Wednesday, May 29, 2024

Atoms and Molecules

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NCERT Solutions for Class 9 Science Chapter 3

  • The way atoms are linked and arranged in a molecule is like a unique recipe. This arrangement determines the molecule’s properties, such as how it behaves and interacts with other substances.

Laws of Chemical Combination: The Rules of the Game

  • When elements participate in a chemical reaction to form a compound, there are some interesting rules they follow:
    • Law of Constant Proportions: This law states that a specific compound always contains the same elements in a fixed ratio by mass. No matter how much or how little of the compound you have, the ratio of elements stays constant. Imagine a recipe that always requires two cups of flour and one cup of sugar, regardless of how many batches you make! This law is explained by the fact that atoms have a definite mass, and they combine in specific whole number ratios during a reaction.The chapter “Atoms and Molecules” in your science book covers the fundamental building blocks of matter: atoms and molecules. Here’s a breakdown of the key concepts:

Atoms:

  • The smallest indivisible particles of an element that can take part in a chemical reaction.
  • Each element has a unique identity defined by its atoms.
  • Atoms have a definite mass, expressed in Atomic Mass Units (amu).

Molecules:

  • Two or more atoms chemically bonded together.
  • Molecules can be formed by atoms of the same element (e.g., oxygen gas, O₂) or different elements (e.g., water, H₂O).
  • The way atoms are arranged in a molecule determines its properties.

Laws of Chemical Combination:

  • Law of Constant Proportions: A compound always contains the same elements in a fixed ratio by mass.
  • This law is explained by the concept of atoms having a definite mass.

Additional Concepts:

  • The chapter also introduces you to ions (charged atoms) and writing chemical formulae to represent elements and compounds.
  • You’ll learn about the concept of mole, a unit for measuring large numbers of atoms or molecules.

By understanding atoms and molecules, you’ll gain a foundation for comprehending how substances interact and form new compounds in future chemistry classes.

Deep Dive into Atoms and Molecules 

This chapter dives deep into the fascinating world of atoms and molecules, the very building blocks of everything around you! Here’s a detailed breakdown:

The Atom: The Tiniest Unit

  • Imagine the tiniest particle possible that can still retain the identity of an element. That’s an atom! It’s the fundamental unit of an element involved in chemical reactions.
  • Each element has its unique fingerprint defined by its atoms. For example, a gold atom is fundamentally different from a hydrogen atom.
  • Atoms are incredibly small. To get a sense of scale, millions of them could fit on the period at the end of this sentence!
  • Interestingly, though tiny, atoms have a measurable mass. This mass is expressed in Atomic Mass Units (amu), a special unit used specifically for atoms.

Molecules: When Atoms Team Up

  • Not all atoms enjoy being loners. Often, two or more atoms join forces through chemical bonds to form molecules.
  • Molecules can be like exclusive clubs:
    • Sometimes, only atoms of the same element are allowed in, forming molecules like oxygen gas (O₂) with two oxygen atoms.
    • Other times, atoms from different elements join the party, like in water (H₂O) where two hydrogen atoms bond with one oxygen atom.

Beyond the Basics

  • The chapter introduces you to ions, which are atoms that have gained or lost electrons, giving them an electrical charge. Ions play a crucial role in many chemical reactions.
  • Chemical formulae are introduced as a shorthand way to represent elements and compounds. Just like a musical note tells a musician which key to play, a chemical formula tells you exactly what kind and how many atoms are present in a substance.
  • The concept of a mole is introduced, a unit used to count vast numbers of atoms or molecules. It’s like having a dozen for eggs, but for atoms and molecules!

Understanding these fundamental concepts about atoms and molecules is like having a backstage pass to the world of chemistry. It paves the way for future chapters where you’ll explore how elements interact, form new compounds, and create the amazing world of materials around us!

NCERT Solutions for Class 9 Science Chapter 3

Q1.  If one mole of carbon atoms weigh 12 grams, what is the mass (in grams) of 1 atom of carbon?

Ans: To find the mass of a single carbon atom, divide the mass of 1 mole of carbon (12 grams) by Avogadro’s number (really big number of atoms in a mole). This gives you roughly 1.99 x 10^-23 grams for a single carbon atom.

Q2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given atomic mass of Na = 23 u, Fe = 56 u)?

Ans: Absolutely! Here’s how to solve it using formulas:

Define the variables:

  • mass_element (grams): Mass of the element (either sodium or iron, both 100 grams in this case)
  • atomic_mass (u): Atomic mass of the element (23 u for Na and 56 u for Fe)
  • Avogadro_constant (atoms/mole): Avogadro’s number (around 6.022 x 10^23 atoms/mole)
  • num_atoms (atoms): Number of atoms to be calculated

Formulas:

Moles (mol):

moles = mass_element / atomic_mass

  1. Number of Atoms (atoms):

    num_atoms = moles * Avogadro_constant

Q3. 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: The compound is 40% boron and 60% oxygen by weight. We can find this by calculating the percentage of one element (boron in this case) and subtracting it from 100%.

Q4. 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: The Law of Constant Proportions dictates a fixed ratio of elements in a compound. In this case, the ratio of CO₂ produced to C reacted remains constant. Even though we have more oxygen, only 3.00 g of carbon will react, forming roughly 11.01 g of CO₂ based on the constant ratio from the first reaction.

Q5. What are polyatomic ions? Give examples.

Ans: Polyatomic ions are groups of covalently bonded atoms acting like a single charged unit. Examples include sulfate (SO₄²⁻) and hydroxide (OH⁻).

Q6.  Write the chemical formulae of the following: 

(a) Magnesium chloride 

(b) Calcium oxide 

(c) Copper nitrate 

(d) Aluminium chloride 

(e) Calcium carbonate.

Ans: The chemical formulas for the given compounds are:

(a) Magnesium chloride – MgCl₂ 

(b) Calcium oxide – CaO 

(c) Copper nitrate – Cu(NO₃)₂ (copper has a +2 charge, so it needs 2 nitrates to balance the charge) 

(d) Aluminium chloride – AlCl₃ (aluminium has a +3 charge, so it needs 3 chlorides to balance the charge) 

(e) Calcium carbonate – CaCO₃

Q7.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 elements present in each compound are:

(a) Quick lime: Calcium (Ca) and Oxygen (O) (Chemical formula: CaO) 

(b) Hydrogen bromide: Hydrogen (H) and Bromine (Br) (Chemical formula: HBr) (c) Baking powder: Primarily Sodium (Na), Hydrogen (H), Carbon 

(C), and Oxygen (O) (Baking powder is a mixture, but the main ingredient is sodium bicarbonate with the formula NaHCO₃) 

(d) Potassium sulphate: Potassium (K), Sulphur (S), and Oxygen (O) (Chemical formula: K₂SO₄)

Q8. Give the names of the elements present in the following compounds: 

(a) Quick lime 

(b) Hydrogen bromide 

(c) Baking powder 

(d) Potassium sulphate.

Ans: Absolutely! Here’s the molar mass calculation for each substance:

Molar mass is the sum of the masses of all the individual atoms present in a molecule or compound.

(a) Ethyne, C₂H₂:

  • Molar mass of C = 12 g/mol (from the periodic table)
  • Molar mass of H = 1 g/mol (from the periodic table)
  • Total molar mass = (2 x 12 g/mol C) + (2 x 1 g/mol H) = 26 g/mol

(b) Sulphur molecule, S₈:

  • Molar mass of S = 32 g/mol (from the periodic table)
  • Total molar mass = 8 x 32 g/mol S = 256 g/mol

(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31):

  • Molar mass of P = 31 g/mol (given)
  • Total molar mass = 4 x 31 g/mol P = 124 g/mol

(d) Hydrochloric acid, HCl:

  • Molar mass of H = 1 g/mol (from the periodic table)
  • Molar mass of Cl = 35.5 g/mol (from the periodic table)
  • Total molar mass = 1 g/mol H + 35.5 g/mol Cl = 36.5 g/mol

(e) Nitric acid, HNO₃:

  • Molar mass of H = 1 g/mol (from the periodic table)
  • Molar mass of N = 14 g/mol (from the periodic table)
  • Molar mass of O = 16 g/mol (from the periodic table)
  • Total molar mass = 1 g/mol H + 14 g/mol N + (3 x 16 g/mol O) = 63 g/mol

Q9.What is the mass of (a) 1 mole of nitrogen atoms? (b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)? (c) 10 moles of sodium sulphite (Na2S03)?

Ans:Here’s the mass for each scenario:

(a) 1 mole of nitrogen atoms:

  • Molar mass of Nitrogen (N) = 14 g/mol (from the periodic table)
  • Mass = 1 mole * 14 g/mol = 14 grams

(b) 4 moles of aluminium atoms:

  • Molar mass of Aluminium (Al) = 27 g/mol (given)
  • Mass = 4 moles * 27 g/mol = 108 grams

(c) 10 moles of sodium sulphite (Na₂SO₃):

  1. We need to find the total molar mass of sodium sulphite (Na₂SO₃) first:
    • Molar mass of Sodium (Na) = 23 g/mol (from the periodic table)
    • Molar mass of Sulphur (S) = 32 g/mol (from the periodic table)
    • Molar mass of Oxygen (O) = 16 g/mol (from the periodic table)
  2. Calculate the total mass:
    • Mass of Na₂SO₃ = (2 x 23 g/mol Na) + (1 x 32 g/mol S) + (3 x 16 g/mol O) = 142 g/mol
  3. Now find the total mass for 10 moles:
    • Mass = 10 moles * 142 g/mol = 1420 grams

Q10. Convert into mole. 

(a) 12 g of oxygen gas 

(b) 20 g of water 

(c) 22 g of Carbon dioxide.

Ans: Here’s the conversion of each mass to moles:

Formula: moles = mass (grams) / molar mass (g/mol)

(a) 12 g of oxygen gas:

  • Molar mass of Oxygen (O₂) = 16 g/mol (since O₂ is a molecule, its molar mass is 2 x atomic mass of Oxygen)
  • Moles of oxygen gas = 12 g / 16 g/mol = 0.75 moles

(b) 20 g of water:

  • Molar mass of Water (H₂O) = 18 g/mol (2 x atomic mass of Hydrogen + 1 x atomic mass of Oxygen)
  • Moles of water = 20 g / 18 g/mol = 1.11 moles

(c) 22 g of Carbon dioxide:

  • Molar mass of Carbon dioxide (CO₂) = 44 g/mol (1 x atomic mass of Carbon + 2 x atomic mass of Oxygen)
  • Moles of carbon dioxide = 22 g / 44 g/mol = 0.5 moles

Q11.  What is the mass of: 

(a) 0.2 mole of oxygen atoms? 

(b) 0.5 mole of water molecules?

Ans: Here’s the mass calculation for the given amounts of atoms and molecules:

(a) 0.2 mole of oxygen atoms:

  • Molar mass of Oxygen (O) = 16 g/mol (atomic mass of oxygen)

Concept to Remember:

  • When dealing with individual atoms, we use the atomic mass of the element.
  • Mass of oxygen atoms = 0.2 mole * 16 g/mol = 3.2 grams

(b) 0.5 mole of water molecules:

  • Molar mass of Water (H₂O) = 18 g/mol (2 x atomic mass of Hydrogen + 1 x atomic mass of Oxygen)

Concept to Remember:

  • When dealing with molecules, we use the molar mass of the molecule.
  • Mass of water molecules = 0.5 mole * 18 g/mol = 9.0 grams

Q12. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Ans: Absolutely! Here’s how to calculate the number of molecules of sulphur (S₈) present in 16 g of solid sulphur:

  1. Moles of Sulphur:
    • First, we need to find the number of moles of sulphur present in 16 grams.
    • Molar mass of Sulphur (S₈) = 256 g/mol (since S₈ is a molecule, its molar mass is 8 x atomic mass of Sulphur)
    • Moles of Sulphur = Mass / Molar mass = 16 g / 256 g/mol = 0.0625 moles
  2. Avogadro’s Number:
    • We know that 1 mole of any substance contains a specific number of particles, called Avogadro’s number (represented by Avogadro’s constant, approximately 6.022 x 10^23 atoms/mole).
  3. Number of Molecules:
    • Now, we can find the number of S₈ molecules by multiplying the number of moles by Avogadro’s number:
    • Number of S₈ molecules = Moles of Sulphur * Avogadro’s number
    • Number of S₈ molecules = 0.0625 moles * 6.022 x 10^23 molecules/mole ≈ 3.76 x 10^22 molecules

Therefore, there are approximately 3.76 x 10^22 molecules of sulphur (S₈) present in 16 grams of solid sulphur.\

Q13.  Calculate the number of aluminium ions present in 0.051 g 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 = 27 u)

Ans: Absolutely! Here’s a simplified solution to find the number of aluminium ions in aluminium oxide:

1. Key Information:

  • Mass of aluminium oxide (Al₂O₃) = 0.051 g
  • Atomic mass of Aluminium (Al) = 27 u (assumed from the hint)

2. Molar Mass Conversion:

  • We need the molar mass of Al₂O₃ in g/mol (grams per mole).
    • Molar mass of Al₂O₃ = (2 x 27 u Al) + (3 x 16 u O) ≈ 102 u (convert u to g/mol by multiplying by 1 g/mol per u)

3. Assumptions (Implicit):

  • The aluminium oxide completely dissociates in water to form aluminium ions (Al³⁺) and oxide ions (O²⁻).
  • We only care about the aluminium ions.

4. Calculation:

a. Moles of Al₂O₃: – Moles of Al₂O₃ = Mass of Al₂O₃ / Molar mass of Al₂O₃ – Moles of Al₂O₃ = 0.051 g / 102 g/mol ≈ 0.0005 mol

b. Aluminium Ions per Molecule: – In one molecule of Al₂O₃, there are 2 aluminium atoms.

c. Total Aluminium Ions: – We can directly multiply to find the total aluminium ions considering the moles, aluminium atoms per molecule, and Avogadro’s constant (N_A) which inherently accounts for the number of particles in a mole.

5. Avogadro’s Constant (Implicit):

  • We don’t explicitly write out the value of Avogadro’s constant (around 6.022 x 10^23 particles/mol) but assume it’s included in the final calculation.

6. Final Answer:

  • Number of Aluminium ions ≈ 0.0005 mol * 2 Al ions/molecule * N_A ≈ 6.022 x 10^17 ions

Therefore, there are approximately 6.022 x 10^17 aluminium ions present in 0.051 g of aluminium oxide.

NCERT Solutions for Class 9 Science Chapter 3

FAQ’s

What topics are covered in Class 9 Science Chapter 3: “Atoms and Molecules”?

Class 9 Science Chapter 3 explores the fundamental concepts of atoms and molecules, covering topics such as the structure of atoms, chemical bonding, and the composition of matter.

How can NCERT solutions for Class 9 Science Chapter 3 help in understanding atoms and molecules?

NCERT solutions offer detailed explanations and analyses of concepts related to atoms and molecules, helping students understand the structure of atoms, types of chemical bonds, and the formation of molecules with clarity.

Where can I find NCERT solutions for Class 9 Science Chapter 3?

NCERT solutions for Class 9 Science Chapter 3 can be found online or in study materials provided by educational platforms or institutes specializing in academic resources.

When should I refer to NCERT solutions for Class 9 Science Chapter 3?

NCERT solutions are beneficial for reinforcing learning, clarifying doubts, and preparing for exams. They can be used alongside regular study or as a revision tool before assessments to enhance understanding of atoms and molecules.

What specific examples are discussed in Class 9 Science Chapter 3 regarding atoms and molecules?

Class 9 Science Chapter 3 discusses various examples of atoms and molecules, such as the structure of water molecules, chemical bonding in compounds like sodium chloride, and the concept of chemical formulas, offering insights into the fundamental building blocks of matter.

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