NCERT Solutions for Class 9 Science (Chemistry) Chapter 3 Atoms and Molecules provides detailed answers for all in-text and exercise Questions. These solutions contain an in-depth explanation of each topic involved in the chapter. All these solutions are prepared by expert teachers and updated for the current academic session.
NCERT Solutions for Class 9 Science(Chemistry) Chapter 3 Atoms and Molecules Pure help students to understand the fundamental concepts given in class 9 Science textbook. We have prepared the answers to all the questions in an easy and well-structured manner. It helps students to grasp the chapter easily.
Class 9 Science Atoms and Molecules Intext Questions and Answers
Page No. 27
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, 0.9g 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
Answer: In a reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon-dioxide, and water.
Mass of sodium carbonate = 5.3g
Mass of ethanoic acid = 6g
Mass of sodium ethanoate = 8.2g
Mass of carbon dioxide = 2.2
Mass of water = 0.9 g
Now, total mass before the reaction = (5.3 + 6) g = 11.3 g
and total mass after the reaction = (8.2 + 2.2 + 0.9) g = 11.3 g
Therefore, Total mass before the reaction = Total mass after the reaction
Hence, the given 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?
Answer: Given that hydrogen and water mix in the ratio 1: 8.
For every 1g of hydrogen, it is 8g of oxygen.
Therefore, for 3g of hydrogen, the quantity of oxygen = 3 x 8 = 24g
Hence, 24g of oxygen would be required for the complete reaction with 3g of hydrogen gas.
Question 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 which is a result of the law of conservation of mass is Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction”.
Question 4: Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer: The postulate of Dalton’s atomic theory which can explain the law of definite proportion is “The relative number and kind of atoms in a given compound remains constant”.
PAGE NO. 30
Question 1: Define atomic mass unit.
Answer: An atomic mass unit is a unit of mass used to express weights of atoms and molecules where one atomic mass is equal to 1/12th the mass of one carbon-12 atom.
Question 2: Why is it not possible to see an atom with naked eyes?
Answer: The size of an atom is so small that it is not possible to see it with naked eyes. Also, atom of an element does not exist independently.
PAGE NO. 34
Question 1. Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Answer: The following are the formulae:
(i) sodium oxide – Na2O
(ii) aluminium chloride – AlCl3
(iii) sodium sulphide – Na2S
(iv) magnesium hydroxide – 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.
Answer: Listed below are the names of the compounds for each of the following formulae
(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?
Answer: The term “chemical formula” refers to a way of representing the composition of a chemical compound using symbols and numbers. From the chemical formula of a compound, we can know the number and kinds of atoms of different elements that constitute the compound. For example, in water (H2O), the formula indicates that each molecule consists of 2 hydrogen atoms (H) and 1 oxygen atom (O). This formula provides a concise way to describe the exact makeup of a compound.
Question 4. How many atoms are present in a
(i) H2S molecule and
(ii) PO43- ion?
Answer: The number of atoms present are as follows:
(i) H2S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in total.
(ii) PO43- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in total.
Page No. 35
Question 1: Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
Answer: Molecular mass of H2 = 2 × Atomic mass of H = 2 × 1 = 2u
Molecular mass of O2 = 2 × Atomic mass of O = 2 × 16 = 32u
Molecular mass of Cl2 = 2 × Atomic mass of Cl = 2 × 35.5 = 71 u
Molecular mass of CO2 = Atomic mass of C + 2 × Atomic mass of O = 12 + 2 × 16 = 44 u
Molecular mass of CH4 = Atomic mass of C + 4 × Atomic mass of H = 12 + 4 × 1 = 16 u
Molecular mass of C2 H6 = 2 × Atomic mass of C + 6 × Atomic mass of H = 2 × 12 + 6 × 1 = 30u
Molecular mass of C2H4 = 2 × Atomic mass of C + 4 × Atomic mass of H = 2 × 12 + 4 × 1 = 28u
Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H = 14 + 3 × 1 =17 u
Molecular mass of CH3OH =Atomic mass of C+4 ×Atomic mass of H + Atomic mass of O = 12 + 4 × 1 + 16 = 32 u
Question 2: Calculate the formula unit masses of ZnO, Na2O, K2CO3, given masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and O = 16u.
Answer: Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O = 65 + 16 = 81 u
Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O = 2 × 23 + 16 = 62u
Formula unit mass of K2 CO3 = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O = 2 × 39 + 12 + 3 × 16 = 138 u
CBSE Class 9 Science Atoms and Molecules Exercise Questions (Solved)
Question 1: A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g if boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer: Given, Mass of boron = 0.096
Mass of oxygen = 0.144g
Mass of sample = 0.24g
Question 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 combinations will govern your answer?
Answer: Carbon + Oxygen → Carbon dioxide
3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt in 50g of oxygen, then 3g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive. In this case also, only 11g of carbon dioxide will be formed. The above answer is governed by the law of constant proportions.
Question 3: What are polyatomic ions? Give examples?
Answer: A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion (𝑁𝐻4+), hydroxide ion (OH−), carbonate ion (𝐶𝑂32−), sulphate ion (𝑆𝑂42− )
Question 4: Write the chemical formula of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer:
(a) Magnesium chloride – MgCl2
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO3)2
(d) Aluminium chloride – AlCl3
(e) Calcium carbonate – CaCO3
Question 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:
| Compound | Chemical formula | Elements present |
| Quick lime | CaO | Calcium, oxygen |
| Hydrogen bromide | HBr | Hydrogen, bromine |
| Baking powder | NaHCO3 | Sodium, hydrogen, carbon, oxygen |
| Potassium sulphate | K2SO4 | Potassium, sulphur, oxygen |
Question 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
Answer:
(a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 28g
(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256g
(c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124g
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5g
(e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63g
Question 7: What is the mass of
(a) 1 mole of nitrogen atoms?
(b) 4 mole of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na2SO3)?
Answer: (a) The mass of 1 mole of nitrogen atoms is 14g.
(b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g
(c) The mass of 10 moles of Sodium Sulphite (Na2SO3) is
=10 × [2 × 23 + 32 + 3 × 16]g
= 10 × 126g
= 1260g
Question 8: Convert into mole.
(a) 12g of oxygen gas
(b) 12g of water
(c) 22g of carbon dioxide
Answer:
(a) 32 g of oxygen gas = 1 mole
Then, 12g of oxygen gas = 12/32 mole = 0.375 mole
(b) 18g of water = 1 mole
Then, 20 g of water = 20/18 mole = 1.11 moles (approx.)
(c) 44g of carbon dioxide = 1 mole
Then, 22g of carbon dioxide = 22/44 mole = 0.5 mole
Question 9: What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer:
(a) Mass of one mole of oxygen atoms = 16g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2g
(b) Mass of one mole of water molecule = 18g
Then, mass of 0.5 mole of water molecules = 0.5 × 18g = 9g
Question 10: Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.
Answer: 1 mole of solid sulphur (S8) = 8 × 32g = 256 g
Question 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. Atomic mass of Al = 27u)
