Some Basic Concepts of Chemistry Class 11 Notes CBSE Chemistry Chapter 1

Section – A (1 Mark Questions)

1. What is the SI unit of time?

Ans. The SI unit of time is second (s).

2. Define atomic mass.

Ans. Atomic mass of an element is defined as the average relative mass of an atom of an element as compared to the mass of an atom of carbon (12C) taken as 12.

3. How many atoms are present in 3 moles of carbon?

Ans. No. of atoms in 1 mole = 6.022×1023

No. of atoms in 3 moles = 3×6.022×1023

= 1.8066×1024 atoms

4. Define Density.

Ans. The density of a material is defined as its mass per unit volume.

5. What is the relation between °C (degree celsius) and °F (degree Fahrenheit) ?

Ans. °F = $\frac{9}{5}$ (°C) + 32

6. What is a homogeneous mixture?

Ans. A mixture in which the components are completely mixed with each other and its composition is uniform throughout is called a homogeneous mixture.

7. What is the molecular mass of ethane?

Ans. Molecular mass of C2H6 = 2 × (Atomic mass of C) 

+ 6 × (Atomic mass of H)

Molecular mass of ethane(C2H6 ) = 2×(12.011u) 

+ 6× (1.008 u) = 30.07 u

8. Define the law of conservation of mass.

Ans. In a chemical reaction the mass of reactants consumed and mass of the products formed is the same, that is mass is conserved.

9. What is the SI unit of weight?

Ans. The SI unit of weight is Newton.

10. Give two examples of molecules having molecular formula same as empirical formula.

Ans. H2O, CO2

Section – B (2 Marks Questions)

11. Calculate the molar mass of the following:

(i) H2O   (ii) NH3

Ans. (i) Molar mass of water = (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)

= (2× 1g mol-1) + (1 × 16 g mol-1)

= 2 + 16

= 18 g mol-1

(ii) Molar mass of NH3 = (Atomic mass of nitrogen) + (3× Atomic mass of hydrogen)

= (14 g mol-1) + (3×1 g mol-1)

= 14+ 3

= 17 g mol-1

12. One atom of an element weighs 4.2 × 10–22 g. What is its atomic mass?

Ans. Mass of 1 atom = 4.2 × 10–22 g

Mass of 6.023 × 1023 atoms = 6.023×1023× 4.2×10–22 g

= 6.023 × 4.2 × 10g

= 252.96 g

Hence, atomic mass of element = 252.96 g/mol.

13. How many significant figures are present in the following?

(i) 0.0657

(ii) 456

(iii) 878.0

Ans. (i) 0.0657 Four significant figures

(ii) 456 Three significant figures

(iii) 878.0 Four significant figures

14. Which set of figures will be obtained after rounding off the following to three significant figures?

(i) 34.216

(ii) 0.06597

(iii) 15.8107

Ans. (i) 34.216 = 34.2

(ii) 0.06597= 0.066

(iii)15.8107 = 15.8

15. 23.72 g of a substance ‘X’ occupies 5.56 cm3. What will be its density measured in correct significant figures?

Ans. $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

Density = $\frac{23.72\text{g}}{5.56\text{cm}^3}$

Density = 4.27 g / cm3

Correct significant figures in case of a multiplication or division is find out by rounding off to the same number of significant figures as possessed by the least precise term in the calculation. Here least precise term is 5.56 so density will also have 3 significant figures.

16. What is difference between mass and weight?

Ans. (i) Mass of a substance is the amount of matter present in it whereas weight is the force exerted by gravity on an object. 

(ii) The mass of a substance is constant whereas its weight may vary from one place to another due to change in gravity.

17. Volume of a gas at STP is 6.57 × 10–8 cm3. Find out the number of molecules in it?

Ans. Given, V =6.57 × 10–8 cm3

∵ 22400 cm3 of the gas at STP has  

= 6.02 × 1023 molecules

∴ 6.57 × 10–8cm3 of the gas at STP will have 

$=\frac{6.01\times10^{23}}{22400}\times6.25\times10^{-8}$

= 1.765 × 1012  molecules

18. If 8.7g of Na2CO3 is added to 18g of CH3COOH  solution, the residue is found to weigh 22g. What is the mass of CO2 released in the reaction?

Ans. Na2CO3 + CH3COOH → CO2 + H2O + CH3COONa

According to law of conservation of mass,

Mass of reactants = mass of products 

8.7 + 18 = 22 + x 

x = 26.7 – 22

x = 4.7g

19. The mass of nitrogen per gram hydrogen in the compound hydrazine is exactly one and half times of the mass of nitrogen in the compound ammonia. Which law is illustrated by the given fact?

Ans. The law of multiple proportions says that, "If two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers".   

The ratio of masses of nitrogen per gram of hydrogen in Hydrazine and Ammonia is 1.5:1 = 3:2 which is simple whole-number ratio. Therefore, it represents the law of multiple proportions.

20. Zinc sulphate solution contains 22.65% zinc and 43.9% of water of crystallization. If the law of constant proportions is true then what is the weight of zinc required to produce 20 g of the zinc sulphate crystals?

Ans. 100g of ZnSO4 is obtained from 

= 22.65g of Zn

20g of ZnSO4 obtained from = $\frac{22.65\times20}{100}$ =4.53 g

PDF Summary of Class 11 Chemistry Chapter 1 Notes - Some Basic Concepts of Chemistry

1. Chemistry

Chemistry (derived from the Egyptian word kēme (chem), which means "earth") is a science that studies the composition, structure and properties of matter and the changes it undergoes during chemical reactions. Chemistry is often referred to as core science because it plays a role in linking physical sciences (including chemistry) with life sciences and applied sciences (such as medicine and engineering).

Chemistry is divided into following branches:

1.1 Physical Chemistry

The branch of chemistry which deals with macroscopic as well as physical phenomena in a universe. It is generally the impact of physical property on the chemical property as well as the structure of a substance.

1.2 Inorganic Chemistry

The branch of chemistry that studies compounds that do not contain carbon and hydrogen atoms is called "inorganic chemistry." Simply put, it is the opposite of organic chemistry. Substances that do not have carbon-hydrogen bonds include metals, salts, and chemicals.

1.3 Organic Chemistry

The discipline which deals with the study of the structure, composition and the chemical properties of organic compounds is known as organic chemistry. It involves the study of Carbon and its compounds

1.4 Biochemistry

Biochemistry is that branch of chemistry that explores the chemical processes in organisms and associated with them. It's a laboratory-based science that connects biology and chemistry. By using chemical knowledge and technology, biochemists can understand and solve biological problems

1.5 Analytical Chemistry

It is the branch of chemistry which uses instruments and analytical techniques to determine structure, functionality and properties of a substance.

2. Matter

Matter is defined as any thing that has some mass and also occupies a certain volume in a space.

Generally matter is classified into three phases:

• Solid- The substance which has a definite shape as well as maintains its volume as per its shape, also they have least freedom of movement. e.g., sugar, iron, gold, wood etc.

• Liquid- A substance is a substance which generally possesses the shape of a container but has a fixed volume. Also liquids have the property to flow or to be poured. E.g., water, milk, oil, mercury, alcohol etc.

• Gas- Substances which do not have a definite volume as well as definite shape. Gases generally completely fill the container they are kept in. E.g., hydrogen, oxygen etc.

The three states are interconvertible by changing the conditions of temperature and pressure as follows:

(Image will be uploaded soon)

3. Classification of Matter at Macroscopic Level

Matter can further be classified into following at bulk or macroscopic level:

(a) Mixtures (b) Pure Substances.

These can be further classified as shown below:

(Image will be uploaded soon) 

(a) Mixtures : A mixture is a substance in which two or more substances are present in any ratio. Primarily, It is of two types: Heterogeneous and Homogeneous mixtures.

• Homogeneous Mixture- Two substances are mixed to form a mixture such that there exist one single uniform phase i.e. composition of substances present is uniform. Sugar solution and air are thus, the examples of homogeneous mixtures.

• Heterogeneous Mixtures- Two or more substances are mixed which result in non-uniform composition throughout the mixture. Some of the examples are suspensions, a mixture of two solids, suppose salt and sugar.

Note: Any distinct portion of matter that is uniform throughout in composition and properties is called a Phase.

(b) Pure Substances:- A material containing only one type of particle is called a pure substance.

Note: In chemistry, forms of matter have constant chemical composition and chemical properties and they cannot be separated into components by physical methods.

Pure substances are further divided as given below:

• Element- An element is defined as a pure substance that contains only one kind of atom and cannot be further broken down. The elements are further split into three classes based on their physical and chemical properties i.e. (1) Metals (2) Non- metals and (3) Metalloids.

• Compound- A compound is a pure material that consists of two or more elements mixed in a defined mass proportion. Furthermore, a compound's qualities are distinct from those of its constituting elements. Moreover, the constituents of a compound cannot be separated into simpler substances by physical methods. They can only be separated by chemical methods.

4. Properties of Matter

Unique or characteristic properties are depicted by every substance.

Physical properties and chemical properties are the two types of properties that are observed.

4.1 Physical Properties

Physical properties are those that may well be measured or observed without affecting the substance's identity or composition. Colour, fragrance, melting point, boiling point, density, and other physical qualities are some of the examples.

4.2 Chemical Properties

Chemical properties are the properties of specific substances that can be observed in chemical reactions. Some of the main chemical properties include flammability, toxicity, heat of combustion, pH, radioactive decay rate, and chemical stability.

5. Measurement

5.1 Physical Quantities

Physical quantities are quantities which we encounter during our scientific study. Any physical quantity can be measured in two parts:

(1) The number, and (2) The unit: Unit is defined as the reference standard chosen to measure any physical quantity.

5.2 S.I. Units

The International System of Units (in French Le Systeme International d’Unités – abbreviated as SI) was established by the eleventh  General Conference on Weights and Measures (CGPM from Conference Generale des Poids at Measures). The CGPM is an intergovernmental treaty organization created by a diplomatic treaty known as the Meter Convention which was signed in Paris in \[1875\].

There are seven base units in SI system as listed below

These units pertain to the seven fundamental scientific quantities. The other physical quantities such as speed, volume, density etc. can be derived from these quantities. The definitions of the SI base units are given below:

Note: The mass standard has been the kilogram since 1889. It has been defined as the mass of a platinum-iridium (Pt-Ir) cylinder that is stored in an airtight jar at the International Bureau of Weights and Measures in Sevres, France. Pt-Ir was chosen for this standard because it is highly resistant to chemical attack and its mass will not change for an extremely long time.

6. Some Important Definition

6.1 Mass and Weight

The mass of a substance is the amount of substance present in it and the weight is the force exerted by gravity on the object. The mass of matter is constant, and due to changes in gravity, its weight can vary from place to place. The SI unit of mass is kilogram (kg). The SI derived unit of weight (the derived unit of the SI base unit) is Newton.

6.2 Volume

Volume is the amount of three-dimensional space surrounded by certain closed boundaries, for example, the space occupied or contained by a substance (solid, liquid, gas, or plasma) or shape. Volume is usually quantified numerically using SI derived units (cubic meters).

6.3 Density

The mass density or density of a material is defined as its mass per unit volume. The density is represented by the symbol  $\rho $ (the lowercase Greek letter rho). The SI unit of density is\[kg\text{ }{{m}^{3}}\].

6.4 Temperature

Temperature is a physical property of matter, which quantitatively expresses the common concepts of heat and cold. There are three common scales for measuring temperature: (Celsius), (Fahrenheit) and (Kelvin).The temperature on two scales is related to each other by the following relationship:

\[{}^\circ F\text{ }=\text{ }9/5\text{ }\left( {}^\circ C \right)\text{ }+\text{ }32\] 

\[K={}^\circ C+273.15\] 

7. Law of Chemical Combination

7.1 Law of Conservation of Mass

“In a chemical reaction the mass of reactants consumed and mass of the products formed is the same, that is mass is conserved.” This is a direct consequence of the law of conservation of atoms. This law was given by Antoine Lavoisier in \[1789\].

7.2 Law of Constant / Definite Proportions

The law of definite proportions states that the mass proportions of the elements in a composite sample are always the same. It was given by a French chemist, Joseph Proust.

7.3 Law of Multiple Proportions

The law of multiple proportions states that when two elements are combined to form more than one compound, the weight of one element is proportional to the fixed weight of the other element as a whole number. This law was proposed by Dalton in \[1803\].

7.4 Law of Reciprocal Proportions

The law states that if two different elements are combined with the fixed mass of the third element, their combined mass ratio is the same, or a simple multiple of their combined mass. This Law was proposed by Richter in \[1792\].

7.5 Gay Lussac’s Law of Gaseous Volumes

The law developed by Gay Lussac in 1808 establishes that "the relationship between the volume of a gaseous reactant and a product can be represented by a simple whole number."

7.6 Avogadro Law

In \[1811\] , Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain an equal number of molecules.

8. Dalton’s Atomic Theory

In 1808, Dalton published ‘A New System of Chemical Philosophy’ in which he proposed the following:

1. Matter consists of indivisible atoms.

2. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass. 

3. Compounds are formed when atoms of different elements combine in a fixed ratio.

4. Chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction.

9. Atom

Atom is defined as the smallest unit that retains the properties of an element as well as participate in a chemical reaction (Note: This definition holds true only for non-radioactive reactions)

9.1 Mass of an Atom

There are two ways to denote the mass of atoms.

9.2 Method 1

Atomic mass can be defined as a mass of a single atom which is measured in atomic mass unit (amu) or unified mass (u) where

\[1\text{ }a.m.u.\text{ }=\text{ }{}^{1}/{}_{12}th\]of the mass of one \[{{C}^{12}}\] atom

9.3 Method 2

Mass of  \[6.022\text{ }\times \text{ }{{10}^{23}}\] atoms of that element taken in grams. This is also known as molar atomic mass.

Note:

• Mass of 1 atom in amu and mass of \[6.022\text{ }\times \text{ }{{10}^{23}}\] atoms in grams are numerically equal.

• When atomic mass is taken in grams it is also called the molar atomic mass.

• \[6.022\text{ }\times \text{ }{{10}^{23}}\] is also called 1 mole of atoms and this  number is also called the Avogadro’s Number.

• Mole is just a number. As \[1\]  dozen\[=\text{ }12\];

\[1\text{ }million\text{ }=\text{ }{{10}^{6}}\text{ };\]

\[1\text{ }mole\text{ }=\text{ }6.022\text{ }\times \text{ }{{10}^{23}}\]

10. Molecules

The smallest particle of a substance that has all the physical and chemical properties of the substance. A molecule consists of one or more atoms. e.g.\[{{H}_{2}},\text{ }N{{H}_{3}}\].

The mass of a molecule is measured by adding the masses of the atoms that compose it. Therefore, the mass of a molecule can also be expressed by the two methods used to measure the mass of an atom i.e. amu and g / mol.

(Image will be uploaded soon) 

11. Chemical Reactions

A process that involves rearranging the molecular or ionic structure of a substance, which is different from a change in physical form or a nuclear reaction.

Points to Remember:

• Equations must be balanced before proceeding for calculations.

• We do not need to conserve the number of molecules in a reaction. e.g.\[{{N}_{2}}+3\text{ }{{H}_{{{2}_{~}}}}\to 2N{{H}_{3}}\]. If we only discuss the rearrangement of atoms in the equilibrium chemical reaction, and the number of molecules is not conserved, then it is clear that the mass of the atoms on the reaction side is equal to the sum of the masses of the atoms on the reaction side. This is the law of conservation of atoms and the law of conservation of mass.

12. Stoichiometry

The study of chemical reactions and related calculations is called stoichiometry. The coefficient used to balance the reaction is called the stoichiometric coefficient.

Points to Remember:

• The stoichiometric coefficient is the ratio of moles of molecules of atoms that reacts, not the mass.

• Only when all reactants are present in a stoichiometric ratio can the stoichiometric ratio be used to predict the number of moles of product formed. The actual quantity of product formed is always less than the quantity predicted by the theoretical calculation

12.1 Limiting Reagent (LR) and Excess Reagent (ER)

If the reagents are not used in a stoichiometric ratio, then the less than the required amount of reagent determines how much product will be formed, which is called the limiting reagent, and the reagent in excess is called the excess reagent. For example, if we burn carbon in the air (it has an unlimited supply of oxygen), the amount of CO2  produced will depend on the amount of carbon inhaled. In this case, Carbon is the limiting reactant and \[{{O}_{2}}\] is the excess reactant.

13. Percent Yield

As mentioned above, due to practical reasons, the amount of product formed by the chemical reaction is less than the amount predicted by the theoretical calculation. When multiplied by 100, the relationship between the amount of product formed and the predicted amount gives the percentage yield \[\text{Percentage Yield=}\dfrac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\] 

14. Reactions in Aqueous Media

The two solids cannot react with each other in the solid phase, so they must be dissolved in the liquid. When solutes dissolve in a solvent, they coexist in a single phase called a solution. Several parameters are used to measure the strength of the solution. The strength of a solution denotes the amount of solute which is contained in the solution.

The parameters used to denote the strength of a solution are:

• Mole dfraction \[X\]: moles of a component / Total moles of solution.

• Mass\[\%\] : Mass of solute (in g) present in \[100g\]  of solution.

• \[{}^{Mass}/{}_{Vol}\] : Mass of solute (in g) present in  \[100mL\]  of solution

•  \[{}^{v}/{}_{v}\] : Volume of solute/volume of solution {only for liq-liq solutions}

• \[{}^{g}/{}_{L}\] : Wt. of solute (g) in \[1L\]  of solution

• ppm : \[\dfrac{\text{mass of solute}}{\text{mass of solution}}\times 10\] 

• Molarity (M) :\[\dfrac{\text{moles of solute}}{\text{volume of solution }\left( \text{L} \right)}\]

• Molality (m) :\[\dfrac{\text{moles of solute}}{\text{mass of solvent }\left( \text{kg} \right)}\] 

Important Relations

1. Relation between molality (\[m\] ) Molarity (\[M\] ), density (\[d\] ) of solution and molar mass of solute (\[{{M}_{O}}\]) 

\[d\] : density in g/mL

\[{{M}_{O}}\]: molar mass in \[g\text{ }mo{{l}^{1}}\] 

Molality, \[m=\dfrac{M\times 1000}{1000d-M{{M}_{O}}}\]

2. Relationship between molality (\[m\]) and mole dfraction (\[{{X}_{B}}\]) of the solute

$m=\dfrac{{{X}_{B}}}{1-{{X}_{B}}}\times \dfrac{1000}{{{M}_{X}}}$ 

$m=\dfrac{1-{{X}_{A}}}{{{X}_{A}}}\times \dfrac{1000}{{{M}_{A}}}$

Points to Remember:

• Molarity is the most common unit of measuring strength of solution.

• The product of Molarity and Volume of the solution gives the number of moles of the solute, \[n=M\times V\] 

• All the formulae of strength have an amount of solute. (weight or moles) in the numerator.

• All the formulae have an amount of solution in the denominator except for molality (\[m\]).

15. Dilution Law

The moles of a solution do not change when it is diluted. If the volume of a solution having a Molarity of \[{{M}_{1}}\] is diluted from \[{{V}_{1}}\] to \[{{V}_{2}}\] by adding more solvent we can write that:

\[{{M}_{1}}{{V}_{1}}=\] moles of solute in the solution \[={{M}_{2}}{{V}_{2}}\] 

16. Effect of Temperature

The volume of solvent increases with increasing temperature. But assuming that the system is closed, that is, there is no mass loss, and it has no effect on the quality of the solute in the solution. The formulae of strength of solutions which do not involve volume of solution are unaffected by changes in temperature.

e.g. molality remains unchanged with temperature. Formulae involving volume are altered by temperature e.g. Molarity.

17. Introduction to Equivalent Concept

The concept of equivalence is a way to understand chemical reactions and processes, and the concept of equivalence is usually used for simplification.

17.1 Equivalent Mass

“The mass of an acid which furnishes \[1\] mol \[H+\] is called its Equivalent mass.”

“The mass of the base which furnishes \[1\] mol \[OH-\] is called its Equivalent mass.”

17.2 Valence Factor

(n-factor(represented by Z))

Valence factor: is the number of  \[H+\] ions supplied by \[1\] molecule or mole of an acid or the number of \[OH-\] ions supplied by \[1\] molecule or \[1\] mole of the base.

Mass Equivalent,$E=\dfrac{\text{Molecular Mass}}{\text{Z}}$ 

17.3 Equivalents

No. of equivalents \[=\dfrac{\text{ }\dfrac{\text{wt}\text{. of acid}}{\text{ base taken}}\text{ }}{\text{Eq}\text{. Wt}}.\]

Note: It should be always remembered that \[1\] equivalent of an acid reacts with \[1\] equivalent of a base.

18. Mixture of Acids and Bases

If we have a mixture of multiple acids and bases, we can use the concept of equivalence to determine whether the resulting solution is acidic or alkaline. For a mixture of multiple acids and bases, find out the equivalents of the acids and bases used, and find out which one is in excess.

19. Law of Chemical Equivalence

According to this law, an equivalent of a reactant is combined with an equivalent of another reactant to obtain an equivalent of each product. For, example in a reaction \[aA+bB~\to cC+dD\] irrespective of the stoichiometric coefficients, \[1\] eq. of \[A\] reacts with \[1\] eq. of \[B\] to give \[1\] eq. each of \[C\] and \[1\]eq of \[D\] 

20. Equivalent Weights of Salts

To calculate the equivalent weight of a compound that is neither an acid nor a base, we need to know the charge of the cation or anion. The mass of the cation divided by the charge on it is called the equivalent mass of the cation, and the mass of the anion divided by the charge on it is called the equivalent mass of the anion. When we add equal masses of anions and cations, it gives us equal amounts of salt. For salts, the valence factor is the total amount of positive or negative charges provided by the 1 mol of salt.

21. Origin of Equivalent Concept

The equivalent weight of an element was originally defined as the weight of the element combined with 1 grams of hydrogen. Subsequently, the definition was modified as follows: the equivalent weight of an element is the weight of the element combined with 8 grams of oxygen. Note: Same element can have multiple equivalent weights depending upon the charge on it. e.g. \[F{{e}^{2+}}\] and \[F{{e}^{3+}}\] 

22. Equivalent Volume of Gases

Equivalent volume of gas is the volume occupied by \[1\] equivalent of a gas at STP.

Equivalent mass of gas\[={}^{molecular\text{ }mass}/{}_{Z}\] .

Since \[1\] mole of gas occupies \[22.4L\]  at STP therefore \[1\] equivalent of a gas will occupy \[{}^{22.4}/{}_{Z}\text{ }L\] at STP. e.g. Oxygen occupies \[5.6L\] , Chlorine and Hydrogen occupy\[11.2L\].

23. Normality

The normality of a solution is the number of equivalents of solute present in \[1\]L of the solution.

\[~N=\dfrac{\text{equivalents of solute }\!\!~\!\!\text{ }}{\text{volume of solution }\left( \text{L} \right)}\]

The number of equivalents of solute present in a solution is given by \[\text{Normality  }\!\!\times\!\!\text{  Volume }\left( \text{L} \right)\text{.}\]

On dilution of the solution the number of equivalents of the solute is conserved and thus, we can apply the formula : \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\]

Caution:

Please note that the above equation can cause a lot of confusion and is a common mistake students make. This is the dilution equation for the conservation of the equivalent number. Now, since one reactant equivalent always reacts with one equivalent of another reactant, similar equations are used in problems involving acid and base titration. Don't extend the same logic to molarity.

Relationship between Normality and Molarity is given by:

\[N=M\times Z\] ; where ‘\[Z\]’ is the Valency factor

CBSE Chemistry Chapter 1 Some Basic Concepts of Chemistry Class 11 Notes 

Chapter 1: Some Basic Concepts of Chemistry Class

This chapter is very important to understand and evolve in Chemistry. Class 11 Chemistry Chapter 1 PDF notes broadly explain some important topics and terms in a simple language for a better understanding of students. These topics are as follows:

• Physical Chemistry 

• Inorganic Chemistry

• Organic Chemistry 

• BioChemistry 

• Analytical Chemistry 

• Different States of Matter 

• Classification of Matter 

• Properties of Matter 

• Measurement 

• Mass

• Weight 

• Volume

• Density

• Law of Multiple Proportions 

• Law of Reciprocal Proportions 

• Avogadro Law

• Dalton’s Atomic Theory

• Molecules 

• Dilution Law

• Mixture of Acids and Bases

• Normality 

• Origin of Equivalent Concept 

• Law of Chemical Equivalence

This is the base of Chemistry, and what must be noted is that a lot of the important objectives - easy marks questions form the base of this chapter appear every year in Class 11 Chemistry Exam.

Class 11th Chemistry Chapter 1 notes cover all these crucial questions with solved problems. Thus Ch 1 Chemistry class 11 notes are essential to refer before attempting any examination, as it fulfils your basic need of studying hard and smart to make your efforts more effective and focussed on quality.

Important Questions

1. What does Avogadro’s law state?

2. There are two oxides of metal that contain 27.6% and 30% of Oxygen. Find the formula of second metal oxide, if the formula for the first metal oxide is given as M3O4.

3. What is the definition of mass. Also, write its SI Unit.

4. What differentiates physical properties from chemical properties?

5. What is the benefit of using Molality over Molarity?

Some Basic Concepts of Chemistry Class 11 Notes - Free PDF Download

Class 11 Chemistry Chapter 1 PDF Notes focuses on providing you with basic concepts of Chemistry explained in an elementary manner which is lucid and easy to comprehend. The students’ grasping power increases as well because Class 11th Chemistry Chapter 1 Notes are detailed and in a simplified format to support better learning. If you memorise these notes, you do not need to refer to any other sources to clear your exam well. Chemistry Class 11 Chapter 1 notes are prepared after analysing previous years question papers to provide students with essential questions that might appear in Class 11 Chemistry exam.

Benefits of Chemistry Class 11 Chapter 1 Notes PDF

Some Basic Concepts of Chemistry Class 11 Notes are made from the updated CBSE syllabus keeping the latest exam pattern in mind to keep students on the right track. Students can download and use these Chemistry Class 11 Chapter 1 Notes to score good marks in exams.

• Vedantu's subject experts explain complex topics in a clear and concise way, using easy-to-understand language.

• Some basic concepts of chemistry notes Class 11 are designed to highlight the most important concepts from the chapter, helping you to focus your studies and avoid getting overwhelmed.

• Some basic concept of chemistry notes class 11 chapter covers essential concepts that form the basis for much of what you will learn in the rest of your chemistry studies. A strong understanding of these concepts is crucial for success in future chapters.

• The concise and well-organized format of the ch 1 chemistry class 11 notes makes it easier to review key information before exams.

• Vedantu’s class 11 chemistry chapter 1  PDF notes include solved problems that demonstrate how to apply the concepts you have learned. This can help you develop your problem-solving skills and boost your confidence for exams.

• The Vedantu Class 11th Chemistry Chapter 1 Notes are free to download as a PDF, making them a cost-effective study resource, you can access the notes anytime, even without an internet connection.

Some Basic Concepts of Chemistry Chapter Related Important Study Materials

It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

Conclusion

The CBSE Chemistry Chapter 1, Some Basic Concepts of Chemistry Class 11 notes introduces fundamental concepts crucial for understanding the subject. It covers essentials like matter, its properties, and measurement units. Focus on comprehending the mole concept, which plays a pivotal role in chemical calculations. Understand the significance of Avogadro's number and its application in stoichiometry. Additionally, pay attention to the concept of empirical and molecular formulas, as they are foundational in chemical equations. Mastering these basics will pave the way for a solid understanding of more complex chemical principles in subsequent chapters. Regular practice and clarity on these fundamentals are key to excelling in chemistry.

Download CBSE Class 11 Chemistry Revision Notes 2024-25 PDF

Also, check CBSE Class 11 Chemistry revision notes for other chapters: