HNO3 = 1 x 1 + 1 x 14 + 3 x 16 = 1 + 14 + 48 = 63 g

Q. What is the mass of:

(a) 1 mole of nitrogen atoms.
(b) 4 moles of aluminium atoms.
(c) 10 moles of sodium sulphite (Na2SO3).
Ans –
(a) 1 mole of nitrogen atoms = 14u = 14gm
(b) 4 moles of aluminium atoms = 4 × 27 = 108u = 108gm
(c) 1 mole of sodium sulphite, Na2SO3 = 2 × 23 + 1 × 32 + 3 × 16 = 126u = 126gm
10 moles of sodium sulphite = 126 x 10 = 1260 u = 1260 gm.

Q. Convert into mole –
(a) 12 gm of oxygen gas         (b) 20 gm of water                  (c) 22 gm of carbon dioxide
Ans-
(a) 32 gm of oxygen gas = 1 mole
12 gm of oxygen gas = 12 / 32 = 0.375 mole.

(b) 18 gm of water = 1 mole
20 gm of water = 20 / 18 = 1.1 mole.

(c) 44 gm of carbon dioxide = 1 mole
22 gm carbon dioxide = 22 / 44 = 0.5 mole.

Q. What is the mass of –
(a) 0.2 mole of oxygen atoms                         (b) 0.5 mole of water molecules
Ans –
(a) 1 mole of oxygen atoms = 16 gm
0.2 moles of oxygen atoms = 16 x 0.2 = 3.2 gm

(b) 1 mole of water molecules = 18 gm
0.5 mole of water molecules = 18 x 0.5 = 9 gm.

Q. Calculate the number of molecules of sulphur (S8) present in 16 gm of solid sulphur.
Ans –   1 mole of S8 = 32 x 8 = 256 g
1 mole of S8 = 6.022 x 1023 molecules
So,    256 gm of S8 = 6.022 x 1023 S8 molecules
Or,    16 gm S8 = (6.022 x 1023 / 256) x 16 = 3.76 x 1022 molecules.

Q. Calculate the number of aluminium ions present in 0.51 gm of aluminium oxide.

Ans – The mass of an ion is the same as that of an atom of the same element.
Atomic mass of Al = 27u.
1 mole of aluminium oxide, Al2O3 = 2 x 27 + 3 x 16 = 102u = 102 gm
That means 102 gm Al2O has 6.022 x 1023 aluminium oxide molecules.
Or, 0.51 gm Al2O3 has = (6.022 x 1023 x 0.51) / 102 = 3.01 x 1021 Al2O3 molecules.
We know, 1 molecule of Al2O3 has 2Al+3 ions.
Hence,      0.51 gm Al2O3   =  2 x 3.01 x 1021 Al+3 ions        =   6.023 x 1021 aluminium ions.