NCERT Science Class 9 Chapter 3 Question Answer Solutions – Atoms and Molecules FREE PDF Download
Page 27-28
Q. 1. In a reaction, 5.3g of sodium carbonate reacted with 6g of acetic acid. The products were 2.2g of carbon dioxide, 0.9g water and 8.2g 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
Answer:-
Verification of the Law of Conservation of Mass ⚖️✨
The Law of Conservation of Mass states that in a chemical reaction, mass is neither created nor destroyed—the total mass of reactants must equal the total mass of products. Let’s check whether this principle holds for the given reaction! 🚀
Given Reaction:
🧪 Sodium Carbonate + Acetic Acid → Sodium Acetate + Carbon Dioxide + Water
Step 1: Find Total Mass of Reactants 🧫
- Mass of Sodium Carbonate (Na₂CO₃) = 5.3g
- Mass of Acetic Acid (CH₃COOH) = 6.0g
- Total Mass of Reactants = 5.3g + 6.0g = 11.3g
Step 2: Find Total Mass of Products 🧪
- Mass of Carbon Dioxide (CO₂) = 2.2g
- Mass of Water (H₂O) = 0.9g
- Mass of Sodium Acetate (CH₃COONa) = 8.2g
- Total Mass of Products = 2.2g + 0.9g + 8.2g = 11.3g
Conclusion: ✅
Since Total Mass of Reactants = Total Mass of Products (11.3g), the observations agree with the Law of Conservation of Mass! This confirms that mass is conserved during the reaction. 🎉🔬
💡 Key Point: In a closed system, chemical reactions obey this fundamental law, making it a crucial concept in chemistry! 🔍✨
Q. 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:-
Mass Calculation Using the Law of Constant Proportions ⚖️💧
The Law of Constant Proportions states that elements combine in fixed ratios by mass to form compounds. In the case of water (H₂O), hydrogen and oxygen always react in the ratio 1:8 by mass. Let’s use this ratio to find the required mass of oxygen! 🚀
Given Information:
🔹 Ratio of hydrogen to oxygen = 1:8
🔹 Mass of hydrogen given = 3g
Step 1: Find Mass of Oxygen Required 🔬
Since oxygen always combines with hydrogen in the ratio 1:8, the mass of oxygen required can be found as:
[Mass of Oxygen] = [ Mass of Hydrogen x 8 times ]
[ = 3g x 8 ]
[ = 24g ]
Thus, 24g of oxygen gas (O₂) is needed to react completely with 3g of hydrogen gas (H₂) to form water! 💧⚡
Final Answer: ✅
Mass of oxygen required = 24g 🌿🔥
This confirms that the reaction follows the Law of Constant Proportions, ensuring mass consistency! 🎯✨
💡 Key Concept: In any chemical combination, elements react in a fixed proportion by mass, making this principle crucial in chemistry! 🧪📚
Q. 3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer:-
Dalton’s Atomic Theory & the Law of Conservation of Mass ⚛️✨
One of Dalton’s postulates directly supports the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. Let’s explore which postulate is responsible! 🚀
Postulate Related to the Law of Conservation of Mass: 🔬
🟢 Atoms are indivisible particles that cannot be created or destroyed in a chemical reaction.
Since atoms remain unchanged during a reaction (only rearranged), the total mass of reactants equals the total mass of products, proving the Law of Conservation of Mass! ⚖️✅
How This Postulate Explains Mass Conservation:
🔹 During a chemical reaction, atoms only rearrange to form new substances.
🔹 No atoms are lost or gained, ensuring that the total mass remains the same before and after the reaction.
🔹 This fundamental idea aligns perfectly with Lavoisier’s Law of Conservation of Mass! 🔍✨
💡 Key Concept: Dalton’s atomic theory laid the foundation for modern chemistry by proving that atoms are the building blocks of matter! 🧪📚
Q. 4. Which postulate of Dalton’s atomic theory can explain the law of definite properties?
Answer:-
Dalton’s Atomic Theory & the Law of Definite Proportions ⚛️✨
The Law of Definite Proportions states that a chemical compound always contains the same elements in a fixed ratio by mass, regardless of the source or method of preparation. This principle aligns perfectly with a key postulate of Dalton’s Atomic Theory! 🚀
Postulate Related to the Law of Definite Proportions: 🔬
🟢 Atoms of a given element have a fixed mass, and compounds are formed by the combination of atoms in simple whole-number ratios.
Since atoms of an element always have the same mass, they combine in definite proportions to form compounds, ensuring constant composition! ✅⚖️
How This Postulate Explains Definite Proportions:
🔹 Every chemical compound consists of specific atoms in a fixed ratio.
🔹 The ratio of atoms remains constant, regardless of the amount or method of preparation.
🔹 For example, in water (H₂O), hydrogen and oxygen always combine in a mass ratio of 1:8! 💧✨
💡 Key Concept: Dalton’s atomic theory laid the foundation for modern chemistry, ensuring that chemical compounds maintain their identity and consistency across different samples! 🧪📚
Page – 30
Q. 1. Define the atomic mass unit.
Answer:-
Definition of Atomic Mass Unit (amu) ⚛️✨
The atomic mass unit (amu), also called the unified mass unit (u), is a standard unit used to express the mass of atoms and subatomic particles. 🚀
Precise Definition:
🟢 1 atomic mass unit (1u) is defined as ( 1/12)th the mass of one atom of carbon-12 (C-12 isotope).
🔹 Mathematically:
[ 1u = 1/12 of the Mass of one C-12 atom]
🔹 Numerical Value:
[ 1u = 1.6605 x 10-27 kg} ]
The atomic mass unit (amu), or dalton (Da), is defined mathematically as one-twelfth (1/12) of the mass of a carbon-12 atom. Specifically, 1 amu is approximately equal to 1.66 x 10-27 kg.
Key Concept: 🧪
💡 Since atoms are extremely small, their masses are difficult to measure in kilograms. Hence, atomic mass unit (amu) is used as a convenient alternative! ✅
📌 Example:
🔹 Hydrogen (H) has an atomic mass of 1u
🔹 Oxygen (O) has an atomic mass of 16u
Final Answer 🏆
✅ The atomic mass unit (amu) is the unit used to measure atomic and molecular masses. It is defined as (1/12)th of the mass of a carbon-12 atom.
Q. 2. Why is it not possible to see an atom with naked eyes?
Answer:-
Why Can’t We See Atoms with the Naked Eye? 🔬👀
Atoms are the basic building blocks of matter, but they are too small to be seen with our eyes. Here’s why! 🚀
Reasons Atoms Are Invisible to the Naked Eye 🧪✨
1️⃣ Extremely Small Size 🏗️
🔹 Atoms have a size in the range of 0.1 to 0.5 nanometers (nm).
🔹 1 nanometer = ( 10^{-9} ) meters, which is a billionth of a meter!
🔹 Our eyes cannot detect objects smaller than 0.1 mm without assistance.
2️⃣ Limitations of Visible Light 🌈⚡
🔹 Human eyes rely on visible light to see objects.
🔹 The wavelength of visible light is around 400–700 nm, much larger than an atom!
🔹 Since atoms are smaller than light waves, they do not reflect light in a way that makes them visible!
3️⃣ Microscopes Needed to See Atoms 🔬📡
✅ To observe atoms, scientists use powerful Electron Microscopes or Scanning Tunneling Microscopes (STM).
✅ These advanced tools help us visualize atomic structures using electron beams instead of visible light!
Final Answer (For Full Marks) 🏆
Atoms cannot be seen with the naked eye because they are extremely small, much smaller than the wavelength of visible light, making them invisible without powerful microscopes!
💡 Fun Fact: Even though we can’t see atoms, everything around us—from air to water to solid objects—is made up of trillions of atoms! ✨🔍
Page – 34
Q. 1. Write down the formulae of
(i) Sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Answer:-
Chemical Formulae of Given Compounds ⚛️✨
Here are the chemical formulae for the given compounds, essential for Class 9 CBSE exams! 🚀
🔹 (i) Sodium Oxide 🧪
Formula: Na₂O
🔹 (ii) Aluminium Chloride ⚗️
Formula: AlCl₃
🔹 (iii) Sodium Sulphide 🧫
Formula: Na₂S
🔹 (iv) Magnesium Hydroxide 💧
Formula: Mg(OH)₂
💡 Key Concept: Each formula represents the fixed composition of elements in a compound, following valency rules for correct chemical bonding! ✅✨
Q. 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:-
Names of Compounds from Given Formulae ⚛️✨
Here are the correct names for the given chemical formulae, essential for Class 9 CBSE exams! 🚀
🔹 (i) Al₂(SO₄)₃ 🧪
Name: Aluminium Sulphate
🔹 (ii) CaCl₂ ⚗️
Name: Calcium Chloride
🔹 (iii) K₂SO₄ 🧫
Name: Potassium Sulphate
🔹 (iv) KNO₃ 💧
Name: Potassium Nitrate
🔹 (v) CaCO₃ 🏗️
Name: Calcium Carbonate
💡 Key Concept:
Each formula represents a specific compound made from fixed elements in a definite proportion! ✅✨
Q. 3. What is meant by the term chemical formula?
Answer:-
Definition of Chemical Formula ⚛️✨
A chemical formula is a symbolic representation of a compound, showing the types and numbers of atoms present in a molecule. 🚀
Key Features of a Chemical Formula 🧪
1️⃣ Uses Element Symbols 🔡
🔹 Each element in the compound is represented by its chemical symbol (e.g., H for hydrogen, O for oxygen).
2️⃣ Shows Atomic Ratio ⚖️
🔹 It indicates the exact number of atoms of each element in the compound using subscripts (small numbers).
🔹 Example: H₂O → Water contains 2 hydrogen atoms and 1 oxygen atom. 💧
3️⃣ Represents a Fixed Composition 🔬
🔹 A chemical formula ensures that a compound always has the same elements in the same proportion.
Examples of Chemical Formulae 📚
🔹 Water: H₂O
🔹 Carbon Dioxide: CO₂
🔹 Sulphuric Acid: H₂SO₄
🔹 Sodium Chloride (Salt): NaCl
💡 Fun Fact: Chemical formulae help scientists identify compounds without writing their full names! 🌍🔍
Final Answer (For Full Marks) 🏆
✅ A chemical formula represents a compound using chemical symbols and subscripts, showing the types and numbers of atoms present in a molecule.
Q. 4. How many atoms are present in a
(I) H2s molecule and
(ii) PO43- ion?
Answer:-
Counting Atoms in Given Molecules & Ions ⚛️✨
Let’s determine the number of atoms present in each! 🚀
(i) H₂S Molecule (Hydrogen Sulphide) 🧪
🔹 Formula: H₂S
🔹 Atoms Present:
- 2 Hydrogen (H) atoms
- 1 Sulphur (S) atom
✅ Total atoms in H₂S = 3 atoms
(ii) PO₄³⁻ Ion (Phosphate Ion) ⚗️
🔹 Formula: PO₄³⁻
🔹 Atoms Present:
- 1 Phosphorus (P) atom
- 4 Oxygen (O) atoms
✅ Total atoms in PO₄³⁻ = 5 atoms
💡 Key Concept:
Every molecule or ion is made up of specific atoms, and counting them helps understand their composition! 🔬✨
Page 35
Q. 1. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
Answer:-
Molecular Mass Calculation ⚛️✨
The molecular mass of a compound is calculated by adding up the atomic masses of all the atoms present in its formula. 🚀
Atomic Masses:
🔹 H (Hydrogen) = 1u
🔹 O (Oxygen) = 16u
🔹 Cl (Chlorine) = 35.5u
🔹 C (Carbon) = 12u
🔹 N (Nitrogen) = 14u
Now, let’s calculate the molecular masses for each! ✅
Molecular Masses of Given Compounds 🧪
🔹 (i) H₂ (Hydrogen Gas) 💨
[ = (2 x 1) = 2u ]
🔹 (ii) O₂ (Oxygen Gas) 🌿
[ = (2 x 16) = 32u ]
🔹 (iii) Cl₂ (Chlorine Gas) 🟢
[ = (2 x 35.5) = 71u ]
🔹 (iv) CO₂ (Carbon Dioxide) 🌍
[ = (1 x 12) + (2 x 16) = 12 + 32 = 44u ]
🔹 (v) CH₄ (Methane) 🔥
[ = (1 x 12) + (4 x 1) = 12 + 4 = 16u ]
🔹 (vi) C₂H₆ (Ethane) 🏗️
[ = (2 x 12) + (6 x 1) = 24 + 6 = 30u ]
🔹 (vii) C₂H₄ (Ethene) 🧪
[ = (2 x 12) + (4 x 1) = 24 + 4 = 28u ]
🔹 (viii) NH₃ (Ammonia) ⚗️
[ = (1 x 14) + (3 x 1) = 14 + 3 = 17u ]
🔹 (ix) CH₃OH (Methanol) 🍶
[ = (1 x 12) + (4 x 1) + (1 x 16) = 12 + 4 + 16 = 32u ]
Final Answer (For Full Marks) 🏆📖
✅ The molecular masses of the given compounds are:
| Compound | Molecular Mass (u) |
|---|---|
| H₂ | 2u |
| O₂ | 32u |
| Cl₂ | 71u |
| CO₂ | 44u |
| CH₄ | 16u |
| C₂H₆ | 30u |
| C₂H₄ | 28u |
| NH₃ | 17u |
| CH₃OH | 32u |
💡 Key Concept: Molecular masses help determine chemical properties and reaction calculations, making them important in chemistry! 🔬✨
Q. 2. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Answer:-
Formula Unit Mass Calculation ⚛️✨
The formula unit mass of an ionic compound is calculated by summing the atomic masses of all atoms present in its formula. 🚀
Given atomic masses:
🔹 Zn (Zinc) = 65u
🔹 Na (Sodium) = 23u
🔹 K (Potassium) = 39u
🔹 C (Carbon) = 12u
🔹 O (Oxygen) = 16u
Now, let’s calculate the formula unit masses! ✅
(i) ZnO (Zinc Oxide) 🧪
[ Formula Unit Mass] = [Mass of Zn + Mass of O] = [ 65u + 16u ] = [81u ]
(ii) Na₂O (Sodium Oxide) ⚗️
[Formula Unit Mass] = [ (2 x Mass of Na) + Mass of O ] = [ (2 x 23u) + 16u ] = [ 46u + 16u ] = [ 62u ]
(iii) K₂CO₃ (Potassium Carbonate) 🏗️
[ Formula Unit Mass ] = [ (2 x Mass of K ) + (Mass of C) + (3 x Mass of O) ] = [2 x 39u) + 12u + (3 x 16u) ] = [ 78u + 12u + 48u ] = [ 138u ]
Final Answer (For Full Marks) 🏆📖
✅ Formula Unit Masses:
| Compound | Formula Unit Mass (u) |
|---|---|
| ZnO | 81u |
| Na₂O | 62u |
| K₂CO₃ | 138u |
💡 Key Concept: Formula unit mass helps determine properties and reactions of ionic compounds, making it important in chemistry! 🔬✨
Back Exercise Question & Answers
Q. 1. A 0.24g sample of 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 wright.
Answer:-
Percentage Composition Calculation ⚛️✨
The percentage composition of a compound is the percentage of each element in the total mass of the compound. 🚀
Given Data:
🔹 Total mass of compound = 0.24g
🔹 Mass of Boron (B) = 0.096g
🔹 Mass of Oxygen (O) = 0.144g
Formula for Percentage Composition 📖
[ Percentage of an element} = [ (Mass of the element) x (Total mass of compound) x100 ]
Step 1: Calculate Percentage of Boron (B) 🧪
[ Percentage of Boron ] = [ 0.096/0.24 x100 ]
[ = 0.4 x 100 ]
[ = 40% ]
✅ Boron = 40%
Step 2: Calculate Percentage of Oxygen (O) ⚗️
[ Percentage of Oxygen ] = [ 0.144/0.24 x 100 ]
[ = 0.6 x 100 ]
[ = 60% ]
✅ Oxygen = 60%
Final Answer (For Full Marks) 🏆📖
The percentage composition of the given compound is:
🔹 Boron (B) = 40%
🔹 Oxygen (O) = 60%
💡 Key Concept: Percentage composition helps determine the relative abundance of elements in a compound, making it useful in chemical analysis! 🔬✨
Q. 2. When 3.0g of carbon is burnt in 8.00 g oxygen, 11.00g 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?
Answer:-
Application of the Law of Constant Proportions ⚛️🔥
The given reaction follows the Law of Constant Proportions, which states that elements always combine in a fixed ratio by mass to form a compound. Let’s analyze the given data! 🚀
Step 1: Understanding the Given Reaction 🔬
Reaction:
[ Carbon + Oxygen ➡️ Carbon Dioxide ]
✅ Given Data for First Case:
🔹 Mass of Carbon (C) = 3.00g
🔹 Mass of Oxygen (O₂) = 8.00g
🔹 Mass of Carbon Dioxide (CO₂) formed = 11.00g
💡 This shows that carbon and oxygen combine in a fixed mass ratio of 3:8 to form CO₂!
Step 2: Mass of CO₂ Formed in Second Case 🧪
✅ Given Data for Second Case:
🔹 Mass of Carbon (C) = 3.00g
🔹 Mass of Oxygen (O₂) = 50.00g
Key Observation:
🔹 Only 8g of oxygen is required to react with 3g of carbon (following the fixed ratio).
🔹 The extra 42g of oxygen will remain unused! ❌
✅ Mass of CO₂ formed will remain the same: 11.00g
Step 3: Governing Law of Chemical Combination 📚
🔹 The answer is based on the Law of Constant Proportions, which states that elements combine in a fixed ratio, regardless of the total quantity available.
🔹 Since carbon and oxygen always react in a 3:8 ratio, adding more oxygen does not increase the amount of CO₂ produced! ⚖️✅
Final Answer (For Full Marks) 🏆📖
✅ When 3.00g of carbon is burnt in 50.00g of oxygen, the mass of carbon dioxide formed remains 11.00g, as per the Law of Constant Proportions.
💡 Key Concept: No matter how much excess oxygen is available, the reaction follows a fixed atomic ratio, ensuring constant composition! 🔬✨
Q. 3. What are polyatomic ions? Give examples.
Answer:-
Polyatomic Ions: Definition & Examples ⚛️✨
A polyatomic ion is a group of two or more covalently bonded atoms that carry a net charge (positive or negative) due to the gain or loss of electrons. These ions behave as a single unit in chemical reactions. 🚀
Key Features of Polyatomic Ions 🔬
🔹 Contain multiple atoms but function as one charged entity
🔹 Can be positively or negatively charged
🔹 Participate in chemical bonding to form compounds
Examples of Common Polyatomic Ions 🧪
| Polyatomic Ion | Formula | Charge |
|---|---|---|
| Hydroxide | OH⁻ | -1 |
| Ammonium | NH₄⁺ | +1 |
| Carbonate | CO₃²⁻ | -2 |
| Sulfate | SO₄²⁻ | -2 |
| Nitrate | NO₃⁻ | -1 |
| Phosphate | PO₄³⁻ | -3 |
| Bicarbonate | HCO₃⁻ | -1 |
💡 Key Concept: Polyatomic ions are widely used in chemical reactions, acid-base chemistry, and biological processes! 🔬✨
Q. 4. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer:-
Chemical Formulae of Given Compounds ⚛️✨
Here are the chemical formulae for the given compounds, essential for Class 9 CBSE exams! 🚀
🔹 (a) Magnesium Chloride 🧪
Formula: MgCl₂
🔹 (b) Calcium Oxide ⚗️
Formula: CaO
🔹 (c) Copper Nitrate 🏗️
Formula: Cu(NO₃)₂
🔹 (d) Aluminium Chloride 🧫
Formula: AlCl₃
🔹 (e) Calcium Carbonate 💎
Formula: CaCO₃
💡 Key Concept: Each formula follows the valency rules of elements to form stable compounds! ✅✨
Q. 5. Give the names of the elements present in the following compounds.
(a) Quick line
(b) Hydrogen bromide
(c) Banking powder
(d) Potassium sulphate
Answer:-
Elements Present in Given Compounds ⚛️✨
Here are the elements present in each compound, essential for Class 9 CBSE exams! 🚀
🔹 (a) Quicklime (CaO) 🏗️
✅ Elements: Calcium (Ca) & Oxygen (O)
🔹 (b) Hydrogen Bromide (HBr) 🧪
✅ Elements: Hydrogen (H) & Bromine (Br)
🔹 (c) Baking Powder (NaHCO₃) 🍞
✅ Elements: Sodium (Na), Hydrogen (H), Carbon (C) & Oxygen (O)
🔹 (d) Potassium Sulphate (K₂SO₄) ⚗️
✅ Elements: Potassium (K), Sulphur (S) & Oxygen (O)
💡 Key Concept: Each compound consists of specific elements in a fixed ratio, forming distinct chemical substances! ✅✨
Q. 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:-
Molar Mass Calculation ⚛️✨
The molar mass of a substance is the mass of one mole of that substance, calculated by summing the atomic masses of all atoms in its formula. 🚀
Given Atomic Masses:
🔹 C (Carbon) = 12u
🔹 H (Hydrogen) = 1u
🔹 S (Sulphur) = 32u
🔹 P (Phosphorus) = 31u
🔹 Cl (Chlorine) = 35.5u
🔹 N (Nitrogen) = 14u
🔹 O (Oxygen) = 16u
Now, let’s calculate the molar masses for each! ✅
(a) Ethyne (C₂H₂) 🔥
[ Molar Mass ] = (2 x 12u) + (2 x 1u) ] = [ 24u + 2u = 26u ]
✅ Molar mass of C₂H₂ = 26u
(b) Sulphur Molecule (S₈) 🌿
[ Molar Mass} = (8 x 32u) ] = [ 256u ]
✅ Molar mass of S₈ = 256u
(c) Phosphorus Molecule (P₄) ⚗️
[ Molar Mass} = (4 x 31u) ] = [ 124u ]
✅ Molar mass of P₄ = 124u
(d) Hydrochloric Acid (HCl) 🏗️
[ Molar Mass ] = (1 x 1u) + (1 x 35.5u) ] = [ 1u + 35.5u = 36.5u ]
✅ Molar mass of HCl = 36.5u
(e) Nitric Acid (HNO₃) 🧪
[ Molar Mass ] = (1 x 1u) + (1 x 14u) + (3 x 16u) ] = [ 1u + 14u + 48u = 63u ]
✅ Molar mass of HNO₃ = 63u
Final Answer (For Full Marks) 🏆📖
| Substance | Molar Mass (u) |
|---|---|
| C₂H₂ (Ethyne) | 26u |
| S₈ (Sulphur molecule) | 256u |
| P₄ (Phosphorus molecule) | 124u |
| HCl (Hydrochloric Acid) | 36.5u |
| HNO₃ (Nitric Acid) | 63u |
💡 Key Concept: Molar mass helps in stoichiometric calculations for chemical reactions, making it crucial in chemistry! 🔬✨