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Basic Concepts of Chemistry - Willer Academy

Complete chapter notes in 3 structured lectures with expandable sections and interactive elements

Lecture 1: Matter & Measurement
Lecture 2: Atomic Structure
Lecture 3: Mole Concept
1

Lecture 1: Matter & Measurement

Classification of Matter

Matter is anything that occupies space and has mass. It can be classified based on physical state or chemical composition.

Pure Substances: Elements and compounds with fixed composition and distinct properties.

Mixtures: Combination of two or more substances mixed physically.

Type Characteristics Examples
Element Pure substance, one type of atom, cannot be broken down Iron (Fe), Oxygen (O₂)
Compound Pure substance, two or more elements chemically combined Water (H₂O), Salt (NaCl)
Homogeneous Mixture Uniform composition, single phase Salt water, Air, Brass
Heterogeneous Mixture Non-uniform composition, multiple phases Sand and water, Salad, Granite

Exercise: Classify the following

1. Sugar dissolved in water
2. Iron filings and sulfur powder
3. Carbon dioxide
4. Milk

Properties of Matter

Physical Properties: Observed without changing composition

  • Color, odor, density, melting point, boiling point, solubility
  • Extensive (depends on quantity: mass, volume) vs Intensive (independent of quantity: density, melting point)

Chemical Properties: Observed during chemical reactions

  • Flammability, reactivity, acidity/basicity, toxicity

States of Matter:

  • Solid: Fixed shape and volume, particles closely packed
  • Liquid: Fixed volume, takes container shape, particles can flow
  • Gas: No fixed shape or volume, particles far apart
  • Plasma: Ionized gas (not covered in basic chemistry)
Density = Mass / Volume

Measurement & Units

The International System of Units (SI) is used for scientific measurements:

Physical Quantity SI Unit Symbol
Mass Kilogram kg
Length Meter m
Time Second s
Temperature Kelvin K
Amount of substance Mole mol

Significant Figures: Meaningful digits in a measured quantity that indicate precision.

Rules:

  • All non-zero digits are significant (e.g., 123 has 3 SF)
  • Zeros between non-zero digits are significant (e.g., 1002 has 4 SF)
  • Leading zeros are not significant (e.g., 0.0023 has 2 SF)
  • Trailing zeros after decimal are significant (e.g., 2.300 has 4 SF)
  • Exact numbers have infinite significant figures (e.g., counting numbers)

Exercise: Determine significant figures

1. 0.00450
2. 1200
3. 2.00 × 10³
4. 100.00

2

Lecture 2: Atomic Structure

Laws of Chemical Combination

Four fundamental laws govern chemical reactions:

1. Law of Conservation of Mass (Lavoisier):

"Mass is neither created nor destroyed in a chemical reaction."

Mass of reactants = Mass of products

2. Law of Definite Proportions (Proust):

"A chemical compound always contains the same elements in the same proportion by mass."

Example: Pure water always has H:O mass ratio of 1:8

3. Law of Multiple Proportions (Dalton):

"When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers."

Example: Carbon forms CO and CO₂ with oxygen. Ratio of oxygen combining with fixed carbon is 1:2

4. Gay Lussac's Law of Gaseous Volumes:

"When gases combine, they do so in volumes that bear a simple ratio to one another and to the product (if gaseous), under the same conditions of temperature and pressure."

Example: 2H₂(g) + O₂(g) → 2H₂O(g) shows volume ratio 2:1:2

Dalton's Atomic Theory

John Dalton proposed the atomic theory in 1808:

  • Matter consists of indivisible atoms
  • All atoms of a given element are identical in mass and properties
  • Atoms of different elements have different masses and properties
  • Compounds form when atoms combine in fixed ratios
  • Atoms are neither created nor destroyed in chemical reactions

Modifications to Dalton's Theory:

  • Atoms are divisible (discovery of subatomic particles: electrons, protons, neutrons)
  • Isotopes: Atoms of same element with different mass numbers
  • Atoms can be transformed in nuclear reactions
  • All atoms of an element are not identical (isotopes)

Exercise: Limitations of Dalton's Theory

Explain how the discovery of isotopes challenges Dalton's theory that all atoms of an element are identical.

Atomic & Molecular Mass

Atomic Mass: Mass of an atom relative to carbon-12 isotope

Atomic mass = (Mass of one atom of the element) / (1/12th mass of one C-12 atom)

Molecular Mass: Sum of atomic masses of all atoms in a molecule

Example: Molecular mass of H₂O = (2 × 1.008) + (1 × 16.00) = 18.016 u

Formula Mass: For ionic compounds that don't exist as discrete molecules

Example: Formula mass of NaCl = Atomic mass of Na + Atomic mass of Cl = 23.0 + 35.5 = 58.5 u

Element Symbol Atomic Mass (u)
Hydrogen H 1.008
Carbon C 12.01
Oxygen O 16.00
Sodium Na 23.0
Chlorine Cl 35.5
3

Lecture 3: Mole Concept

Mole & Avogadro's Number

The mole is the SI unit for amount of substance that contains as many elementary entities as there are atoms in exactly 12g of carbon-12.

Avogadro's Number (NтВР):

1 mole = 6.022 × 10²³ particles (atoms, molecules, ions)

Molar Mass: Mass of one mole of a substance in grams (numerically equal to atomic/molecular mass in u)

Example: Molar mass of H₂O = 18.016 g/mol

Concept Formula Example
Number of moles (n) n = Mass / Molar mass Moles in 36g H₂O: 36/18 = 2 moles
Number of particles N = n × NтВР Molecules in 2 moles H₂O: 2 × 6.022×10²³
Mass from moles Mass = n × Molar mass Mass of 0.5 mole O₂: 0.5 × 32 = 16g

Exercise: Mole Calculations

1. How many moles are in 25g of calcium carbonate (CaCO₃)?
2. Calculate the number of oxygen atoms in 1g of oxygen gas (O₂).

Stoichiometry

Calculation of reactants and products in chemical reactions based on balanced equations.

Steps for Stoichiometric Calculations:

  1. Write balanced chemical equation
  2. Convert given quantities to moles
  3. Use mole ratios from balanced equation
  4. Convert moles to required units

Example: How many grams of O₂ are needed to burn 36g of carbon?

C + O₂ → CO₂

Mole ratio: 1 mol C : 1 mol O₂

Moles of C = 36g / 12g/mol = 3 moles

Moles of O₂ needed = 3 moles

Mass of O₂ = 3 × 32 = 96g

Limiting Reagent: The reactant that is completely consumed and determines the amount of product formed.

Excess Reagent: The reactant present in greater quantity than needed.

Concentration Terms

Ways to express concentration of solutions:

Term Formula Unit
Mass Percentage (Mass of solute / Mass of solution) × 100 %
Mole Fraction (x) n₁ / (n₁ + n₂ + ...) Dimensionless
Molarity (M) Moles of solute / Liters of solution mol/L (M)
Molality (m) Moles of solute / kg of solvent mol/kg

Important Relationship:

Molarity (M) and density (d):

M = (Mass percentage × d × 10) / Molar mass of solute

Exercise: Concentration Calculations

1. Calculate the molarity of a solution containing 5g of NaOH in 250mL solution.
2. What is the molality of a solution with 20g glucose in 500g water?

Willer Academy Chemistry Notes | CBSE English Medium

Complete Chapter Notes in 3 Lectures | © 2023 Willer Academy

Chemistry Notes - CBSE English & Hindi - Willer Academy

Complete chapter notes in 3 structured lectures with bilingual content (English and Hindi)

Show Hindi Content
Lecture 1: Matter & Measurement
Lecture 2: Atomic Structure
Lecture 3: Mole Concept
1

Lecture 1: Matter & Measurement

Classification of Matter

Matter is anything that occupies space and has mass. It can be classified based on physical state or chemical composition.

Pure Substances: Elements and compounds with fixed composition and distinct properties.

Mixtures: Combination of two or more substances mixed physically.

Type Characteristics Examples
Element Pure substance, one type of atom, cannot be broken down Iron (Fe), Oxygen (O₂)
Compound Pure substance, two or more elements chemically combined Water (H₂O), Salt (NaCl)
Homogeneous Mixture Uniform composition, single phase Salt water, Air, Brass
Heterogeneous Mixture Non-uniform composition, multiple phases Sand and water, Salad, Granite

рдкрджाрд░्рде рдХा рд╡рд░्рдЧीрдХрд░рдг

рдкрджाрд░्рде рдХुрдЫ рднी рд╣ै рдЬो рд╕्рдеाрди рдШेрд░рддा рд╣ै рдФрд░ рдЬिрд╕рдоें рдж्рд░рд╡्рдпрдоाрди рд╣ोрддा рд╣ै। рдЗрд╕े рднौрддिрдХ рдЕрд╡рд╕्рдеा рдпा рд░ाрд╕ाрдпрдиिрдХ рд╕ंрд░рдЪрдиा рдХे рдЖрдзाрд░ рдкрд░ рд╡рд░्рдЧीрдХृрдд рдХिрдпा рдЬा рд╕рдХрддा рд╣ै।

  • рд╢ुрдж्рдз рдкрджाрд░्рде: рддрдд्рд╡ рдФрд░ рдпौрдЧिрдХ рдЬिрдирдХी рдиिрд╢्рдЪिрдд рд╕ंрд░рдЪрдиा рдФрд░ рд╡िрд╢िрд╖्рдЯ рдЧुрдг рд╣ोрддे рд╣ैं
  • рдоिрд╢्рд░рдг: рджो рдпा рджो рд╕े рдЕрдзिрдХ рдкрджाрд░्рдеों рдХा рднौрддिрдХ рд░ूрдк рд╕े рдоिрд╢्рд░рдг

рдЙрджाрд╣рд░рдг:

  • рддрдд्рд╡: рд▓ोрд╣ा (Fe), рдСрдХ्рд╕ीрдЬрди (O₂)
  • рдпौрдЧिрдХ: рдкाрдиी (H₂O), рдирдордХ (NaCl)
  • рд╕рдоांрдЧी рдоिрд╢्рд░рдг: рдирдордХीрди рдкाрдиी, рд╡ाрдпु, рдкीрддрд▓
  • рд╡िрд╖рдоांрдЧी рдоिрд╢्рд░рдг: рд░ेрдд рдФрд░ рдкाрдиी, рд╕рд▓ाрдж, рдЧ्рд░ेрдиाрдЗрдЯ

Exercise: Classify the following

1. Sugar dissolved in water
2. Iron filings and sulfur powder
3. Carbon dioxide
4. Milk

Properties of Matter

Physical Properties: Observed without changing composition

  • Color, odor, density, melting point, boiling point, solubility
  • Extensive (depends on quantity: mass, volume) vs Intensive (independent of quantity: density, melting point)

Chemical Properties: Observed during chemical reactions

  • Flammability, reactivity, acidity/basicity, toxicity

States of Matter:

  • Solid: Fixed shape and volume, particles closely packed
  • Liquid: Fixed volume, takes container shape, particles can flow
  • Gas: No fixed shape or volume, particles far apart
  • Plasma: Ionized gas (not covered in basic chemistry)
Density = Mass / Volume

рдкрджाрд░्рде рдХे рдЧुрдг

рднौрддिрдХ рдЧुрдг: рд╕ंрд░рдЪрдиा рдмрджрд▓े рдмिрдиा рджेрдЦे рдЬा рд╕рдХрдиे рд╡ाрд▓े рдЧुрдг

  • рд░ंрдЧ, рдЧंрдз, рдШрдирдд्рд╡, рдЧрд▓рдиांрдХ, рдХ्рд╡рдердиांрдХ, рдШुрд▓рдирд╢ीрд▓рддा
  • рд╡िрд╕्рддाрд░ी рдЧुрдг (рдоाрдд्рд░ा рдкрд░ рдиिрд░्рднрд░: рдж्рд░рд╡्рдпрдоाрди, рдЖрдпрддрди) vs рдЧрд╣рди рдЧुрдг (рдоाрдд्рд░ा рд╕े рд╕्рд╡рддंрдд्рд░: рдШрдирдд्рд╡, рдЧрд▓рдиांрдХ)

рд░ाрд╕ाрдпрдиिрдХ рдЧुрдг: рд░ाрд╕ाрдпрдиिрдХ рдЕрднिрдХ्рд░िрдпाрдУं рдХे рджौрд░ाрди рджेрдЦे рдЬाрддे рд╣ैं

  • рдЬ्рд╡рд▓рдирд╢ीрд▓рддा, рдЕрднिрдХ्рд░िрдпाрд╢ीрд▓рддा, рдЕрдо्рд▓ीрдпрддा/рдХ्рд╖ाрд░ीрдпрддा, рд╡िрд╖ाрдХ्рддрддा

рдкрджाрд░्рде рдХी рдЕрд╡рд╕्рдеाрдПँ:

  • рдаोрд╕: рдиिрд╢्рдЪिрдд рдЖрдХाрд░ рдФрд░ рдЖрдпрддрди, рдХрдг рд╕рдШрди рд░ूрдк рд╕े рд╡्рдпрд╡рд╕्рдеिрдд
  • рдж्рд░рд╡: рдиिрд╢्рдЪिрдд рдЖрдпрддрди, рдкाрдд्рд░ рдХा рдЖрдХाрд░ рд▓ेрддा рд╣ै, рдХрдг рдк्рд░рд╡ाрд╣िрдд рд╣ो рд╕рдХрддे рд╣ैं
  • рдЧैрд╕: рдиिрд╢्рдЪिрдд рдЖрдХाрд░ рдпा рдЖрдпрддрди рдирд╣ीं, рдХрдг рджूрд░ рд╣ोрддे рд╣ैं
рдШрдирдд्рд╡ = рдж्рд░рд╡्рдпрдоाрди / рдЖрдпрддрди

Measurement & Units

The International System of Units (SI) is used for scientific measurements:

Physical Quantity SI Unit Symbol
Mass Kilogram kg
Length Meter m
Time Second s
Temperature Kelvin K
Amount of substance Mole mol

Significant Figures: Meaningful digits in a measured quantity that indicate precision.

Rules:

  • All non-zero digits are significant (e.g., 123 has 3 SF)
  • Zeros between non-zero digits are significant (e.g., 1002 has 4 SF)
  • Leading zeros are not significant (e.g., 0.0023 has 2 SF)
  • Trailing zeros after decimal are significant (e.g., 2.300 has 4 SF)
  • Exact numbers have infinite significant figures (e.g., counting numbers)

рдоाрдкрди рдПрд╡ं рдоाрдд्рд░рдХ

рд╡ैрдЬ्рдЮाрдиिрдХ рдоाрдкрди рдХे рд▓िрдП рдЕंрддрд░्рд░ाрд╖्рдЯ्рд░ीрдп рдоाрдд्рд░рдХ рдк्рд░рдгाрд▓ी (SI) рдХा рдЙрдкрдпोрдЧ рдХिрдпा рдЬाрддा рд╣ै:

  • рдж्рд░рд╡्рдпрдоाрди: рдХिрд▓ोрдЧ्рд░ाрдо (kg)
  • рд▓ंрдмाрдИ: рдоीрдЯрд░ (m)
  • рд╕рдордп: рд╕ेрдХंрдб (s)
  • рддाрдкрдоाрди: рдХेрд▓्рд╡िрди (K)
  • рдкрджाрд░्рде рдХी рдоाрдд्рд░ा: рдоोрд▓ (mol)

рд╕ाрд░्рдердХ рдЕंрдХ: рдоाрдкी рдЧрдИ рдоाрдд्рд░ा рдоें рд╕ाрд░्рдердХ рдЕंрдХ рдЬो рдкрд░िрд╢ुрдж्рдзрддा рдХो рджрд░्рд╢ाрддे рд╣ैं।

рдиिрдпрдо:

  • рд╕рднी рд╢ूрди्рдпेрддрд░ рдЕंрдХ рд╕ाрд░्рдердХ рд╣ोрддे рд╣ैं (рдЬैрд╕े 123 рдоें 3 рд╕ाрд░्рдердХ рдЕंрдХ)
  • рд╢ूрди्рдпेрддрд░ рдЕंрдХों рдХे рдмीрдЪ рдХे рд╢ूрди्рдп рд╕ाрд░्рдердХ рд╣ोрддे рд╣ैं (рдЬैрд╕े 1002 рдоें 4 рд╕ाрд░्рдердХ рдЕंрдХ)
  • рдЕрдЧ्рд░ рд╢ूрди्рдп рд╕ाрд░्рдердХ рдирд╣ीं рд╣ोрддे (рдЬैрд╕े 0.0023 рдоें 2 рд╕ाрд░्рдердХ рдЕंрдХ)
  • рджрд╢рдорд▓рд╡ рдХे рдмाрдж рдХे рдЕрдиुрдЧाрдоी рд╢ूрди्рдп рд╕ाрд░्рдердХ рд╣ोрддे рд╣ैं (рдЬैрд╕े 2.300 рдоें 4 рд╕ाрд░्рдердХ рдЕंрдХ)
  • рдиिрд╢्рдЪिрдд рд╕ंрдЦ्рдпाрдУं рдоें рдЕрдиंрдд рд╕ाрд░्рдердХ рдЕंрдХ рд╣ोрддे рд╣ैं (рдЬैрд╕े рдЧिрдирддी рдХी рд╕ंрдЦ्рдпाрдПँ)

Exercise: Determine significant figures

1. 0.00450
2. 1200
3. 2.00 × 10³
4. 100.00

2

Lecture 2: Atomic Structure

Laws of Chemical Combination

Four fundamental laws govern chemical reactions:

1. Law of Conservation of Mass (Lavoisier):

"Mass is neither created nor destroyed in a chemical reaction."

Mass of reactants = Mass of products

2. Law of Definite Proportions (Proust):

"A chemical compound always contains the same elements in the same proportion by mass."

Example: Pure water always has H:O mass ratio of 1:8

3. Law of Multiple Proportions (Dalton):

"When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers."

Example: Carbon forms CO and CO₂ with oxygen. Ratio of oxygen combining with fixed carbon is 1:2

рд░ाрд╕ाрдпрдиिрдХ рд╕ंрдпोрдЧ рдХे рдиिрдпрдо

1. рдж्рд░рд╡्рдпрдоाрди рд╕ंрд░рдХ्рд╖рдг рдХा рдиिрдпрдо (рд▓ाрд╡ॉрдЬ़िрдпрд░):

"рд░ाрд╕ाрдпрдиिрдХ рдЕрднिрдХ्рд░िрдпा рдоें рдж्рд░рд╡्рдпрдоाрди рди рддो рдЙрдд्рдкрди्рди рд╣ोрддा рд╣ै рдФрд░ рди рд╣ी рдирд╖्рдЯ рд╣ोрддा рд╣ै।"

рдЕрднिрдХाрд░рдХों рдХा рдж्рд░рд╡्рдпрдоाрди = рдЙрдд्рдкाрджों рдХा рдж्рд░рд╡्рдпрдоाрди

2. рдиिрд╢्рдЪिрдд рдЕрдиुрдкाрдд рдХा рдиिрдпрдо (рдк्рд░ाрдЙрд╕्рдЯ):

"рдПрдХ рд░ाрд╕ाрдпрдиिрдХ рдпौрдЧिрдХ рдоें рддрдд्рд╡ рд╕рджैрд╡ рдж्рд░рд╡्рдпрдоाрди рдХे рдЕрдиुрд╕ाрд░ рд╕рдоाрди рдЕрдиुрдкाрдд рдоें рдЙрдкрд╕्рдеिрдд рд╣ोрддे рд╣ैं।"

рдЙрджाрд╣рд░рдг: рд╢ुрдж्рдз рдЬрд▓ рдоें рд╣ाрдЗрдб्рд░ोрдЬрди : рдСрдХ्рд╕ीрдЬрди рдХा рдж्рд░рд╡्рдпрдоाрди рдЕрдиुрдкाрдд рд╕рджैрд╡ 1:8 рд╣ोрддा рд╣ै

3. рдЧुрдгिрдд рдЕрдиुрдкाрдд рдХा рдиिрдпрдо (рдбाрд▓्рдЯрди):

"рдЬрдм рджो рддрдд्рд╡ рдПрдХ рд╕े рдЕрдзिрдХ рдпौрдЧिрдХ рдмрдиाрддे рд╣ैं, рддो рдПрдХ рддрдд्рд╡ рдХा рдж्рд░рд╡्рдпрдоाрди рдЬो рджूрд╕рд░े рддрдд्рд╡ рдХे рдиिрд╢्рдЪिрдд рдж्рд░рд╡्рдпрдоाрди рд╕े рд╕ंрдпोрдЧ рдХрд░рддा рд╣ै, рдЫोрдЯे рдкूрд░्рдг рд╕ंрдЦ्рдпाрдУं рдХे рдЕрдиुрдкाрдд рдоें рд╣ोрддा рд╣ै।"

рдЙрджाрд╣рд░рдг: рдХाрд░्рдмрди рдСрдХ्рд╕ीрдЬрди рдХे рд╕ाрде CO рдФрд░ CO₂ рдмрдиाрддा рд╣ै। рдиिрд╢्рдЪिрдд рдХाрд░्рдмрди рдХे рд╕ाрде рд╕ंрдпुрдХ्рдд рдСрдХ्рд╕ीрдЬрди рдХा рдЕрдиुрдкाрдд 1:2 рд╣ै

Dalton's Atomic Theory

John Dalton proposed the atomic theory in 1808:

  • Matter consists of indivisible atoms
  • All atoms of a given element are identical in mass and properties
  • Atoms of different elements have different masses and properties
  • Compounds form when atoms combine in fixed ratios
  • Atoms are neither created nor destroyed in chemical reactions

Modifications to Dalton's Theory:

  • Atoms are divisible (discovery of subatomic particles: electrons, protons, neutrons)
  • Isotopes: Atoms of same element with different mass numbers
  • Atoms can be transformed in nuclear reactions
  • All atoms of an element are not identical (isotopes)

рдбाрд▓्рдЯрди рдХा рдкрд░рдоाрдгु рд╕िрдж्рдзांрдд

рдЬॉрди рдбाрд▓्рдЯрди рдиे 1808 рдоें рдкрд░рдоाрдгु рд╕िрдж्рдзांрдд рдк्рд░рд╕्рддाрд╡िрдд рдХिрдпा:

  • рдкрджाрд░्рде рдЕрд╡िрднाрдЬ्рдп рдкрд░рдоाрдгुрдУं рд╕े рдоिрд▓рдХрд░ рдмрдиा рд╣ोрддा рд╣ै
  • рдХिрд╕ी рддрдд्рд╡ рдХे рд╕рднी рдкрд░рдоाрдгु рдж्рд░рд╡्рдпрдоाрди рдФрд░ рдЧुрдгों рдоें рд╕рдоाрди рд╣ोрддे рд╣ैं
  • рднिрди्рди рддрдд्рд╡ों рдХे рдкрд░рдоाрдгुрдУं рдХे рднिрди्рди рдж्рд░рд╡्рдпрдоाрди рдФрд░ рдЧुрдг рд╣ोрддे рд╣ैं
  • рдпौрдЧिрдХ рддрдм рдмрдирддे рд╣ैं рдЬрдм рдкрд░рдоाрдгु рдиिрд╢्рдЪिрдд рдЕрдиुрдкाрдд рдоें рд╕ंрдпोрдЧ рдХрд░рддे рд╣ैं
  • рд░ाрд╕ाрдпрдиिрдХ рдЕрднिрдХ्рд░िрдпाрдУं рдоें рдкрд░рдоाрдгु рди рддो рдмрдирддे рд╣ैं рдФрд░ рди рдирд╖्рдЯ рд╣ोрддे рд╣ैं

рдбाрд▓्рдЯрди рдХे рд╕िрдж्рдзांрдд рдоें рд╕ंрд╢ोрдзрди:

  • рдкрд░рдоाрдгु рд╡िрднाрдЬ्рдп рд╣ैं (рдЙрдкрдкрд░рдоाрдгुрдХ рдХрдгों рдХी рдЦोрдЬ: рдЗрд▓ेрдХ्рдЯ्рд░ॉрди, рдк्рд░ोрдЯॉрди, рди्рдпूрдЯ्рд░ॉрди)
  • рд╕рдорд╕्рдеाрдиिрдХ: рдПрдХ рд╣ी рддрдд्рд╡ рдХे рдкрд░рдоाрдгु рдЬिрдирдХे рдж्рд░рд╡्рдпрдоाрди рд╕ंрдЦ्рдпा рднिрди्рди рд╣ोрддी рд╣ै
  • рдкрд░рдоाрдгु рдиाрднिрдХीрдп рдЕрднिрдХ्рд░िрдпाрдУं рдоें рдкрд░िрд╡рд░्рддिрдд рд╣ो рд╕рдХрддे рд╣ैं
  • рдПрдХ рддрдд्рд╡ рдХे рд╕рднी рдкрд░рдоाрдгु рд╕рдоाрди рдирд╣ीं рд╣ोрддे (рд╕рдорд╕्рдеाрдиिрдХ)

Exercise: Limitations of Dalton's Theory

Explain how the discovery of isotopes challenges Dalton's theory that all atoms of an element are identical.

Atomic & Molecular Mass

Atomic Mass: Mass of an atom relative to carbon-12 isotope

Atomic mass = (Mass of one atom of the element) / (1/12th mass of one C-12 atom)

Molecular Mass: Sum of atomic masses of all atoms in a molecule

Example: Molecular mass of H₂O = (2 × 1.008) + (1 × 16.00) = 18.016 u

Formula Mass: For ionic compounds that don't exist as discrete molecules

Example: Formula mass of NaCl = Atomic mass of Na + Atomic mass of Cl = 23.0 + 35.5 = 58.5 u

рдкрд░рдоाрдгु рдПрд╡ं рдЖрдгрд╡िрдХ рдж्рд░рд╡्рдпрдоाрди

рдкрд░рдоाрдгु рдж्рд░рд╡्рдпрдоाрди: рдХाрд░्рдмрди-12 рд╕рдорд╕्рдеाрдиिрдХ рдХे рд╕ाрдкेрдХ्рд╖ рдкрд░рдоाрдгु рдХा рдж्рд░рд╡्рдпрдоाрди

рдкрд░рдоाрдгु рдж्рд░рд╡्рдпрдоाрди = (рддрдд्рд╡ рдХे рдПрдХ рдкрд░рдоाрдгु рдХा рдж्рд░рд╡्рдпрдоाрди) / (рдПрдХ C-12 рдкрд░рдоाрдгु рдХे рдж्рд░рд╡्рдпрдоाрди рдХा 1/12)

рдЖрдгрд╡िрдХ рдж्рд░рд╡्рдпрдоाрди: рдПрдХ рдЕрдгु рдоें рд╕рднी рдкрд░рдоाрдгुрдУं рдХे рдкрд░рдоाрдгु рдж्рд░рд╡्рдпрдоाрдиों рдХा рдпोрдЧ

рдЙрджाрд╣рд░рдг: H₂O рдХा рдЖрдгрд╡िрдХ рдж्рд░рд╡्рдпрдоाрди = (2 × 1.008) + (1 × 16.00) = 18.016 u

рд╕ूрдд्рд░ рдж्рд░рд╡्рдпрдоाрди: рдЖрдпрдиिрдХ рдпौрдЧिрдХों рдХे рд▓िрдП рдЬो рдЕрд▓рдЧ-рдЕрд▓рдЧ рдЕрдгुрдУं рдХे рд░ूрдк рдоें рдирд╣ीं рд╣ोрддे

рдЙрджाрд╣рд░рдг: NaCl рдХा рд╕ूрдд्рд░ рдж्рд░рд╡्рдпрдоाрди = Na рдХा рдкрд░рдоाрдгु рдж्рд░рд╡्рдпрдоाрди + Cl рдХा рдкрд░рдоाрдгु рдж्рд░рд╡्рдпрдоाрди = 23.0 + 35.5 = 58.5 u

Element Symbol Atomic Mass (u)
Hydrogen H 1.008
Carbon C 12.01
Oxygen O 16.00
Sodium Na 23.0
Chlorine Cl 35.5
3

Lecture 3: Mole Concept

Mole & Avogadro's Number

The mole is the SI unit for amount of substance that contains as many elementary entities as there are atoms in exactly 12g of carbon-12.

Avogadro's Number (NтВР):

1 mole = 6.022 × 10²³ particles (atoms, molecules, ions)

Molar Mass: Mass of one mole of a substance in grams (numerically equal to atomic/molecular mass in u)

Example: Molar mass of H₂O = 18.016 g/mol

рдоोрд▓ рдПрд╡ं рдЖрд╡ोрдЧाрдж्рд░ो рд╕ंрдЦ्рдпा

рдоोрд▓ рдкрджाрд░्рде рдХी рдоाрдд्рд░ा рдХी SI рдЗрдХाрдИ рд╣ै рдЬिрд╕рдоें рдЙрддрдиे рд╣ी рдоूрд▓ рдЗрдХाрдЗрдпाँ рд╣ोрддी рд╣ैं рдЬिрддрдиे рдкрд░рдоाрдгु рдаीрдХ 12g рдХाрд░्рдмрди-12 рдоें рд╣ोрддे рд╣ैं।

рдЖрд╡ोрдЧाрдж्рд░ो рд╕ंрдЦ्рдпा (NтВР):

1 рдоोрд▓ = 6.022 × 10²³ рдХрдг (рдкрд░рдоाрдгु, рдЕрдгु, рдЖрдпрди)

рдоोрд▓рд░ рдж्рд░рд╡्рдпрдоाрди: рдХिрд╕ी рдкрджाрд░्рде рдХे рдПрдХ рдоोрд▓ рдХा рдЧ्рд░ाрдо рдоें рдж्рд░рд╡्рдпрдоाрди (рд╕ंрдЦ्рдпाрдд्рдордХ рд░ूрдк рд╕े u рдоें рдкрд░рдоाрдгु/рдЖрдгрд╡िрдХ рдж्рд░рд╡्рдпрдоाрди рдХे рдмрд░ाрдмрд░)

рдЙрджाрд╣рд░рдг: H₂O рдХा рдоोрд▓рд░ рдж्рд░рд╡्рдпрдоाрди = 18.016 g/mol

Concept Formula Example
Number of moles (n) n = Mass / Molar mass Moles in 36g H₂O: 36/18 = 2 moles
Number of particles N = n × NтВР Molecules in 2 moles H₂O: 2 × 6.022×10²³
Mass from moles Mass = n × Molar mass Mass of 0.5 mole O₂: 0.5 × 32 = 16g

Exercise: Mole Calculations

1. How many moles are in 25g of calcium carbonate (CaCO₃)?
2. Calculate the number of oxygen atoms in 1g of oxygen gas (O₂).

Stoichiometry

Calculation of reactants and products in chemical reactions based on balanced equations.

Steps for Stoichiometric Calculations:

  1. Write balanced chemical equation
  2. Convert given quantities to moles
  3. Use mole ratios from balanced equation
  4. Convert moles to required units

Example: How many grams of O₂ are needed to burn 36g of carbon?

C + O₂ → CO₂

Mole ratio: 1 mol C : 1 mol O₂

Moles of C = 36g / 12g/mol = 3 moles

Moles of O₂ needed = 3 moles

Mass of O₂ = 3 × 32 = 96g

рд░рд╕рд╕рдоीрдХрд░рдгрдоिрддि (рд╕्рдЯॉрдЗрдХिрдпोрдоेрдЯ्рд░ी)

рд╕ंрддुрд▓िрдд рд╕рдоीрдХрд░рдгों рдХे рдЖрдзाрд░ рдкрд░ рд░ाрд╕ाрдпрдиिрдХ рдЕрднिрдХ्рд░िрдпाрдУं рдоें рдЕрднिрдХाрд░рдХों рдФрд░ рдЙрдд्рдкाрджों рдХी рдЧрдгрдиा।

рд░рд╕рд╕рдоीрдХрд░рдгрдоिрддीрдп рдЧрдгрдиा рдХे рдЪрд░рдг:

  1. рд╕ंрддुрд▓िрдд рд░ाрд╕ाрдпрдиिрдХ рд╕рдоीрдХрд░рдг рд▓िрдЦें
  2. рджी рдЧрдИ рдоाрдд्рд░ाрдУं рдХो рдоोрд▓ рдоें рдмрджрд▓ें
  3. рд╕ंрддुрд▓िрдд рд╕рдоीрдХрд░рдг рд╕े рдоोрд▓ рдЕрдиुрдкाрдд рдХा рдЙрдкрдпोрдЧ рдХрд░ें
  4. рдоोрд▓ рдХो рдЖрд╡рд╢्рдпрдХ рдЗрдХाрдЗрдпों рдоें рдмрджрд▓ें

рдЙрджाрд╣рд░рдг: 36g рдХाрд░्рдмрди рдХो рдЬрд▓ाрдиे рдХे рд▓िрдП рдХिрддрдиे рдЧ्рд░ाрдо O₂ рдХी рдЖрд╡рд╢्рдпрдХрддा рд╣ोрдЧी?

C + O₂ → CO₂

рдоोрд▓ рдЕрдиुрдкाрдд: 1 рдоोрд▓ C : 1 рдоोрд▓ O₂

C рдХे рдоोрд▓ = 36g / 12g/mol = 3 рдоोрд▓

рдЖрд╡рд╢्рдпрдХ O₂ рдХे рдоोрд▓ = 3 рдоोрд▓

O₂ рдХा рдж्рд░рд╡्рдпрдоाрди = 3 × 32 = 96g

Concentration Terms

Ways to express concentration of solutions:

Term Formula Unit
Mass Percentage (Mass of solute / Mass of solution) × 100 %
Mole Fraction (x) n₁ / (n₁ + n₂ + ...) Dimensionless
Molarity (M) Moles of solute / Liters of solution mol/L (M)
Molality (m) Moles of solute / kg of solvent mol/kg

рд╕ांрдж्рд░рддा рдкрдж

рд╡िрд▓рдпрди рдХी рд╕ांрдж्рд░рддा рд╡्рдпрдХ्рдд рдХрд░рдиे рдХे рддрд░ीрдХे:

  • рдж्рд░рд╡्рдпрдоाрди рдк्рд░рддिрд╢рдд: (рд╡िрд▓ेрдп рдХा рдж्рд░рд╡्рдпрдоाрди / рд╡िрд▓рдпрди рдХा рдж्рд░рд╡्рдпрдоाрди) × 100
  • рдоोрд▓ рдк्рд░рднाрдЬ (x): n₁ / (n₁ + n₂ + ...)
  • рдоोрд▓рд░рддा (M): рд╡िрд▓ेрдп рдХे рдоोрд▓ / рд╡िрд▓рдпрди рдХे рд▓ीрдЯрд░
  • рдоोрд▓рд▓рддा (m): рд╡िрд▓ेрдп рдХे рдоोрд▓ / рд╡िрд▓ाрдпрдХ рдХे kg

Exercise: Concentration Calculations

1. Calculate the molarity of a solution containing 5g of NaOH in 250mL solution.
2. What is the molality of a solution with 20g glucose in 500g water?

Willer Academy Chemistry Notes | CBSE English & Hindi Medium

Complete Chapter Notes in 3 Lectures | © 2023 Willer Academy

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