NCERT Solutions for Class 9 Science Chapter 3 - Atoms and Molecules

NCERT Solutions for Class 9 Science Chapter 3 – Atoms and Molecules. Furthermore, here we’ve provided you with the latest solution for Class 9 Science Chapter 3 – Atoms and Molecules. As a result here you’ll find solutions to all the exercises. This NCERT Class 9 solution will help you to score good marks in your exam.

Students can refer to our solution for NCERT Class 9 Science Chapter 3 – Atoms and Molecules. The Chapter 3 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the latest solution.

NCERT Solutions for Class 9 Science Chapter 3 – Atoms and Molecules Pdf Download

Here you’ll find out the answer to questions on page no. 32 of Ncert Class 9 Science Chapter 3 – Atoms and Molecules.

Question 1: In a reaction, 5.3 g 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 ethanoate. Show that these observations are in agreement with the law of conservation of mass.

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

Solution:

Law of conservation of mass states that mass is neither created nor destroyed during a chemical reaction.
It means the mass remains the same. So, we add the mass of the reactants on LHS and add the mass of all products on RHS

LHS = 5.3 g + 6 g = 11.3 g

RHS = 8.2g + 2.2g + 0.9g = 11.3g

LHS = RHS

So, the observations are in agreement with the law of conservation of mass.

Question 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 3g of hydrogen gas?

Solution: ‘Law of constant proportions’ states that composition of a compound is always fixed.

Applying this

∵ 1g of hydrogen gas combines with oxygen = 8g.

∴ 3g of hydrogen gas will combine with oxygen = 8 x 3 = 24g

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

Solution: Following postulate of Dalton’s atomic theory is the result of the law of conservation of mass. ‘Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.’

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

Solution: Following postulate of Dalton’s atomic theory can explain the ‘law of definite proportions’. ‘The relative number and kinds of atoms are constant in a given compound.’

Here you’ll find out the answer to questions on page no. 35 of Ncert Class 9 Science Chapter 3 – Atoms and Molecules.

Question 1: Define the atomic mass unit.

Solution: One atomic mass unit (amu) is a mass unit equal to exactly one- twelfth (1/12th) the mass of one atom of carbon-12.’ The relative atomic masses of all the elements have been found with respect to an atom of carbon-12.

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

Solution: As an atom is extremely small in size, it is not possible to see it with naked eyes. Generally, radius of an atom is of the order of nanometres. For example, atomic radius of hydrogen atom is 10-10m (or 10^-1nm).

Here you’ll find out the answer to questions on page no. 35 of Ncert Class 9 Science Chapter 3 – Atoms and Molecules.

Question 1: Write down the formulae of

sodium oxide
aluminium chloride
sodium sulphide
magnesium hydroxide Solution:

Na₂O
NH₄Cl
Na₂SO₄
Mg(OH)₂

Question 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:

Aluminium sulphate
Calcium chloride
Potassium sulphate
Potassium nitrate
Calcium carbonate.

Question 3: What is meant by the term chemical formula? Solution:

Answer: Chemical formula of a compound (or element) is the symbolic representation of its composition. It represents

The number and kind of atoms present per molecule of the compound,
One mole of the compound,
Molar mass of the compound.

Question 4: How many atoms are present in a

(i) H₂S molecule and (ii) PO₄−ion?

Solution:

2 atom of hydrogen + 1 atom of sulphur

= three (3) atoms (in a H₂S molecule).

1 atom of phosphorus + 4 atoms of oxygen

= five (5) atoms (in a PO₄−ion).

Here you’ll find out the answer to questions on page no. 40 of Ncert Class 9 Science Chapter 3 – Atoms and Molecules.

Question 1: Calculate the molecular masses of

H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

Solution:

Molecular mass of H₂(hydrogen)

= Atomic mass of hydrogen × 2

= 1 × 2 = 2u

Molecular mass of O₂ (oxygen)

=Atomic mass of oxygen × 2

= 16 × 2 = 32u

Molecular mass of CI₂(chlorine)

= Atomic mass of chlorine × 2

= 35.5 × 2 = 71u

Molecular mass of CO₂(carbon dioxide)

= (Atomic mass of carbon × 1)+ (Atomic mass of oxygen × 2)

= 12 + (16 × 2) = 12 + 32 = 44 u

Molecular mass of CH₄(methane)

= (Atomic mass of carbon × 1) + (Atomic mass of hydrogen × 4)

= 12 + (1 × 4) = 12 + 4 = 16 u

Molecular mass of C₂H₆(ethane)

= (Atomic mass of carbon × 2) + (Atomic mass of hydrogen × 6)

= (12 × 2) + (1 × 6) = 24 + 6 = 30 u

Molecular mass of C₂H₄(ethene)

= (Atomic mass of carbon × 2) + (Atomic mass of hydrogen × 4)

= (12 × 2) + (1 × 4) = 24 + 4 = 28 u

Molecular mass of NH₃(ammonia)

= (Atomic mass of nitrogen × 1) + (Atomic mass of hydrogen × 3)

= (14 × 1) + (1 × 3) = 14 + 3 = 17 u

Molecular mass of CH₃OH (methanol or methyl alcohol)

= (Atomic mass of carbon × 1) + (Atomic mass of hydrogen × 3)+ (Atomic mass of oxygen × 1) + (Atomic mass of hydrogen × 1)

= 12 + 3 + 16 + 1 = 32 u

Question 2: Calculate the formula unit masses of ZnO, Na₂O, and K₂CO₃. Given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u and 0 = 16 u. Solution:

Formula unit mass of ZnO (zinc oxide) = 65 + 16 = 81 u
Formula unit mass of Na2O (sodium oxide) = (23 × 2) + (16 × 1) = 46 + 16 = 62 u
Formula unit mass of K2CO3(potassium carbonate) = (39 × 2) + (12 × 1) + (16 × 3) = 78 + 12+ 48 = 138 u

Here you’ll find out the answer to questions on page no. 42 of Ncert Class 9 Science Chapter 3 – Atoms and Molecules.

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

Solution:

1 mole carbon atom = 6.022 x 10²³ atoms.

Molar atomic mass = 12 g

6.022 x 10²³ carbon atoms weigh = 12 g

1 carbon atom weighs = 12/6.022 x 10²³ =1.99 x 10^-23g

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

Solution:

Molar mass of sodium = 23 g

1 mole atom = 6.022 x 10²³ atoms

23g sodium contains = 6.022 x 10²³ atoms

1g sodium contains = 6.022 x 10²³ atoms ÷ 23

= 2.618 x 10²² atoms

100g sodium contains = 6.022 x 10²³ atoms * 100

= 2.618 x 10²⁴ atoms

By the above method or by formula we find number of atoms in 100 g Fe;

Number of atoms of an element in a given mass Given mass

Hence, 100 g of sodium has more number of atoms as compared to 100 g of iron.

Exercises for Ncert Class 9 Science

Question 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 weight.

Solution:

Mass of the compound = 0.24g.
Mass of boron = 0.096g
Mass of oxygen = 0.144g
Mass of boron 0.096g

Percentage of boron = (Mass of boron / Mass of compound) x 100

= 0.096/0.240 x 100 = 40%

Percentage of oxygen = (Mass of oxygen / Mass of compound) x 100

=( 0.144 / 0.240) x 100 = 60%

Alternative method

Percentage of oxygen = 100 percentage of boron

=100 – 40 = 60%

Question 2: When 3.0 g 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 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Solution:

First, we find the proportion of mass of carbon and oxygen in carbon dioxide.

In CO₂ , C:O = 12:32 or 3 : 8

In other words, we can say that

12.00 g carbon reacts with oxygen = 32.00 g

3.00 g carbon will react with oxygen = 32 x 3 = 8g

Therefore, 3.00 g of carbon will always react with 8.00 g of oxygen to form CO₂ (11g), even if large amount (50.00 g) of oxygen is present. This answer will be governed by ‘the law of constant proportions’.

Question 3: What are polyatomic ions? Give examples.

Solution: The group of atoms which carry a fixed charge (either positive or negative) on them and behave as ions are called polyatomic ions.

Example

(i) Carbonate ion (ii) Sulphate ion

(iii) Ammonium ion (iv) Phosphate ion

Question 4: Write the chemical formulae of the following.

(i) Magnesium chloride (ii) Calcium oxide

(iii) Copper nitrate (iv) Aluminium chloride

(v) Calcium carbonate Solution:

Answer:

(i) Formula = MgCl₂ (Magnesium chloride)

(ii) Formula = CaO (Calcium oxide)

(iii) Formula = Cu(NO₃)₂ (Copper nitrate)

(iv) Formula = AICI₃ (Aluminium Chloride)

(v) Formula = CaCO₃ (Calcium carbonate)

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

(a) Quick lime (b) Hydrogen bromide

Solution:

Quick lime- Calcium oxide – CaO Elements–Calcium, oxygen.

Hydrogen bromide- HBr Elements- Hydrogen, bromine.

Baking powder- Sodium hydrogen carbonate -NaHCO3 Elements- Sodium, hydrogen, carbon, oxygen.

Potassium sulphate- K2SO4 Elements- Potassium, sulphur, oxygen.

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

Ethyne, C₂H₂
Sulphur molecule, S₈
Phosphorus molecule, P4(Atomic mass of phosphorus = 31)
Hydrochloric acid, HCl
Nitric acid, HNO₃ Solution:
Molar mass of C₂H₂

= (2 × Atomic mass of C) + (2 × Atomic mass of H)

= (2 × 12) + (2 × 1)

= 26 u

Molar mass of S₈

= (8 × Atomic mass of S)

= 8 × 32 = 256 u

Molar mass of P₄

= 4 × Atomic mass of P

= 4 × 31 = 124u

Molar mass of HCI

= Atomic mass of hydrogen + Atomic mass of CI

= 1 + 35.5 = 36.5 u

Molar mass of HNO₃

= Atomic mass of H + Atomic mass of N + (3 × Atomic mass of 0)

= 1 + 14 + (3 × 16) = 15 + 48 = 63 u

Question 7: What is the mass of

1 mole of nitrogen atoms?
4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
10 moles of sodium sulphite (Na₂SO₃)?

Solution:

Molar mass of N atom = Atomic mass of N.

Mass of 1 mol of N atoms = 14g

Mass of 1 mole AI atoms = 27g

Mass of 4 moles of AI atoms = 27 × 4 = 108g

= 46 + 32 + 48 = 126g

Mass of 10 moles of Na₂SO₃ = 126 × 10 = 1260 g

Question 8: Convert into mole.

12g of oxygen gas
20g of water
22g of carbon dioxide.

Solution:

(O₂)

Molar mass of oxygen (O₂) = 16 x 2 = 32 g

32g oxygen gas = 1 mol

12g of oxygen gas = (1 x 12g/32) = 0.375 mol

(H₂O)

Molar mass of water (H₂O) = 2 + 16 = 18g
18g water = 1 mol

20g water = (1 x 20g/18g) = 1.11 mol

22g of Carbon Dioxide(CO₂)

Molar mass of carbon dioxide (CO₂) =12 + 32 = 44g

44g of CO₂ = 1 mol

22g of CO₂ =(1x 22g/44) = 0.5 mol

Question 9: What is the mass of

0.2 mole of oxygen atoms?
0.5 mole of water molecules?

Solution:

Mass of 1 mole O-atoms = 16 g

Mass of 0.2 mole O-atoms = 16 × 0.2 = 3.2 g

Mass of 1 mole of H₂O molecules = 18 g

Mass of 0.5 mole of H₂O molecules = 18 × 0.5 = 9.0 g

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

Solution:

Molar mass of sulphur (S₈) = 32 x 8 = 256g

Number of S₈ molecules in 256g of solid sulphur = 6.022 x 10²³
Number of S₈ molecules in 16g of solid sulphur=(6.022 x10²³x 16g)/256 =3.76 x 10²³ molecules

Question 11: Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. [HintThe mass of an ion is the same as that of an atom of the same element. Atomic mass of AI= 27 u.]

Solution: