NCERT Solutions for Class 9 Science Chapter 3 – Atoms and Molecules (updated Pattern)
What you have learnt
- During a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is known as the Law of Conservation of Mass.
- In a pure chemical compound, elements are always present in a definite proportion by mass. This is known as the Law of Definite Proportions.
- An atom is the smallest particle of the element that cannot usually exist independently and retain all its chemical properties.
- A molecule is the smallest particle of an element or a compound capable of independent existence under ordinary conditions. It shows all the properties of the substance.
- A chemical formula of a compound shows its constituent elements and the number of atoms of each combining element.
- Clusters of atoms that act as an ion are called polyatomic ions. They carry a fixed charge on them.
- The chemical formula of a molecular compound is determined by the valency of each element.
- In ionic compounds, the charge on each ion is used to determine the chemical formula of the compound.
In Chapter Exercise- 1
1. In a reaction, 5.3g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of 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 + acetic acid → Sodium acetate + carbon dioxide + water
5.3g 6g 8.2g 2.2g 0.9g
As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of products.
As per the above reaction, L.H.S. = R.H.S. i.e., 5.3g + 6g = 2.2g + 0.9 g + 8.2 g = 11.3 g
Hence, the observations are in agreement with the law of conservation of mass.
2. Hydrogen and oxygen combine in a 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 hydrogen and water mix in a 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.
3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Solution: The relative number and types of atoms are constant in a given composition, according to Dalton’s atomic theory, which is based on the rule of conservation of mass. “Atoms cannot be created nor be destroyed in a chemical reaction.”
4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Solution: The postulate of Dalton’s atomic theory that can explain the law of definite proportions is that the relative number and kinds of atoms are equal in given compounds.
NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure (Updated Pattern)
In Chapter Exercise- 2
1. Define the atomic mass unit.
Solution: An atomic mass unit is a unit of mass used to express the weights of atoms and molecules where one atomic mass is equal to 1/12th the mass of one carbon-12 atom.
2. Why is it not possible to see an atom with the naked eye?
Solution: Firstly, atoms are minuscule, measured in nanometers. Secondly, except for atoms of noble gases, they do not exist independently. Hence, an atom cannot be visible to the naked eye.
In Chapter Exercise- 3
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: 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?
Solution: Chemical formulas are used to describe the different types of atoms and their numbers in a compound or element. Each element’s atoms are symbolised by one or two letters. A collection of chemical symbols that depicts the elements that make up a compound and their quantities. For example, the chemical formula of hydrochloric acid is HCl.
4. How many atoms are present in a
(i) H2S molecule and
(ii) PO43- ion?
Solution: The number of atoms present is 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.
In Chapter Exercise- 4
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: Given:
The atomic mass of Zn = 65u
The atomic mass of Na = 23u
The atomic mass of K = 39u
The atomic mass of C = 12u
The atomic mass of O = 16u
The formula unit mass of ZnO= Atomic mass of Zn + Atomic mass of O = 65u + 16u = 81u
The formula unit mass of Na2O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u + 16u = 46u + 16u = 62u
The formula unit mass of K2CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u
Back Exercise
1. A 0.24g sample of a 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 the 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 – the 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: 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.
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 is 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
