NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

 

NCERT solutions for class 9 science chapter 3 Atoms and Molecules: If you are a class 9 student, then you must have come across this chapter and surely you must be looking for solutions for NCERT class 9 science chapter 3 Atoms and Molecules. Through this article, you will get complete exercise solutions and topic wise solutions mentioned in between the chapter. These NCERT solutions are designed by experts. In Class 9 Science Chapter 3 Atoms and Molecules, you will also study that during a chemical reaction, the sum of the masses of reactants and products remain unchanged. This is known as the Law of Conservation of Mass. In a pure chemical compound, elements are always present in a definite proportion by mass. This is known as the Law of Definite Proportions. An atom is the smallest particle of the element that cannot usually exist independently and retain all its chemical properties. All these facts are explained beautifully in solutions of NCERT for class 9 chapter 3 atoms and molecules. After going through this chapter, you must be able to get the CBSE NCERT solutions for class 9 science chapter 3 Atoms and Molecules which are mentioned below:

Here are the important Topics of NCERT Solutions for Class 9 Science Chapter 3 - Atoms and Molecules mentioned below:

3.1 Laws of Chemical Combination

3.1.1 Law of Conservation of Mass

3.1.2 Law of Constant Proportions

3.2 What Is an Atom?

3.2.1 What Are the Modern Day Symbols of Atoms of Different Elements?

3.2.2 Atomic Mass

3.2.3 How Do Atoms Exist?

3.3 What Is a Molecule?

3.3.1 Molecules of Elements

3.3.2 Molecules of Compounds

3.3.3 What Is an Ion?

3.4 Writing Chemical Formulae

3.4.1 Formulae of Simple Compounds

3.5 Molecular Mass and Mole Concept

3.5.1 Molecular Mass

3.5.2 Formula Unit Mass

3.5.3 Mole Concept

NCERT solutions for class 9 science chapter 3 Atoms and Molecules 

Topic 3.1 Laws of chemical combination

Q 1.   In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.

sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water

Answer:

Given, the reaction 

 sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water

        5.3g                       6g                     8.2g                     2.2g               0.9g

Now,

Total Mass on the Left Hand Side = 5.3g + 6g = 11.3g

Total Mass on the Right Hand Side = 8.2g + 2.2g + 0.9g = 11.3g

As Mass on the LHS is equal to RHS, the observation is in agreement with the law of conservation of mass. 

Q 2.   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?

Answer:

Given:

Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water.

Now,

For every 1 g of Hydrogen, 8 g of oxygen is needed for the reaction to take place. 

Therefore, for 3 g of Hydrogen, the mass of oxygen needed = 8 * 3 = 24 g.

Hence, 24 g of Oxygen is needed to complete a reaction with 3 g of Hydrogen.

Q 3.  Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer:

The postulate of Dalton’s atomic theory is the result of the law of conservation of mass is

"Atoms can neither be created nor can be destroyed".

Q 4.  Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Answer:

The postulate of Dalton’s atomic theory can explain the law of definite proportions is

"The relative number and kinds of atoms are equal in given compounds."

CBSE NCERT solutions for class 9 science chapter 3 Atoms and Molecules

Topic 3.2 What is an atom?

Q 1.   Define the atomic mass unit.

Answer:

An atomic mass unit is a unit of mass used to express the weight subatomic particles, where one unit is equal to exactly one-twelfth the mass of a carbon-12 atom.

Q 2.   Why is it not possible to see an atom with naked eyes?

Answer:

We can't see the atom with the naked eye because they are minuscule in nature. they are measured in the nanometres. Also, except for noble gases, all-atom do not exist independently.they exist in the form of any compound. 

NCERT free solutions for class 9 science chapter 3 Atoms and Molecules 

Topic 3.4 Writing chemical formulae

Q 1.  Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium suphide

(iv) magnesium hydroxide

Answer:

The Formula of given compounds are : 

(i) sodium oxide :

 \Rightarrow Na_2O

(ii) aluminium chloride:

\Rightarrow AlCl_3

(iii) sodium sulphide:

\Rightarrow Na_2S

 (iv) magnesium hydroxide:

\Rightarrow Mg(OH)_2

Q 2.  Write down the names of  compounds represented by the following formulae:

(i)     Al_{2}(SO_{4})_{3}

(ii)     CaCl_{2}

(iii)     K_{2}SO_{4}

(iv)     KNO_{3}

(v)       CaCO_{3}           

Answer:

The Names of the following compounds are :

 (i)  Al_{2}(SO_{4})_{3} 

        =Aluminium Sulphate

  (ii)     CaCl_{2}

           =Calcium Chloride

 (iii)     K_{2}SO_{4}

          =Potessium Sulphate

  (iv)     KNO_{3}

          = Potessium Nitrate

  (v)       CaCO_{3}

           = Calcium Carbonate.  

Q 3.    What is meant by the term chemical formula?

Answer:

The chemical formula of a compound is a symbolic representation of its composition. For example, The chemical formula for common salt is NaCl as it is made up of Sodium (Na) and Chlorine (Cl).

Q 4.(i)  How many atoms are present in a

Answer:

 H_{2}S molecule has 2 atoms of Hydrogen and 1 atom of Sulphur.

and hence,

Total of 3 atoms are present in H_{2}S molecule .

Q 4.(ii)   How many atoms are present in a 

Answer:

  PO_{4}^{3-} ion has one atom of Phosphorus and 4 atoms of Oxygen.

And Hence,

A total of 5 atoms are present in   PO_{4}^{3-} ion.

Solutions for NCERT class 9 science chapter 3 Atoms and Molecules Excercise:

Topic 3.5 Molecular mass and mole concept

Q 1.   Calculate the molecular masses of H_{2}, O_{2},Cl_{2},CO_{2},CH_{4},C_{2}H_{6}, C_{2}H_{4}, NH_{3},CH_{3}OH

Answer:

The molecular mass of H_2 :

= 2 * Atomic mass of Hydrogen 

= 2 * 1u

=2u.

The molecular mass of O_2 :

= 2 * Atomic mass of Oxygen

= 2 * 16u

= 32u.

The molecular mass of Cl_2 :

= 2 * Atomic mass of Chlorine

= 2 * 35.5uu

= 71u.

The molecular mass of CO_2 :

= Atomic mass of Carbon +2 * Atomic mass of Oxygen

= 12u + 2 * 16u

= 44u.

The molecular mass of CH_4 :

= Atomic mass of Carbon +4 * Atomic mass of Hydrogen

= 12u + 4 * 1u

= 16u.

The molecular mass of C_2H_6 :

= 2 * Atomic mass of Carbon + 6 * Atomic mass of Hydrogen

= 2*12u+ 6 * 1u

=24u+ 6u

= 30u.

The molecular mass of C_2H_4 :

= 2 * Atomic mass of Carbon + 64* Atomic mass of Hydrogen

= 2*12u+ 4* 1u

=24u+ 4u

= 28u.

The molecular mass of NH_3 :

= Atomic mass of Nitrogen + 3 * Atomic mass of Hydrogen

= 14u+ 3 * 1u

= 17u.

The molecular mass of CH_3OH :

= Atomic mass of Carbon +4 * Atomic mass of Hydrogen + Atomic mass of Oxygen 

= 12u+ 4 * 1u + 16u

= 32u.

Q 2.  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

Answer:

Given, 

the atomic mass of Zn = 65 u,

the atomic mass of Na = 23 u,

the atomic mass of K = 39 u,

the atomic mass of C = 12 u,

and the atomic mass of O = 16 u

Now,

Formula unit mass of ZnO = Atomic mass of Zinc + Atomic mass of O

                                          =   65 u + 16 u

                                          =   81 u.

Formula unit mass of Na2O = 2 * Atomic mass of Na+ Atomic mass of O

                                          =    2 * 23 u + 16 u

                                          =    62 u.

Formula unit mass of K2CO3 = 2 * Atomic mass of K+ Atomic mass of C + 3 * Atomic mass of O

   =    2 * 39 u + 12 u + 3 * 16 u

                                                =    138 u.

NCERT solutions for class 9 science chapter 3 Atoms and Molecules

Topic 3.5.3 Mole concept

Q 1.  If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?

Answer:

As we know,

1\: mole=6.022\times10^{23}\:atoms

Now, 

the mass of 1 mole or 6.022\times10^{23}\:atoms  = 12  g

The mass of 1 atom :

 =\frac{12}{6.022\times10^{23}}=1.99\times 10^{-23}\:g

Hence the mass of 1 atom of Carbon is  1.99\times 10^{-23}\:g .

Q 2.   Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, the atomic mass of Na = 23 u, Fe = 56 u)?

Answer:

The number of moles of 100 g Na atoms :

=\frac{100}{23}=4.35\:moles

The number of moles in 100 g of Fe atoms :

=\frac{100}{56}=1.76\:moles

As we know, the one-mole atoms contain 6.022\times10^{23} atoms. 

So,  more the number of moles, more the number of atoms. and hence 100 g of Na atom has a greater number of atoms than 100 g Fe.

NCERT solutions for class 9 science chapter 3 Atoms and Molecules - Exercise solutions

Q 1.   A 0.24 g sample of a 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.

Answer:

The total mass of the compound = 0.24 g

Mass of boron in the compound = 0.096 g

Mass of oxygen in the compound = 0.144 g

Now, As we know,

The percentage of an element in the compound :

 =\frac{total\:mass\:of\:element}{total\:mass\:of\:compound}\times 100

So,

The percentage of Boron in the compound by weight :

=\frac{0.096}{0.24}\times 100

=40\%

The percentage of Oxygen in the compound by weight :

=\frac{0.144}{0.24}\times 100

=60 \%

Q 2.  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:

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.

So, by the law of definite proportions,

When 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8.00 g of oxygen will be used to produce 11.00 gram of carbon dioxide. The remaining 42.00 g of oxygen will remain unreacted. Law of constant proportion is Held.

Q 3. What are polyatomic ions? Give examples.

Answer:

Polyatomic ions are ions that contain more than one atom. These atoms can be of the same type or of a different type.

Some examples of polyatomic ions are NH4+, OH-, SO42-, and SO32-.

Q 4.   Write the chemical formulae of the following.

 (a) Magnesium chloride

 (b) Calcium oxide

 (c) Copper nitrate

 (d) Aluminium chloride

 (e) Calcium carbonate

Answer:

The chemical formula of Given compounds are :

(a) Magnesium chloride

=MgCl_2

 (b) Calcium oxide

=CaO

(c) Copper nitrate

=Cu(NO_3)_2

(d) Aluminium chloride

=AlCl_3

 (e) Calcium carbonate

=CaCO_3

Q 5.   Give the names of the elements present in the following compounds.

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate.

Answer:

The names of the elements present in the given compounds are :

 (a) Quick lime :

      Cao = Calcium and Oxygen.

 (b) Hydrogen bromide:

       HBr = Hydrogen and Bromine.

 (c) Baking powder:

      NaHCO_3 = Sodium, Hydrogen, Carbon, and Oxygen.

 (d) Potassium sulphate:

      K_2SO_4 = Potassium, Sulphur, and oxygen.

Q 6.(a)   Calculate the molar mass of the following substances.

Answer:

The molecular mass of ethyne, C_{2}H_{2}  =  2 * Atomic Mass of C + 2 * Atomic Mass of H 

                                                               =  2 * 12u + 2 * 1u

                                                               =  24u + 2u

                                                               =   26u 

Q 6. (b)  Calculate the molar mass of the following substances.

Answer:

The molecular mass of Sulphur molecule, =  8 * Atomic Mass of S 

                                                             =  8 * 32u

                                                 =  256u 

Q 6.(c) Calculate the molar mass of the following substances

Answer:

The molecular formula of Baking soda is NaHCO3 

So the molecular mass will be some of the individual atoms involved in the formula.

Molecular Mass of baking soda = Molecular mass (Na + H + C + 3xO)

                                                 = 23 + 1 + 12 + 3x16

                                                = 84 g/mol

Q 6.(d)  Calculate the molar mass of the following substances.

Answer:

The molecular mass of Hydrochloric acid, HCl  =  1 * Atomic Mass of Cl + 1 * Atomic Mass of H 

                                                                           =  35.5 u +   1u

                                                                           =  36.5u

Q 6.(e) Calculate the molar mass of the following substances.

Answer:

The molecular mass of  Nitric acid, HNO3  =  1 * Atomic Mass of N + 1 * Atomic Mass of H + 3 * Atomic Mass of O

                                                                    =  1 * 14u + 1 * 1u + 3 * 16u

                                                                    =  14u + 1u + 48u

                                                                    =   63u  

Q 7.(a)  What is the mass of—

Answer:

 Atomic Mass of Nitrogen atom = 14 u.

 Mass of one mole of nitrogen atoms = molecular mass of nitrogen atoms in grams 

                                                           = 14 g

Q 7.(b)   What is the mass of-

Answer:

The atomic mass of Aluminium = 27 u.

Mass of 4 moles of aluminium atoms = 4 x Mass of 1 mole of Al atoms

                                                            = 4 x molecular mass of aluminum atoms in grams

                                                            = 4 x 27

                                                            = 108 g

Q 7. (c)  What is the mass of— 

Answer:

The molecular mass of Sodium sulphite (Na2SO3 ) = 2 * 23 + 32 + 3 * 16 

                                                                                 = 126u 

Mass of 10 moles of sodium sulphite = 10 x Mass of 1 mole of Na2SO3 

                                                           = 10 x molecular mass of Na2SO3 in grams

                                                           = 10 x 126 g 

                                                           = 1260 g.

Q 8.(a)   Convert into a mole.

Answer:

The molecular Mass of the Oxygen = 32 g

Now,

Since 32 g Oxygen = 1 mole 

The number of moles in 12 g Oxygen:

=\frac{1}{32}\times 12\:moles

=0.375\:moles

Q 8.(b)    Convert into mole. 

Answer:

Given, Mass of water = 20 g

The Molecular mass of the water  = 2 * Mass of Hydrogen + Mass of Oxygen

                                                      = 2 * 1 + 16

                                                      =  18 g.

Now, Number of moles :

=\frac{given\:mass}{molar\:mass}=\frac{20}{18}=1.11\:moles

Q 8.(c)  Convert into mole.

Answer:

Given Mass of the carbon dioxide = 22 g.

The molecular mass of Carbon dioxide in grams  =  mass of Carbon + 2 * mass of Oxygen

                                                                               =  12u + 2 * 16u

                                                                               =  44u

The number of mole :

=\frac{given\:mass}{molecular\:mass}=\frac{22}{44}=0.5\:moles

Q 9.(a)  What is the mass of:

Answer:

 The mass of 1 mole of oxygen atoms = 16 g

 The mass of 0.2 mole of oxygen atoms = 0.2 x 16 g

                                                                = 3.2 g

Q 9.(b)    What is the mass of:

 0.5 mole of water molecules?

Answer:

The mass of 1 mole of water molecules = 18 g

Thus,

the mass of 0.5 mole of water molecules = 18 x 0.5

                                                                  = 9.0 g

Q 10.  Calculate the number of molecules of sulphur (S8 ) present in 16 g of solid sulphur.

Answer:

Given the mass of Sulphur = 16 g

The molecular mass of the Sulphur molecule = 8 x 32 = 256 g.

The number of moles of Sulphur molecule :

=\frac{16}{256}

=0.0625\:moles

Now 

Number of molecules in 1 mole = 6.022\times 10^{23} molecules

So,

The number of molecules in 0.0625 moles = 0.0625\times6.022\times 10^{23} molecules.

                                                                    =  3.76375\times 10^{22} Molecules.

Hence there are 3.76375\times 10^{22} Sulphur molecules in  16 g of solid sulphur.

Q 11.  Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

Answer:

Give, the mass of Aluminum Oxide = 0.051 g.

The molecular mass of Aluminium Oxide = 2 x mass of Aluminium + 3 x mass of Oxygen.

                                                                  = 2 x 27 + 3 x 16

                                                                  = 102 g

Number of Moles of Aluminium Oxide :

\frac{0.051}{102}=0.0005\:moles

Now, Since 1 mole of Aluminium Oxide contain 2 moles of Aluminium ion,

The Number of Moles of Aluminium ion = 2 x number of moles of Aluminium Oxide 

                                                               =  2 x 0.0005

                                                               = 0.001 moles.

Now, As the number of ion in 1 mole = 6.022\times10^{23}  ions.

The number of ions in 0.001 moles of Aluminium ion = 0.001\times6.022\times10^{23} ions

                                                                                    = 6.022\times10^{20} ions.

Hence there are 6.022\times10^{20} Aluminium ions in 0.051 g of aluminium oxide.

NCERT Solutions for Class 9 Science- Chapter Wise

Chapter No.

Chapter Name

Chapter 1

Solutions for NCERT class 9 science chapter 1 Matter in Our Surroundings

Chapter 2

NCERT free solutions for class 9 science chapter 2 Is Matter Around Us Pure

Chapter 3

NCERT solutions for class 9 science chapter 3 Atoms and Molecules

Chapter 4

NCERT textbook solutions for class 9 science chapter 4 Structure of The Atom

Chapter 5

NCERT solutions for class 9 science chapter 5 The Fundamental Unit of Life

Chapter 6

CBSE NCERT solutions for class 9 science chapter 6 Tissues

Chapter 7

NCERT free solutions for class 9 science chapter 7 Diversity in Living Organisms

Chapter 8

NCERT textbook solutions for class 9 science chapter 8 Motion

Chapter 9

Solutions for NCERT class 9 science chapter 9 Force and Laws of Motion

Chapter 10

NCERT solutions for class 9 science chapter 10 Gravitation

Chapter 11

NCERT free solutions for class 9 science chapter 11 Work and Energy

Chapter 12

CBSE NCERT solutions for class 9 science chapter 12 Sound

Chapter 13

Solutions for NCERT class 9 science chapter 13 Why Do We Fall ill?

Chapter 14

NCERT Textbook solutions for class 9 science chapter 14 Natural Resources

Chapter 15

NCERT solutions for class 9 science chapter 15 Improvement in Food Resources

NCERT Solutions for Class 9 - Subject Wise

NCERT Solutions for Class 9 Maths
NCERT Solutions for Class 9 Science

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