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

Page No. 27

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 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
Ans: 

Mass of reactants = 5.3 g + 6 g = 11.3 g
Mass of products = 2.2 g + 0.9 g + 8.2 g = 11.3 g
Mass of reactants = Mass of products
Therefore, the Law of conservation of mass is proved.

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 3g of hydrogen gas?
Ans: Since hydrogen and oxygen combine in the ratio of 1:8 by mass, 3g of hydrogen gas will react completely with 24 g of oxygen gas.

Q3. Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?
Ans: Dalton’s postulate that “Atoms can neither be created nor destroyed,” is a result of the law of conservation of mass.

Q4. Which postulate of Dalton's atomic theory can explain the law of definite proportions?
Ans: Atoms combine in a fixed ratio to form compounds, which can explain the law of definite proportions.

Q1. Define atomic mass unit.
Ans: It is defined as equal to 1/12th of the mass of 1 atom of C-12. It is called unified mass denoted by 'u’ these days.

Q2. Why is it not possible to see an atom with naked eyes?
Ans: The size of an atom is so small that it is not possible to see it with naked eyes. Also, the atom of an element does not exist independently.

Q1. Write down the formulae of 
(a) sodium oxide
(b) aluminium chloride
(c) sodium sulphide
(d) magnesium hydroxide
Ans:
(a) Formula of Sodium Oxide

(b) Formula of Aluminium Chloride

(c) Formula of Sodium Sulphide

(d) Formula of Magnesium Hydroxide

Q2. Write down the names of compounds represented by the following formulae:
(a) Al₂(SO₄)₃
(b) CaCl₂
(c) K₂SO₄
(d) KNO₃
(e) CaCO₃
Ans: 
(a) Aluminium sulphate 
(b) Calcium chloride 
(c) Potassium sulphate 
(d) Potassium nitrate 
(e) Calcium carbonate

Q3. What is meant by the term chemical formula?
Ans: The chemical formula of a compound is a symbolic representation of its composition.
Chemical Formula of Water

Q4. How many atoms are present in
(a) H₂S molecule and
(b) PO₄^3- ion?
Ans: 
(i) H₂S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in totality.
(ii) PO₄^3- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in totality.

Q1. Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.
Ans: 
Molecular mass of H₂ = 2 × Atomic mass of H
= 2 × 1
= 2 u
Molecular mass of O₂ = 2 × Atomic mass of O
= 2 × 16
= 32 u
Molecular mass of Cl₂ = 2 × Atomic mass of Cl
= 2 × 35.5
= 71 u
Molecular mass of CO₂ = Atomic mass of C₂ × Atomic mass of O
= 12+ (2+16) = (12 + 32)u 
= 44 u
Molecular mass of CH₄= Atomic mass of C4 × Atomic mass of H
= 12+ (4 x 1)u = (12 + 4)u 
= 16 u
Molecular mass of C₂H₆ = 2× Atomic mass of C₆× Atomic mass of H
= (2 x 12 + 6 x 1)u = (24 + 6)u 
= 30 u
Molecular mass of C₂H₄ = 2 x Atomic mass of C₄ × Atomic mass of H
= (2 x 12 + 4 x 1)u = (24 + 4)u 
= 28 u
Molecular mass of NH₃ = Atomic mass of N₃ × Atomic mass of H
= (14 + 3 x 1)u = (14 + 3)u 
= 17 u
Molecular mass of CH₃OH = Atomic mass of C₃ × Atomic mass of H Atomic mass of O Atomic mass of H
= (12 + 3 x 1 + 16 + 1)u = (12 + 3 + 17)u 
= 32 u

Q2. 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:
(i) Formula unit mass of ZnO 
= 65 + 16 = 81 u 
(ii) Formula unit mass of Na₂O
= 2 x 23 + 16 = 46 + 16 = 62 u
(iii) Formula unit mass of K₂CO₃ 
= 2 x 39 + 12 + 3 x 16
= 78 + 12 + 48 = 138 u

Q1. A 0.24 g sample of a 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.
Ans: 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%

Q2. 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:  When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.
Given that
3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.
Find out
We need to find out the mass of carbon dioxide that will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen.
Solution
First, let us write the reaction taking place here.
C + O₂ → CO₂
As per the given condition, when 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.
3g + 8g →11 g ( from the above reaction)
The total mass of reactants = mass of carbon + mass of oxygen
= 3g+8g
= 11g
The total mass of reactants = Total mass of products
Therefore, the law of conservation of mass is proved.
Then, it also depicts that carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.
Thus, it further proves the law of constant proportions.
3 g of carbon must also combine with 8 g of oxygen only.
This means that (50−8)=42g of oxygen will remain unreacted.
The remaining 42 g of oxygen will be left un-reactive. In this case, too, only 11 g of carbon dioxide will be formed
The above answer is governed by the law of constant proportions.

Q3. What are polyatomic ions? Give examples.
Ans: Polyatomic ions are ions that contain more than one atom, but they behave as a single unit.
Example: CO₃^2-, H₂PO₄^– 

Q4. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Ans: 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

Q5. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Ans:  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)

Q6. 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
Ans:  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