NCERT solutions of class 9 science chapter3-Atoms and molecules - Future Study Point

NCERT solutions of class 9 science chapter3-Atoms and molecules

atoms and molecules

Science NCERT Solutions of the chapter3-Atoms and Molecules of chemistry part

The NCERT solutions of the chapter 3-Atoms and molecules are solutions of chemistry part of class 9 NCERT science textbook, these NCERT solutions are the initiations to get to know about the composition of matter, these class 9 science NCERT solutions of chapter 3-Atoms and molecules of chemistry part reveal the information how and why the elements or compounds react with each other and forms another compound. In chapter 3-Atoms and molecules are the particles that participate in chemical reactions? All NCERT solutions are solved by an expert teacher as per the CBSE norms.

atoms and molecules


NCERT solutions of class 9 science chapter3-Atoms and molecules

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 Q1- In a reaction, 5.3 g of sodium –  carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon oxide. 0.9 g of water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. 

Sodium carbonate +ethanoic acid →sodium ethanoate +carbon dioxide +water

Ans-The mass of reactants = mass of Sodium carbonate + mass of ethanoic acid

= 5.3 g + 6 g = 11.3 g

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

= 2.2 g + 0.9 g + 8.2 g

=11.3 g

So, observations are in agreement with the law of conservation of mass because in this chemical reaction the observation follows the law of conservation of mass as mentioned below.

Mass of reactants = Mass of products

   Q2-  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-According to the law of constant proportion all the elements in a compound are always in the fixed ratio,so let 3 g of hydrogen reacts with X grams of oxygen.


(1/8 ) = 3/X

X  =24

So, 24 g oxygen will be required to react with 3 g of hydrogen in the formation of water.

Q3-  Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Ans-2 nd postulate of Dalton’s atomic theory is the result of the law of conservation of mass because it states that atoms are indivisible particles, which can not be created or destroyed in a chemical reaction.

 Q4-  Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Ans- 6 th postulate of Dalton’s theory explains the law of definite proportions because it states that relative number and kinds of atoms are constant in a given compound.

Q5-Write down the formulae of

           (i)  Sodium oxide

           (ii)Aluminium chloride

           (iii) Sodium sulphide

           (iv) Magnesium hydroxide


(i) Sodium chloride – NaCl

(ii)  Aluminium chloride – AlCl3

(iii)  Sodium sulphide – Na2S

(iv)  Magnesium hydroxide – Mg(OH)2

Q6-Write down the names of the compounds represented by the following formulae;

(i)  Al2(SO4)3

(ii) CaCl2

(iii) K2SO4

(iv) KNO3

(v) CaCO3


(i)  Al2(SO4)3 – Aluminium sulphate

(ii)  CaCl2 – Calcium chloride

(iii)  K2SO4 – Potassium sulphate

(iv)  KNO3 – Potassium nitrate

(v)  CaCO3 – Calcium carbonate

Q7-What is meant by the term chemical formula?

Ans-Chemical formula of a compound shows what is its composition, from the chemical formula we get the combining capacity of each atom which is known as valency. The

way of writing chemical formula is i.e aluminium sulphate.

Valancy of aluminium 3 means it has capacity to hold 3 ions of sulphate and sulphate

has the charge 2 has capacity to hold 2 aluminium atoms.

So its chemical formula will be – Al2(SO4)3 

Q9-How many atoms are present in a

(i)  H2S molecule and

(ii)  PO43- ions


(i)In H2S, Atoms of H =2, sulphur atom =1, so total atoms = 3

(ii) In PO43- ions, phosphorus atom = 1, oxygen atoms= 4, total atoms = 5

Q10-Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, NH3, CH3OH .

Ans-Molecular masses of the given substances are as follows.

H2 – 2 X 1 = 2 u

O2 – 2 X 16 = 32 u

Cl2 – 2 x 35 = 70 u

CO2 – 12 + 2 x 16 = 12 +32 = 44 u

CH4 – 12 + 4 x 1 = 12 + 4 = 16 u

C2H4 – 2 x 12 + 4 x 1 = 24 + 4 = 28 u

NH3 – 14 + 3 x 1 = 14 +3 = 17 u

CH3OH – 12 + 3 x 1 + 16 + 1 = 32 u 

Q11-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).

Ans-ZnO – 65 + 16 = 81 u

Na2O – 23 x 2 + 16 =62 u

K2CO3 – 39 x 2 + 12 + 16x 3 = 138 u                       


Q12-If one mole of carbon atoms weighs 12 grams. What is the mass (in grams) of 1 atom of carbon ? 

Ans- 1 mole of carbon atoms = 6.022 x 1023 atoms

6.022 x 1023 atoms weighs = 12 grams

So 1 atom of carbon will weigh =12/(6.022 x 1023)

= 1.99 x 10-23 gram

Q13-Which has more number of atoms. 1OO grams of sodium or 100 grams of iron (given atomic mass of Na = 23 u, Fe = 56 u) ?


Number of Na atoms

= (given mass of Na/molar mass of Na)x N0

atomic mass of Na = 23 u, so molar mass of Na= 23 g

given mass of Na = 100 g

N0 =  6.022 x 1023

Number of Na atoms = (100/23) x6.022 x 1023

= 2.62 x 1024

On the same way number of iron atoms in 100 g of iron = (100/56) x 6.022 x 1023

= 1.08 x 1024

In 100 g of iron and sodium number of Na atoms are more than the number of Fe atoms. 


Q14.  A 0.24 g sample of compound 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.


% of boron in the compound

= (quantity of boron )/(mass of sample)x 100

=( 0.096/0.24) x 100

= 40 %

Similarly % of oxygen = (0.144/0.24) x 100

= 60 %

Q15.   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 ?

Ans.If 3.0 g of carbon is burnt in 8.00 g oxygen as a result of  11 g of carbon dioxide is formed. According to law of constant proportion, all elements are in fixed proportion in a compound.So 3.00 g of carbon will react only with 8 g of oxygen and rest of the oxygen 50 – 8 = 42 mg will be freed or unreacted, So hear law of definite proportion is followed.

 Q16 –   What are polyatomic ions ? Give examples.

Ans-Polyatomic are group of atoms that have a net charge on them as an example  (sulphate ion)SO42-, (nitrate ion)NO3 etc.

Q17-   Write the chemical formulae of the following.

(a) Magnesium chloride

(b)Calcium oxide

 (c) Copper nitrate

 (d)Aluminium chloride

(e)Calcium carbonate

Ans- Valency of Mg, Ca , Cu  and Al are 2, 2, (1,2) and 3 respectively.

Q18-Give the names of the elements present in the following compounds.

 (a)Quick lime

(b)Hydrogen bromide

(c) Baking powder

(d)Potassium sulphate


(a)Quick lime – CaO( Calcium, oxygen)

(b) Hydrogen bromide – HBr( Hydrogen, Bromine)

(c)Baking powder – NaHCO3(Sodium, Hydrogen, Carbon, Oxygen)

(d) Potassium sulphate – K2SO4( Potassium, Sulphor, Oxygen)

Q18-Calculate the molar mass of the following substances.

(a)Ethyne, C2H2

(b)The sulphur molecule, S8

(c)Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)

(d)Hydrochloric acid, HCl

(e)Nitric acid, HNO3


(a)  Molecular mass of C2H2 = 2 x 12 + 2 x 1 =26 u

So molar mass of C2H2 = 26 g

(b)Molecular mass of sulphur molecule, S8 = 8 x 32 =256 u

Molar mass of S8 = 256 g

(c)molecular mass of P4 =4 x31 = 124 u

Molar mass of P4 =124 g

(d)Molecular mass of HCl = 1 + 35 = 36 u

Molar mass of HCl = 36 g

(e)Molecular mass of HNO3 = 1 + 14 + 16 x 3 =63 u

Molar mass of HNO= 63 g


Q19-What is the mass of –

(a)1 mole of nitrogen atoms?

(b)4 moles of aluminium atoms (Atomic mass of aluminium =27)

(c)10 moles of sodium sulphite(Na2SO3)?


(a)Mass of 1 mole nitrogen atom

= its molar mass =14 

Q20- Convert into mole.

(a)12 g of oxygen gas

(b)20 g of water

(c)22 g of carbon dioxide


(a)Molecular mass of oxygen gas, O2= 16 x2 = 32 u

Molar mass of O2 = 32 g

No. of moles in 12 g of O2 = m/M

Given mass of O2, m= 12 g

Molar mass of O2, M =32 g

So, no. of moles of O2 =12/32 = 0.375

(b)  Molecular mass of water, H2O = 2×1 +16 =18 u

Molar mass of water, M = 18 g

Given mass , m= 20 g

No. of moles of water =m/M = 20/18 = 1.11

(C) Molecular mass of CO2 = 12 + 16 x 2 = 44 u

Molar mass of CO2, M =44 g

Given mass of CO2, m =22 g

No. of moles of CO2 = m/M = 22/44 = 0.5

Q21-What is the mass of :

(a)0.2 moles of oxygen atoms ?

(b)0.5 moles of water molecules ?

Ans –

Atomic mass of oxygen = 16 u

So, molar mass (M )of ,O = 16 g

Given mass of O =m

No. of moles of O =m/M

0.2 = m/16

0.2 x 16 = m, m= 3.2 g

(b)Molecular mass of water, H2O = 18 u

Molar mass, M of water = 18 g

No. of moles of H2O = m/M

So, given mass, m = 0.5 x 18 = 9 g

Q22-Calculate the number of molecules of sulphur(S8) present in 16 g of solid sulphur.

Ans- Molecular mass of S8 = 8 x 32 =256 u

Molar mass ,M of S8 = 256 g

Given mass, m = 16 g

No. of molecules of sulphur = (m/M) x N0

N0 = 6.022 x 1023

No. of molecules of sulphur = (16/256) x 6.022 x 1023

= 0.376 x 1023

= 3.76 x 1022

Q23-Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

(HINT: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al =27 u)


Molecular mass of aluminium oxide ,Al2O3 = 27 x2 + 16 x 3 =54 + 48 = 102 u

Molar mass of aluminium oxide ,M= 102 g

No. of molecules of ,Al2O3 = (m/M) x6.022 x 1023

Given mass , m = 0.051 g

No. of molecules of ,Al2O3 = (0.051/102) x6.022 x 1023

= 0.0030 x 1023

=3.0 x 1020

No. of ions in Al2O3 = no. of atoms Al2O3 = 2

So,total number of ions present in 0.051 g of Al2O3

= 3.0 x 2 x 1020

 = 6 x 1020

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