Friday, July 18, 2025

RBSE Class 9 Science Chapter 3 Atoms and Molecules Solutions

This post provides you a complete solution of intext and End exercise questions. This chapter provides information about atoms and molecule and you will learn structure of atom in the next chapter.

RBSE Class 9 Science Chapter 3  Atoms and Molecules    Solutions


This will help you complete your home assignments and secure good marks in examination.

 

 

Chapter 3Atoms and Molecules

 

Intext Questions

Answers of Questions on Page 27 -28

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

Answer

    


Q.2 Hydrogen and oxygen combine in the ration 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     

         


 

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

Answer      

In every chemical reaction, total masses of all the reactants, is equal to the masses of all of the products. Atoms cannot be created nor destroyed in a chemical reaction.

 

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

Answer     

In a given chemical compound, the proportions by mass of the elements that compose it are constant.

 

Answers of Questions on Page 30

 

Q.1 Define the atomic mass unit.

Answer  

One atomic mass unit is equal to exactly one twelfth (1/12)th the mass of one atom of C-12.

 

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

Answer   

Atoms are very small to be seen with naked eyes. Their size is measured in nanometers.

           (1 nm = 10-9 m).

 

Answers of Questions on Page 34

Q.1 Write down the formulae of:

 (i)   Sodium oxide      (ii) Aluminium chloride

(iii) Sodium sulphide  (iv) Magnesium hydroxide

Answer          

 


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

 (i) Al2(SO4)3    (ii) CaCl2   (iii) K2SO4   (iv) KNO3    

(v)   CaCO3

Answer  

(i) Aluminium Sulphate                    

(ii) Calcium Chloride

(iii) Potassium Sulphate              

(iv) Potassium Nitrate

(v)  Calcium Carbonate

 

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

Answer    

The chemical formula of the compound is symbolic representation of its composition. Eg- Chemical formula of Sodium oxide is Na2O

 

Q.4 How many atoms are present in a    

 (i)  H2S molecule and     (ii) PO4-3 ion ?

Answer    

(i)   H2S → 3atoms are present                

(ii) PO4-3 → 5 atoms are present

 

Answers of Questions on Page 35


Q.1 Calculate the molecular masses of H2, O2 , Cl2 ,CO2, CH4 ,C2H6 , C2H4 , NH3 ,CH3OH    

Answer

               


                        

 Q.2Calculate the formula unit masses of ZnO, Na2O, and K2CO3 (Given atomic mass: Zn = 65u, Na = 23u, K= 39u, C= 12u and O = 16u)

Answer                                     

     


 

Answers of End Exercise Questions

 

Q.1 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 mass.

Answer


Percentage of Boron = `\frac\text{mass of Boron}{Mass    of    compounds}\times 100`


  =`\frac{0.096}{0.24}\times100 =40.0`

 

Percentage of oxygen=`\frac\text{mass of oxygen}{Mass    of    compounds}\times 100`


=`\frac{0.144}{0.24}\times100 =60.0`




 Hence, the mass percentage of boron and oxygen in the given compound are 40% and 60% respectively.

 

Q.2 When 3.0 g of carbon is burnt in 8.0 g of oxygen, 11.0 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.0 g of oxygen? Which law of chemical combination will govern your answer?

Answer          

              


            

For first case:

 3.0 g of carbon is burnt in 8.0 g of oxygen to form 11.0 g of CO2.

For second case:

 3.0 g of carbon must also combine with 8.0 g of oxygen only. This means that (50-8) = 42 g of oxygen will remain un-reacted.

 The mass of CO2 in this case must also be 11 g. So, law of constant proportion is also valid here.

 

Q.3 What are polyatomic ions? Give examples.

Answer    

The ions which contain more than one atoms and behave as a single unit are called polyatomic ion e.g.,  PO42-, CO35-, SO42- , OH-  etc.


Q.4 Write the chemical formulae of the following:

 (a) Magnesium chloride            

 (b)  Calcium oxide

 (c)  Copper nitrate                   

(d)  Aluminium chloride

Answer                                   

          


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   

(a)  Quick lime is calcium oxide

     Elements – Calcium and Oxygen

(b).  Hydrogen bromide

      Elements – Hydrogen and bromine

( c). Baking powder is Sodium hydrogen carbonate

       Elements – Sodium, hydrogen, carbon and oxygen

 (d).   Potassium sulphate

       Elements – Potassium, suphur, oxygen

 

Q.6 Calculate the molar mass of the following substances:

    (a)  Ethyene, C2H2

     (b)  Sulphur molecule, S8

  (c)  Phosphorous molecule, P4 (Atomic mass of phosphorous = 31)

     (d)  Hydrochloric acid, HCl

      (e) Nitric acid, HNO3.

Answer   

         

No comments:

Post a Comment