NCERT Solution for class 9 Atoms and Molecules

NCERT Solution Class 9 Science Chapter 3- (Atoms and Molecules)

In this post, you are going to get a detailed NCERT solution of atoms and molecules of class 9 prepared by the experts of chemistry. So here we will cover all the NCERT questions and answer related to atoms and molecules class 9. If you want to study the notes of atoms and molecules class 9 so please click here-  Atoms and Molecules class 9 notes.

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Exercise 1 of Atoms and Molecules Class 9    Page no:- 32 

1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The Product 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 

Solution:- 

Sodium Carbonate (5.3 g) + acetic acid (6 g) → Sodium acetate (8.2 g) + carbon dioxide (2.2 g) + water (0.9 g)

According to the law of conservation of mass, the mass of the reactant is always equal to the mass of its product. And in this experiment here we proved that the mass of the reactant is equalled to the mass of its Product. 5.3g + 6g  =  8.2g + 2.2g + 0.9g 

                           L.H.S =  R.H.S 

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?

Solution:-

We know that Hydrogen and oxygen combine in the ratio of 1:8 

So, 1 g of Hydrogen reacts with 8 g of oxygen.

Therefore,  3 g of Hydrogen will react with 3x8= 24g of Oxygen.

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

Solution:-

"Atoms can neither be created nor destroyed in the chemical reaction" is the result of the law of conservation of mass.

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

Solution:-

The relative number and kind of atoms are constant in a given compound explained the law of definite proportions. 

 Exercise 2 of Atoms and Molecules Class 9    Page no:- 35

1. Define the atomic mass unit?

Solution:- 

An Atomic mass unit is a unit that is used to measure the mass of the atoms and molecules.

One Atomic mass unit equals to mass of 1/12th the mass of one carbon-12 atom.

The atomic unit mass is symbolized as AMU or amu.

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

Solution:-

We can not see an atom with our naked eyes because atoms are very small that why we measure atoms in nanometers.

 Exercise 3 of Atoms and Molecules Class 9    Page no:- 39

1. Write down the formulae of 

(i) sodium oxide 

(ii) aluminium chloride 

(iii) sodium sulphide 

(iv) magnesium hydroxide

Solution:-

The following are the formulae: 

(i) sodium oxide – Na2O 

(ii) aluminium chloride – AlCl3 

(iii) sodium sulphide – Na2S 

(iv) magnesium hydroxide – Mg (OH)2

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

(i) Al2(SO4)3 

(ii) CaCl2 

(iii) K2SO4 

(iv) KNO3 

(v) CaCO3.

Solution:-

The following are the names of the compound represented by these formulas 

(i) Al2(SO4)3 – Aluminium sulphate 

(ii) CaCl2 – Calcium chloride 

(iii) K2SO4 – Potassium sulphate 

(iv) KNO3 – Potassium nitrate 

(v) CaCO3 – Calcium carbonate

3. What is meant by the term chemical formula?

Solution:-

The chemical formula is the Symbolic representation of its composition. For example, the chemical formula of Water is H2O it means the one water molecule contains 2 hydrogen atoms and one oxygen atom. 

4. How many atoms are present in a 

(i) H2S molecule and
(ii) P043- ion?

Solution:- 

(i) H2S molecule:- In this molecule 2 atoms of hydrogen and 1 atom of sulphur is present. Therefore totally this molecule is composed of 3 atoms.

(ii) P043- ion? :- In this molecule 1 atom of phosphorus and 4 atoms of oxygen are present. Therefore totally this molecule is composed of 5 atoms.

 Exercise 4 of Atoms and Molecules Class 9    Page no:- 41

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

Solution:-

The following are the molecular masses:

The molecular mass of H2 – 2 x atoms atomic mass of H = 2 x 1u = 2u

The molecular mass of O2 – 2 x atoms atomic mass of O = 2 x 16u = 32u

The molecular mass of Cl2 – 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u

The molecular mass of CO2 – atomic mass of C + 2 x atomic mass of O = 12 + ( 2×16)u = 44u

The molecular mass of CH4 – atomic mass of C + 4 x atomic mass of H = 12 + ( 4 x 1)u = 16u

The molecular mass of C2H6– 2 x atomic mass of C + 6 x atomic mass of H = (2 x 12) +

(6 x 1)u=24+6=30u

The molecular mass of C2H4– 2 x atomic mass of C + 4 x atomic mass of H = (2x 12) +

(4 x 1)u=24+4=28u

The molecular mass of NH3– atomic mass of N + 3 x atomic mass of H = (14 +3 x 1)u= 17u

The molecular mass of CH3OH – atomic mass of C + 3x atomic mass of H + atomic mass of O + atomic mass of H = (12 + 3×1+16+1)u=(12+3+17)u = 32u

2. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65u,  Na = 23 u,  K=39u, C = 12u, and O=16u.

Solution:- 

The formula unit mass of 

(i) ZnO = 65 u + 16 u = 81 u 

(ii) Na2O = (23 u x 2) + 16 u = 46 u + 16 u = 62 u 

(iii) K2C03 = (39 u x 2) + 12 u + 16 u x 3 = 78 u + 12 u + 48 u = 138 u

Exercise 5 of Atoms and Molecules Class 9    Page no:- 42

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

Solution:- 

Given: 1 mole of carbon weighs 12g

1 mole of carbon atoms = 6.022 X 1023

Molecular mass of carbon atoms = 12g = an atom of carbon mass

Therefore, mass of 1 carbon atom = 12 / 6.022 X 1023  = 1.99 X 10-23g

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

Solution:- 

Basically, 100g of Sodium will have more atoms as compared to 100g of iron because the atomic mass of Sodium is less than the atomic mass of iron.

Main exercise from NCERT textbook 

1. A 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.

Solution:

Given: Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g

To calculate percentage composition of the compound:

Percentage of boron = mass of boron / mass of the compound x 100

= 0.096g / 0.24g x 100  = 40%

Percentage of oxygen = 100 – percentage of boron

= 100 – 40 = 60%

2. When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen?  Which law of chemical combination will govern your answer?

Solution:

11.00g of carbon dioxide is formed when 3.00g carbon is burnt in 8.00g of oxygen.

Carbon and oxygen are combined in the ratio 3:8 to give carbon dioxide using up all the carbon and

oxygen

Hence, for 3g of carbon and 50g of oxygen, 8g of oxygen is used and 11g of carbon is formed, the

left oxygen is unused i.e., 50-8=42g of oxygen is unused.

This depicts the law of definite proportions – The combining elements in compounds are present in

definite proportions by mass.

3. What are polyatomic ions? Give examples.

Solution:

Polyatomic ions are ions that contain more than one atom but they behave as a single unit

Example: CO32-, H2PO4–

4. Write the chemical formula of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

Solution:

The following are the chemical formula of the above-mentioned list:

(a) Magnesium chloride – MgCl2

(b) Calcium oxide – CaO

(c) Copper nitrate – Cu(NO3)2

(d) Aluminium chloride – AlCl3

(e) Calcium carbonate – CaCO3

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

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate.

Solution:

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

(a) Quick lime – Calcium and oxygen (CaO)

(b) Hydrogen bromide – Hydrogen and bromine (HBr)

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

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

6. 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

Solution:

Listed below is the molar mass of the following substances:

(a) Molar mass of Ethyne C2H2= 2 x Mass of C+2 x Mass of H = (2×12)+(2×1)=24+2=26g

(b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8  x 32 = 256g

(c) Molar mass of  Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124g

(d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g

(e) Molar mass of Nitric acid, HNO3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+

3×16 = 63g

7. 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)?

Solution:

The mass of the above-mentioned list is as follows:

(a) Atomic mass of nitrogen atoms = 14u

Mass of 1 mole of nitrogen atoms= Atomic mass of nitrogen atoms

Therefore, the mass of 1 mole of nitrogen atom is 14g

(b) Atomic mass of aluminium =27u

Mass of 1 mole of aluminium atoms = 27g

1 mole of aluminium atoms = 27g, 4 moles of aluminium atoms = 4 x 27 = 108g

(c) Mass of 1 mole of sodium sulphite Na2SO3 = Molecular mass of sodium sulphite = 2 x Mass of Na + Mass of S + 3 x Mass of O =  (2 x 23) + 32 +(3x 16) = 46+32+48 = 126g

Therefore, mass of 10 moles of Na2SO3  = 10 x 126 = 1260g

8. Convert into mole.

(a) 12g of oxygen gas

(b) 20g of water

(c) 22g of carbon dioxide

Solution:

Conversion of the above-mentioned molecules into moles is as follows:

(a) Given: Mass of oxygen gas=12g

Molar mass of oxygen gas = 2 Mass of Oxygen =  2 x 16 = 32g

Number of moles = Mass given / molar mass of oxygen gas = 12/32 =  0.375 moles

(b) Given: Mass of water = 20g

Molar mass of water = 2 x Mass of Hydrogen + Mass of Oxygen = 2 x 1 + 16 = 18g

Number of moles = Mass given / molar mass of water

= 20/18 = 1.11 moles

(c) Given: Mass of carbon dioxide = 22g

Molar mass of carbon dioxide = Mass of C + 2 x Mass of Oxygen = 12 + 2x 16 = 12+32=44g

Number of moles = Mass given/ molar mass of carbon dioxide = 22/44 = 0.5 moles

9. What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Solution:

The mass is as follows:

(a) Mass of 1 mole of oxygen atoms = 16u, hence it weighs 16g

Mass of 0.2 moles of oxygen atoms = 0.2 x 16 = 3.2u

(b) Mass of 1 mole of water molecules = 18u, hence it weighs 18g

Mass of 0.5 moles of water molecules = 0.5 x 18 = 9u

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

Solution:

To calculate the molecular mass of sulphur:

Molecular mass of Sulphur (S8) = 8xMass of Sulphur = 8×32 = 256g

Mass given = 16g

Number of moles = mass given/ molar mass of sulphur

= 16/256 = 0.0625 moles

To calculate the number of molecules of sulphur in 16g of solid sulphur:

Number of molecules = Number of moles x Avogadro number

= 0.0625 x 6.022 x 10²³ molecules

= 3.763 x 1022 molecules

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

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

Solution:

To calculate the number of aluminium ions in 0.051g of aluminium oxide:

1 mole of aluminium oxide = 6.022 x 1023 molecules of aluminium oxide

1 mole of aluminium oxide (Al2O3) = 2 x Mass of aluminium + 3 x Mass of Oxygen

= (2x 27) + (3 x16) = 54 +48 = 102g

1 mole of aluminium oxide = 102g = 6.022 x 1023 molecules of aluminium oxide

Therefore, 0.051g of aluminium oxide has = 0.051 x 6.022 x 1023 / 102

     = 3.011 x 1020 molecules of aluminium oxide.

One molecule of aluminium oxide has 2 aluminium ions, hence the number of aluminium ions present in 0.051g of aluminium oxide = 2 x 3.011x 1020 molecules of aluminium oxide

= 6.022 x 1020.

NCERT solutions for class 9 science chapter 3.

This chapter has a value of 23 marks in the final exam. The questions are not specific and can be asked from any topic. The students are therefore advised to go through the entire chapter thoroughly. The chemical formulas and the numerical must be practised well.

This chapter includes all these topics:-

1. Laws of Chemical Combination 
2. What is an Atom? 
3. What is a Molecule? 
4. Writing Chemical Formulae
5. Molecular Mass and Mole Concept

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