Chapter 3: Atoms and Molecules - NCERT Solutions
In-Text Questions (Page No- 27)
Q1. In a reaction, 5.3 g 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.
Answer - According to the law of conservation of mass, the total mass of reactants should be equal to the total mass of products.
Mass of reactants:
Sodium carbonate = 5.3 g
Acetic acid = 6.0 g
Total = 11.3 g
Mass of products:
Carbon dioxide = 2.2 g
Water = 0.9 g
Sodium acetate = 8.2 g
Total = 11.3 g
Since the total mass of reactants equals the total mass of products (11.3 g), the law of conservation of mass is verified.
Mass of reactants:
Sodium carbonate = 5.3 g
Acetic acid = 6.0 g
Total = 11.3 g
Mass of products:
Carbon dioxide = 2.2 g
Water = 0.9 g
Sodium acetate = 8.2 g
Total = 11.3 g
Since the total mass of reactants equals the total mass of products (11.3 g), the law of conservation of mass is verified.
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 3 g of hydrogen gas?
Answer - Given: Hydrogen and oxygen combine in a 1:8 mass ratio.
If 1 g of hydrogen requires 8 g of oxygen,
then 3 g of hydrogen will require:
3 × 8 = 24 g of oxygen.
Answer: 24 g of oxygen is required to react completely with 3 g of hydrogen.
If 1 g of hydrogen requires 8 g of oxygen,
then 3 g of hydrogen will require:
3 × 8 = 24 g of oxygen.
Answer: 24 g of oxygen is required to react completely with 3 g of hydrogen.
Q3. 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 that reflects the law of conservation of mass is:
"Atoms can neither be created nor destroyed in a chemical reaction."
This means the total number and mass of atoms remain the same during a chemical change.
"Atoms can neither be created nor destroyed in a chemical reaction."
This means the total number and mass of atoms remain the same during a chemical change.
Q4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer - The relevant postulate of Dalton’s atomic theory is:
"Atoms of a given element are identical in mass and chemical properties and combine in fixed ratios to form compounds."
This explains why a particular compound always contains the same elements in the same fixed proportions by mass.
"Atoms of a given element are identical in mass and chemical properties and combine in fixed ratios to form compounds."
This explains why a particular compound always contains the same elements in the same fixed proportions by mass.
In-Text Questions (Page No- 30)
Q1. Define the atomic mass unit.
Answer - One atomic mass unit (1 u or 1 amu) is defined as one-twelfth (1/12th) the mass of a carbon-12 atom. It is used to express the relative masses of atoms and molecules.
Q2. Why is it not possible to see an atom with naked eyes?
Answer - Atoms are extremely small in size, typically around 0.1 nanometers (1 × 10−10 m). Due to their tiny size, they cannot reflect or absorb visible light in a way that our eyes can detect, which is why we cannot see them with the naked eye.
In-Text Questions (Page No- 34)
Q1. Write down the formulae of:
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide
Answer - (i) Sodium oxide – Na2O
(ii) Aluminium chloride – AlCl3
(iii) Sodium sulphide – Na2S
(iv) Magnesium hydroxide – Mg(OH)2
(ii) Aluminium chloride – AlCl3
(iii) Sodium sulphide – Na2S
(iv) Magnesium hydroxide – Mg(OH)2
Q2. Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3
(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3
Answer - (i) Aluminium sulphate
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate
Q3. What is meant by the term chemical formula?
Answer - A chemical formula is a symbolic representation of a compound, showing the elements present and the number of atoms of each element. For example, H2O represents water, which contains 2 atoms of hydrogen and 1 atom of oxygen.
Q4. How many atoms are present in:
(i) H2S molecule
(ii) PO43− ion?
(i) H2S molecule
(ii) PO43− ion?
Answer - (i) H2S contains 3 atoms (2 hydrogen and 1 sulfur).
(ii) PO43− contains 5 atoms (1 phosphorus and 4 oxygen).
(ii) PO43− contains 5 atoms (1 phosphorus and 4 oxygen).
In-Text Questions (Page No- 35)
Q1. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
Answer - Using atomic masses: H = 1 u, O = 16 u, Cl = 35.5 u, C = 12 u, N = 14 u
H2: 2 × 1 = 2 u
O2: 2 × 16 = 32 u
Cl2: 2 × 35.5 = 71 u
CO2: 12 + 2 × 16 = 12 + 32 = 44 u
CH4: 12 + 4 × 1 = 12 + 4 = 16 u
C2H6: 2 × 12 + 6 × 1 = 24 + 6 = 30 u
C2H4: 2 × 12 + 4 × 1 = 24 + 4 = 28 u
NH3: 14 + 3 × 1 = 14 + 3 = 17 u
CH3OH: 1 × 12 + 4 × 1 + 1 × 16 = 12 + 4 + 16 = 32 u
H2: 2 × 1 = 2 u
O2: 2 × 16 = 32 u
Cl2: 2 × 35.5 = 71 u
CO2: 12 + 2 × 16 = 12 + 32 = 44 u
CH4: 12 + 4 × 1 = 12 + 4 = 16 u
C2H6: 2 × 12 + 6 × 1 = 24 + 6 = 30 u
C2H4: 2 × 12 + 4 × 1 = 24 + 4 = 28 u
NH3: 14 + 3 × 1 = 14 + 3 = 17 u
CH3OH: 1 × 12 + 4 × 1 + 1 × 16 = 12 + 4 + 16 = 32 u
Q2. Calculate the formula unit masses of ZnO, Na2O, K2CO3. Given: Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, O = 16 u.
Answer - ZnO: 65 + 16 = 81 u
Na2O: 2 × 23 + 16 = 46 + 16 = 62 u
K2CO3: 2 × 39 + 12 + 3 × 16 = 78 + 12 + 48 = 138 u
Na2O: 2 × 23 + 16 = 46 + 16 = 62 u
K2CO3: 2 × 39 + 12 + 3 × 16 = 78 + 12 + 48 = 138 u
Exercises
Q1. 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.
Answer - Total mass of the compound = 0.24 g
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g
Percentage of boron = (0.096 / 0.24) × 100 = 40%
Percentage of oxygen = (0.144 / 0.24) × 100 = 60%
So, the compound contains 40% boron and 60% oxygen by weight.
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g
Percentage of boron = (0.096 / 0.24) × 100 = 40%
Percentage of oxygen = (0.144 / 0.24) × 100 = 60%
So, the compound contains 40% boron and 60% oxygen by weight.
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?
Answer - From the given data:
3.0 g carbon + 8.0 g oxygen → 11.0 g carbon dioxide
Even when 50.0 g of oxygen is available, only 8.0 g will react with 3.0 g carbon (as per the fixed ratio), and the rest will remain unused.
So, mass of CO2 formed = 3.0 g + 8.0 g = 11.0 g
This is governed by the Law of Constant Proportions, which states that a chemical compound always contains its elements in a fixed ratio by mass.
3.0 g carbon + 8.0 g oxygen → 11.0 g carbon dioxide
Even when 50.0 g of oxygen is available, only 8.0 g will react with 3.0 g carbon (as per the fixed ratio), and the rest will remain unused.
So, mass of CO2 formed = 3.0 g + 8.0 g = 11.0 g
This is governed by the Law of Constant Proportions, which states that a chemical compound always contains its elements in a fixed ratio by mass.
Q3. What are polyatomic ions? Give examples.
Answer - Polyatomic ions are ions that are made up of two or more atoms combined together, carrying a positive or negative charge as a whole.
Examples:
• Ammonium: NH4+
• Sulphate: SO42–
• Nitrate: NO3–
• Carbonate: CO32–
Examples:
• Ammonium: NH4+
• Sulphate: SO42–
• Nitrate: NO3–
• Carbonate: CO32–
Q4. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
(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
(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.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Answer - (a) Quick lime – Calcium and Oxygen
(b) Hydrogen bromide – Hydrogen and Bromine
(c) Baking powder – Sodium, Hydrogen, Carbon, and Oxygen (NaHCO3)
(d) Potassium sulphate – Potassium, Sulphur, and Oxygen
(b) Hydrogen bromide – Hydrogen and Bromine
(c) Baking powder – Sodium, Hydrogen, Carbon, and Oxygen (NaHCO3)
(d) Potassium sulphate – Potassium, Sulphur, and Oxygen
Q6. Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3
Answer - Using atomic masses: C = 12 u, H = 1 u, S = 32 u, P = 31 u, Cl = 35.5 u, N = 14 u, O = 16 u
(a) C2H2: 2 × 12 + 2 × 1 = 24 + 2 = 26 g/mol
(b) S8: 8 × 32 = 256 g/mol
(c) P4: 4 × 31 = 124 g/mol
(d) HCl: 1 + 35.5 = 36.5 g/mol
(e) HNO3: 1 + 14 + 3 × 16 = 1 + 14 + 48 = 63 g/mol
(a) C2H2: 2 × 12 + 2 × 1 = 24 + 2 = 26 g/mol
(b) S8: 8 × 32 = 256 g/mol
(c) P4: 4 × 31 = 124 g/mol
(d) HCl: 1 + 35.5 = 36.5 g/mol
(e) HNO3: 1 + 14 + 3 × 16 = 1 + 14 + 48 = 63 g/mol