assignment for class 11 chemistry chapter 1

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• Class 11 - Chemistry
• Chapter 1: Basic Concepts Chemistry

Important Questions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry.

Class 11 chemistry important questions with answers are provided here for Chapter 1 Some Basic Concepts of Chemistry. These important questions are based on the CBSE board curriculum and correspond to the most recent Class 11 chemistry syllabus. By practising these Class 11 important questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 11 Annual examinations as well as other entrance exams such as NEET and JEE.

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Class 11 Some Basic Concepts of Chemistry Important Questions with Answers

Short answer type questions.

Q1. What will be the mass of one atom of C-12 in grams?

The mass of 1 mole of C-12 atoms = 12 g

1 mole of C-12 atoms = 6.022 × 10 23 atoms

The mass of 1 atom of C-12 = 12 / (6.022 × 10 23 )

= 1.99 × 10 –23 g

Q2. How many significant figures should be present in answer to the following calculation?

The number of significant figures that should be present in the calculation \(\begin{array}{l}\frac{2.5\times 1.25\times 3.5}{2.01}\end{array} \) is 2.

Q3. What is the symbol for the SI unit of the mole? How is the mole defined?

The symbol for the SI unit of the mole is mol.

One mole is defined as the amount of a substance containing the same number of particles or entities as there are atoms in exactly 12 g (0.012 kg) of the C – 12 isotope.

Q4. What is the difference between molality and molarity?

Q5. Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate Ca 3 (PO 4 ) 2

Molecular formula of calcium phosphate is Ca 3 (PO 4 ) 2

Its molar mass will be – 3(40) + 2(31) + 8(16) = 310 g/mol.

• Mass percent of calcium

Percentage of Calcium = 38.71%

• Mass percent of phosphorus

Percentage of Phosphorus =20%

• Mass percent of oxygen

Percentage of Oxygen = 41.29%

Q6. 45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below:

2N 2 (g) + O 2 (g) → 2N 2 O(g)

Which law is being obeyed in this experiment? Write the statement of the law.

The volumes of dinitrogen and dioxygen that combine (i.e., 45.4 L and 22.7 L) have a simple ratio of 2: 1. As a result, it follows Gay Lussac’s law of gaseous volumes.

According to this law, “when gases combine or are produced in a chemical reaction, they do so in a simple volume ratio provided all gases are at the same temperature and pressure.”

Q7. If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in a whole-number ratio.

• Is this statement true?
• If yes, state according to which law?
• Give one example related to this law.

If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in a whole-number ratio.

• Yes, the statement is true.
• According to the law of multiple proportions.
• Consider the example, H 2 + O 2 → H 2 O H 2 + O 2 → H 2 O 2

Here masses of oxygen, (i.e., 16g in H 2 O and 32g in H 2 O 2 ) which combine with a fixed mass of hydrogen (2g) are in the simple ratio of 16 : 32 or 1 : 2.

Q8. Calculate the average atomic mass of hydrogen using the following data:

The average atomic mass is given by the following formula-

Hence, the average atomic mass will be 1.00015 μ.

Q9. Hydrogen gas is prepared in the laboratory by reacting dilute HCI with granulated zinc.

The following reaction takes place.

Zn + 2HCl → ZnCl 2 + H 2

Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCI. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of Zn = 65.3 u.

65.3 g of Zn reacts with HCl to form 22.7 of H 2 gas.

∴ 32.65 g of Zn at STP reacts with HCl to form =

Q10. The density of the 3 molal solution of NaOH is 1.110 g mL –1 . Calculate the molarity of the solution.

Molality is 3 m

Also, number of moles=mass/molar mass

Density of solution ⍴ = 1.110 g/mol

Mass of 1 mL solution = 1.11g = m solute + m solvent

n = 2.97 × 10 –3 mol

Q11. Volume of a solution changes with change in temperature, then, will the molality solution be affected by temperature? Give reason for your answer.

The temperature has no effect on the molality of the solution because molality is expressed in mass, and mass remains constant as temperature changes.

Q12. If 4 g of NaOH dissolves in 36 g of H 2 O, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1g ml -1 )

Mass of NaOH = 4 g

Number of moles of NaOH = 4g/40g = 0.1 mol

Mass of H 2 O = 36 g

Number of moles of H 2 O = 36g/18g = 2 mol

Mass of solution = mass of water + mass of NaOH = 36g+4 g = 40 g

Volume of solution =40×1=40 mL.( Since specific gravity of solution is =1 g mL –1 )

Q13. The reactant which is entirely consumed in the reaction is known as a limiting reagent.

In the reaction 2A + 4B → 3C + 4D, when 5 moles of A react with 6 moles of B, then

(i) Which is the limiting reagent?

(ii) Calculate the amount of C formed?

2A + 4B ⟶ 3C + 4D

According to the above equation, 2 moles of ‘A’ require 4 moles of ‘B’ for the reaction. As a result, the moles of ‘B’ required for 5 moles of ‘A’ are 10 moles.

(i) The limiting agent is B, as 5 moles of A requires 10 moles of B but only 6 moles are present.

(ii) The amount of ‘C’ formed will be determined by the amount of ‘B’ formed. Since 4 moles of ‘B’ produce 3 moles of ‘C’. As a result, 6 moles of ‘B’ will produce

Matching Type Questions

Q1. Match the following:

Q2. Match the following physical quantities with units

Assertion and Reason Type Questions

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Q1. Assertion (A): The empirical mass of ethene is half of its molecular mass.

Reason (R): The empirical formula represents the simplest whole-number ratio of various atoms present in a compound.

(i) Both A and R are true and R is the correct explanation of A.

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false.

Correct Option is (i) Both A and R are true and R is the correct explanation of A.

Molecular Formula = n × Empirical formula

Empirical formula of Ethene = C 2 H 4

Empirical Formula Mass = 14 amu= ½ Molecular Mass of Ethene

The ratio of Carbon and Hydrogen in the empirical formula is 1: 2.

Q2. Assertion (A): One atomic mass unit is defined as one-twelfth of the mass of one carbon-12 atom.

Reason (R): Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as the standard.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

Since C-12 is used as the standard atom, one atomic mass unit is defined as one-twelfth of the mass of one carbon – 12 atom. This is due to the fact that it has an equal number of protons and neutrons (6) and makes up the majority of matter.

Carbon-12 is the most abundant isotope of carbon.

Q3. Assertion (A): Significant figures for 0.200 are 3 whereas for 200 it is 1.

Reason (R): Zero at the end or right of a number is significant provided they are not on the right side of the decimal point.

(1) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not a correct explanation of A.

Correct Option is (iii) A is true but R is false.

Zero at the end or to the right of a number is significant if it is on the right side of the decimal point. For example, 0.200 has 3 significant figures.

Q4. Assertion (A): Combustion of 16 g of methane gives 18 g of water.

Reason (R): In the combustion of methane, water is one of the products.

(i) Both A and R are true but R is not the correct explanation of A.

Correct Option is (iii) A is false but R is true.

CH 4 + 2O 2 → CO 2 + 2H 2 O

Water is produced during the combustion of methane, but 16 g of methane on complete combustion gives 36 g of water.

Long Answer Type Questions

Q1. A vessel contains 1.6 g of dioxygen at STP (273.15K, 1 atm pressure). The gas is now transferred to another vessel at a constant temperature, where pressure becomes half of the original pressure. Calculate

(i) volume of the new vessel.

(ii) number of molecules of dioxygen.

(i) P 1 = 1atm, P 2 = ½ = 0.5 atm, T 1 = 273.15, V 2 = ?, V 1 = ?

32 g of dioxygen occupies = 22.4 L volume at STP

V 1 = 1.12 L

From Boyle’s law (as T is constant)

P 1 V 1 = P 2 V 2

V 2 = 22.4 L

(ii) Number of moles of dioxygen = Mass of dioxygen/Molar mass of dioxygen

1 mol of dioxygen = 6.022 × 10 23 molecule of dioxygen

∴ 0.05 mol of dioxygen = 6.022 × 10 23 × 0.5 molecules of dioxygen

= 0.3011 × 10 23 molecules

= 3.011 × 10 22 molecules.

Q2. Calcium carbonate reacts with aqueous HCl to give CaCl 2 and CO 2 according to the reaction given below:

CaCO 3 (s) + 2HCl (aq) → CaCl 2 (aq) + CO 2 (g) + H 2 O(l)

What mass of CaCl 2 will be formed when 250 mL of 0.76 M HCI reacts with 1000 g of CaCO 3 ? Name the limiting reagent. Calculate the number of moles of CaCl 2 formed in the reaction.

Mass of CaCO 3 = 1000 g

Number of moles of CaCO 3 = 1000 g/100g = 10 mol

According to the equation, 1 mol of CaCO 3 requires 2 mol of HCl.

Hence, for the reaction of 10 mol of CaCO 3 number of moles of HCl required would be:

There is only 0.19 mol of HCl available, hence HCl is a limiting reagent.

Therefore, the amount of CaCl 2 formed will depend on the amount of HCl would give-

Or 0.095 × molar mass of CaCl 2 = 0.095 × 111 = 10.54 g.

Q3. Define the law of multiple proportions. Explain it with two examples. How does this law point to the existence of atoms?

Dalton first studied the law of multiple proportions in 1803, and it can be stated as follows.

When two elements combine to form two or more chemical compounds, the masses of one of the elements combine with a fixed mass of the other in a simple ratio.

For example, hydrogen reacts with oxygen to form two compounds: water and hydrogen peroxide.

In this case, the masses of oxygen (i.e. 16g and 32g) that combine with a fixed mass of hydrogen (2g) have a simple ratio, i.e. 16:32 or 1:2.

As we all know, when elements are mixed in different proportions, they form different compounds. For example, when hydrogen is mixed with a different proportion of oxygen, it forms water or hydrogen peroxide.

It demonstrates that there are constituents that combine in a specific manner. These constituents could be atoms. As a result, the law of multiple proportions demonstrates the existence of atoms that combine to form molecules.

Q4. A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labelled as B, each weighing 5 grams. Consider the combinations AB, AB 2 , A 2 B and A 2 B 3 , and show that a law of multiple proportions is applicable.

When two elements combine to form two or more compounds, the different masses of one element that combine with a fixed mass of the other bear a simple ratio to one another, according to the law of multiple proportions.

The mass of B when combined with a fixed mass of A (say 1g) is 2.5g, 5g, 1.25g, and 3.75g. They have a 2:4:1 ratio, which is a simple whole-number ratio. Hence. The multiple proportions law is applicable.

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how to calculate the atomic mass of isotopes?

The atomic mass of isotopes are calculated by finding the no.of protons and neutrons and multiplying by 1amu

the sum of the numbers of protons and neutrons in the nucleus is called the mass number. This is because each proton and each neutron weigh one atomic mass unit (amu). By adding together the number of protons and neutrons and multiplying by 1 amu, you can calculate the mass of the atom. it may help u

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Some Basic Concepts of Chemistry Class 11 Chapter 1 Notes is an introductory chapter but very crucial to understand for students as it forms the basis of Chemical reactions happens around us. To understand these concepts, Class 11th Chemistry Chapter 1 Notes are prepared by subject experts in well-defined and easy language. The goal of Chemistry Class 11 Chapter 1 notes is to study just one topic and make short, readable notes to help you quickly remember the core points before the test. This is how you won't feel like panicking before the exam.

Class 11th Chemistry Chapter 1 Notes are available in PDF file on the official website of Vedantu and its app to download for free by the students. Students can use Chemistry ch 1 Class 11 notes in PDF file both offline and online to understand the critical terms and concepts in easy to understand language with examples, diagrammatic explanations and solved problems.

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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 ( 12 C) taken as 12.

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

Ans. No. of atoms in 1 mole = 6.022×10 23

No. of atoms in 3 moles = 3×6.022×10 23

= 1.8066×10 24 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 C 2 H 6 = 2 × (Atomic mass of C) 

+ 6 × (Atomic mass of H)

Molecular mass of ethane(C 2 H 6 ) = 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. H 2 O, CO 2

Section – B (2 Marks Questions)

11. Calculate the molar mass of the following:

(i) H 2 O   (ii) NH 3

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 )

= 18 g mol -1

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

= (14 g mol -1 ) + (3×1 g mol -1 )

= 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 × 10 23 atoms = 6.023×10 23 × 4.2×10 –22 g

= 6.023 × 4.2 × 10g

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

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

(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?

(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 cm 3 . 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 / cm 3

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 cm 3 . Find out the number of molecules in it?

Ans. Given, V =6.57 × 10 –8 cm 3

∵ 22400 cm 3 of the gas at STP has  

= 6.02 × 10 23 molecules

∴ 6.57 × 10 –8 cm 3 of the gas at STP will have 

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

= 1.765 × 10 12  molecules

18. If 8.7g of Na 2 CO 3 is added to 18g of CH 3 COOH  solution, the residue is found to weigh 22g. What is the mass of CO 2 released in the reaction?

Ans. Na 2 CO 3 + CH 3 COOH → CO 2 + H 2 O + CH 3 COONa

According to law of conservation of mass,

Mass of reactants = mass of products 

8.7 + 18 = 22 + x 

x = 26.7 – 22

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 ZnSO 4 is obtained from 

= 22.65g of Zn

20g of ZnSO 4 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.

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.

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:

Matter consists of indivisible atoms.

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

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

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

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.

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.

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.

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 CO 2   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

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

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

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 : \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\]

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 

Weight 

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

What does Avogadro’s law state?

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 M 3 O 4 .

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

What differentiates physical properties from chemical properties?

What is the benefit of using Molality over Molarity?

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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 2023-24 PDF

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

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Notes of Chapter 1 Chemistry Class 11 are beneficial to clear fundamentals of basic concepts of Chemistry to clear any scientific or research-related exam. Also, this chapter helps students understand the basic Chemistry and Class 11 Chemistry Chapter 1 notes make this process easy for students.

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Organic Chemistry and some basic principle and technique  

Redox Reaction

Equilibrium

Thermodynamics

Chemical Bonding and Molecular Structure 

State of matter

4. What are some basic concepts covered in Chapter 1 of Chemistry of Class 11?

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5. What is the name of Chapter 1 of Chemistry of Class 11?

The Chapter 1 of Chemistry of Class 11 is ‘Basic concepts of Chemistry’. This chapter gives an introduction to Chemistry and why it is an important part of science. Students can find NCERT Revision notes of Chapter 1 of Chemistry of Class 11 from Vedantu website as well as Vedantu Mobile app. The online notes for Chapter 1 of Chemistry of Class 11 will help students to understand the chapter and fundamentals of Chemistry. They can understand the important topics to be covered in Chemistry in the complete syllabus.

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7. What are the different branches of chemistry?

There are two main branches:

Inorganic Chemistry: Deals with compounds that don't contain carbon-hydrogen bonds, like metals, salts, and many industrial chemicals.

Organic Chemistry: Focuses on the structure, composition, and properties of compounds containing carbon and hydrogen atoms.

8. What is Dalton's Atomic Theory?

This theory, proposed by John Dalton in 1808, laid the foundation for modern atomic theory. It suggests that:

Matter is made of tiny indivisible particles called atoms.

Atoms of a specific element are identical in properties and mass.

Atoms of different elements have different masses.

Compounds form when atoms of different elements combine in fixed ratios.

Chemical reactions involve rearrangement of atoms, which are neither created nor destroyed.

CBSE Study Materials for Class 11