NCERT Exemplar Solutions-Basic Concepts of Chemistry

The NCERT Exemplar Chemistry Class 11 Chapter Some Basic Concepts of Chemistry is necessary to study for students to score good marks in Class 11 and entrance examinations.

NCERT Exemplar Class 11 Chemistry Chapter Basic Concepts of Chemistry has important questions with answers, exercise solutions, numerical, MCQs, HOTS, worksheets and questions from previous years’ question papers and sample papers.

NCERT Exemplar Solutions Class 11 Chemistry Chapter Some Basic Concepts of Chemistry

Multiple Choice Questions (Type-1)

1. Two students performed the same experiment separately and each one of

them recorded two readings of mass which are given below. The correct reading

of mass is 3.0 g. Based on given data, mark the correct option out of the

following statements.

Student Readings

 (i)(ii)
A3.012.99
B3.052.95

(i) Results of both the students are neither accurate nor precise.

(ii) Results of student A are both precise and accurate.

(iii) Results of student B are neither precise nor accurate.

(iv) Results of student B are both precise and accurate.

Solution:

Option (ii) is the answer.

2. A measured temperature on the Fahrenheit scale is 200 °F. What will this reading

be on a Celsius scale?

(i) 40 °C

(ii) 94 °C

(iii) 93.3 °C

(iv) 30 °C

Solution:

Option (iii) is the answer.

3. What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per

500 mL?

(i) 4 mol L-1

(ii) 20 molL-1

(iii) 0.2 molL-1

(iv) 2molL-1

Solution:

Option (iii) is the answer.

4. If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of

the solution obtained?

(i) 1.5 M

(ii) 1.66 M

(iii) 0.017 M

(iv) 1.59 M

Solution:

Option (ii) is the answer.

5. The number of atoms present in one mole of an element is equal to Avogadro

number. Which of the following element contains the greatest number of

atoms?

(i) 4g He

(ii) 46g Na

(iii) 0.40g Ca

(iv) 12g He

Solution:

Option (iv) is the answer.

6. If the concentration of glucose (C6H12O6) in the blood is 0.9 g L-1, what will be the

molarity of glucose in the blood?

(i) 5 M

(ii) 50 M

(iii) 0.005 M

(iv) 0.5 M

Solution:

Option (iii) is the answer.

7. What will be the molality of the solution containing 18.25 g of HCl gas in

500 g of water?

(i) 0.1 m

(ii) 1 M

(iii) 0.5 m

(iv) 1 m

Solution:

Option (iv) is the answer.

8. One mole of any substance contains 6.022 × 1023 atoms/molecules. Number

of molecules of H2SO4 present in 100 mL of 0.02M H2SO4 solution is ______.

(i) 12.044 × 1020 molecules

(ii) 6.022 × 1023 molecules

(iii) 1 × 1023 molecules

(iv) 12.044 × 1023molecules

Solution:

Option (i) is the answer.

9. What is the mass per cent of carbon in carbon dioxide?

(i) 0.034%

(ii) 27.27%

(iii) 3.4%

(iv) 28.7%

Solution:

Option (ii) is the answer.

10. The empirical formula and molecular mass of a compound are CH2O and

180 g respectively. What will be the molecular formula of the compound?

(i) C9H18O9

(ii) CH2O

(iii) C6H12O6

(iv) C2H4O2

Solution:

Option (iii) is the answer.

11. If the density of a solution is 3.12 g mL-1, the mass of 1.5 mL solution in

significant figures is _______.

(i) 4.7g

(ii) 4680 × 10 -3g

(iii) 4.680g

(iv) 46.80g

Solution:

Option (i) is the answer

12. Which of the following statements about a compound is incorrect?

(i) A molecule of a compound has atoms of different elements.

(ii) A compound cannot be separated into its constituent elements by

physical methods of separation.

(iii) A compound retains the physical properties of its constituent elements.

(iv) The ratio of atoms of different elements in a compound is fixed.

Solution:

Option (iii) is the answer.

13. Which of the following statements is correct about the reaction given below:

4Fe(s) + 3O2(g) → 2Fe2O3(g)

(i) The total mass of iron and oxygen in reactants = total mass of iron and

oxygen in product therefore it follows the law of conservation of mass.

(ii) The total mass of reactants = total mass of product; therefore, the law of multiple

proportions is followed.

(iii) Amount of Fe2O3 can be increased by taking any one of the reactants

(iron or oxygen) in excess.

(iv) Amount of Fe2O3 produced will decrease if the amount of any one of the

reactants (iron or oxygen) is taken in excess.

Solution:

Option (i) is the answer.

14. Which of the following reactions is not correct according to the law of

conservation of mass.

(i) 2Mg(s) + O2(g) →2MgO(s)

(ii) C3H8(g) + O2(g) → CO2(g) + H2O(g)

(iii) P4(s) + 5O2(g) → P4O10(s)

(iv) CH4(g) + 2O2(g) → CO2(g) + 2H2O (g)

Solution:

Option (ii) is the answer.

15. Which of the following statements indicates that the law of multiple proportions is

being followed.

(i) Sample of carbon dioxide taken from any source will always have carbon

and oxygen in the ratio 1:2.

(ii) Carbon forms two oxides namely CO2 and CO, where masses of oxygen

which combine with a fixed mass of carbon are in the simple ratio 2:1.

(iii) When magnesium burns in oxygen, the amount of magnesium taken

for the reaction is equal to the amount of magnesium in magnesium

oxide formed.

(iv) At constant temperature and pressure, 200 mL of hydrogen will combine

with 100 mL oxygen to produce 200 mL of water vapour.

Solution:

Option (ii) is the answer.

Multiple Choice Questions (Type-11)

In the following questions, two or more options may be correct.

16. One mole of oxygen gas at STP is equal to _______.

(i) 6.022 × 1023 molecules of oxygen

(ii) 6.022 × 1023 atoms of oxygen

(iii) 16 g of oxygen

(iv) 32 g of oxygen

Solution:

Option (i) and (iv) are the answers.

17. Sulphuric acid reacts with sodium hydroxide as follows:

H2SO4 + 2NaOH → Na2SO4+ 2H2O

When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M

sodium hydroxide solution, the amount of sodium sulphate formed and its

molarity in the solution obtained is

(i) 0.1 mol L-1

(ii) 7.10 g

(iii) 0.025 mol L-1

(iv) 3.55 g

Solution:

Option (ii) and (iii) are the answers.

18. Which of the following pairs have the same number of atoms?

(i) 16 g of O2(g) and 4 g of H2(g)

(ii) 16 g of O2 and 44 g of CO2

(iii) 28 g of N2 and 32 g of O2

(iv) 12 g of C(s) and 23 g of Na(s)

Solution:

Option (iii) and (iv) are the answers.

19. Which of the following solutions have the same concentration?

(i) 20 g of NaOH in 200 mL of solution

(ii) 0.5 mol of KCl in 200 mL of solution

(iii) 40 g of NaOH in 100 mL of solution

(iv) 20 g of KOH in 200 mL of solution

Solution:

Option (i) and (ii) are the answers.

20. 16 g of oxygen has the same number of molecules as in

(i) 16 g of CO

(ii) 28 g of N2

(iii) 14 g of N2

(iv) 1.0 g of H2

Solution:

Option (iii) and (iv) are the answers.

21. Which of the following terms is unitless?

(i) Molality

(ii) Molarity

(iii) Mole fraction

(iv) Mass per cent

Solution:

Option (iii) and (iv) are the answers.

22. One of the statements of Dalton’s atomic theory is given below:

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

ratio”

Which of the following laws is not related to this statement?

(i) Law of conservation of mass

(ii) Law of definite proportions

(iii) Law of multiple proportions

Solution:

Option (i) and (iv)

III. Short Answer Type

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

Solution:

1 mole of carbon atom = 12g= 6.022 × 1023 atoms.

24. How many significant figures should be present in the answer to the following

calculations?

2.5 1.25 3.5/2.01

Solution:

Two significant figures should be present in this.

Since the least number of significant figures from the given figure is 2 (in 2.5 and 3.5).

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

Solution:

The symbol for the SI unit of the mole is mol. A mole is defined as the amount of substance that contains as many entities as there are atoms in 12g carbon.

26. What is the difference between molality and molarity?

Solution:

Molarity is the number of moles of solute dissolved in 1 litre of the solution. Molality is the number of moles of solute present in 1kg of the solvent.

27. Calculate the mass percent of calcium, phosphorus and oxygen in calcium

phosphate Ca3(PO4)

Solution:

Molecular mass of Ca3(PO4) = 3*40+2*31+8*16 =310

Mass per cent of Ca = 3*40/310*100 = 38.71%

Mass per cent of P = 2*31/310*100 = 20%

Mass per cent of O = 8*16/310 = 41.29%

28. 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:

2N2(g) + O2(g) → 2N2O(g)

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

Solution:

The above experiment proves Gay-Lussac’s law which states that gases combine or produced in a chemical reaction in a simple whole-number ratio by volume provided that all gases are the same temperature and pressure.

29. 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 whole-number ratio.

(a) Is this statement true?

(b) If yes, according to which law?

(c) Give one example related to this law.

Solution:

(a) Yes, the statement is true.

(b) According to the law of multiple proportions

(c), Hydrogen and oxygen react to form water and hydrogen peroxide

H2 + 1/2O2 → H2O

H2 + O2 → H2O2

Masses of oxygen which combine the fixed mass of hydrogen are in the ratio 16:32 or 1

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

Isotope %Natural abundanceMolar mass
1H99.9851
2H0.0152

Solution:

Average atomic mass = 99.985*1+0.015*2/100

=099.985*1+0.015*2/100

=1.00015u

31. Hydrogen gas is prepared in the laboratory by reacting dilute HCl with

granulated zinc. Following reaction takes place.

Zn + 2HCl → ZnCl2 + H2

Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc

reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of

Zn = 65.3 u.

Solution:

1 mol of gas occupies = 22.7L Volume at STP atomic mass of Zn = 65.3u

From the above equation,

65.3g of Zn when reacts with HCl produces = 22.7L H2 at STP

Therefore, 32.65g of Zn when reacts with HCl will produce = 22.7 * 32.65/65.3 =11.35L of H2 at STP

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

of the solution.

Solution:

3 molal solution of NaOH = 3 moles of NaOH dissolved in 1000g water

3 mole of NaOH = 3*40g = 120g

Density of solution = 1.110gmL-1

Volume = mass/density = 1120g/1.110gmL-1 =1.009L

Molarity of the solution = 3/1.009 = 2.97M

33. The volume of a solution changes with change in temperature, then, will the molality

of the solution be affected by temperature? Give a reason for your answer.

Solution:

Mass does not change as the temperature changes. Therefore, the molality of a solution does not change.

Molality = moles of solute/ weight of solvent (in g) *1000

34. If 4 g of NaOH dissolves in 36 g of H2O, calculate the mole fraction of each

component in the solution. Also, determine the molarity of the solution (specific

gravity of solution is 1g mL–1).

Solution:

Mole fraction of H2O = No; of moles of H2O/ Total no: of moles (H2O+NaOH)

No: of moles of H2O = 36/18=2moles

No: of moles of NaOH = 4/40=0.1mol

Total no: of moles = 2+0.1= 2.1

Mole fraction of H2O = 2/2.1 = 0.952

Mole fraction of NaOH = 0.1/2.1 = 0.048

Mass of solution = Mass of H2O + Mass of NaOH = 36+4=40G

Volume of the solution = 40/1 = 40mL

Molarity = 0.1/0.04 = 2.5M

35. The reactant which is entirely consumed in the reaction is known as 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?

Solution;

(i) B will be the limiting reagent as it gives a lesser amount of product.

(ii) Let B is completely consumed

4 mol B gives 3 mol C

6 mol B will give 3/4 *6 mol C =4.5 mol C

Match The Following Type

36.

(i) 88 g of CO2 (ii) 6.022 ×1023 molecules of H2O (iii) 5.6 litres of O2 at STP (iv) 96 g of O2 (v) 1 mol of any gas(a) 0.25 mol (b) 2 mol (c) 1 mol (d) 6.022 × 1023 molecules (e) 3 mol

Solution:

A → b

B → c

C → a

D → e

E → d

37. Match the following

Physical quantityUnit
(i) Molarity (ii) Mole fraction (iii) Mole (iv) Molality (v) Pressure (vi) Luminous intensity (vii) Density (viii) Mass(a) g mL–1 (b) mol (c) Pascal (d) Unitless (e) mol L–1 (f) Candela (g) mol kg–1 (h) Nm–1 (i) kg

Solution:

(i → e)

(ii → d)

(iii → b)

(iv → g)

(v → c)

(vi → f)

(vii → a)

(viii → i)

V. Assertion and Reason Type

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 are given below each question.

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

Reason (R): The empirical formula represents the simplest whole-number

the 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.

Solution:

Option (i) is correct.

39. 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.

(i) 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 the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Solution:

Option (ii) is correct. Carbon-12 is considered a standard for defining the atomic and molecular mass.

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

Reason (R): Zero at the end or right of a number are significantly provided

they are not on the right side of the decimal point.

(i) 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.

(iii) A is true but R is false.

(iv) Both A and R are false.

Solution:

Option (iii) is correct. Significant figures for 0.200 = 3 and for 200 =1

Zero at the end of a number without decimal point may or may not be significant depending on the accuracy of the measurement.

41. 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.

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false.

Solution:

Option (iii) is correct.

16g of CH4 on complete combustion will give 36g of water.

Long Answer Type Question

42. 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

the pressure becomes half of the original pressure. Calculate

(i) the volume of the new vessel.

(ii) a number of molecules of dioxygen.

Solution:

(i) Moles of oxygen = 1.6/32 = 0.05mol

At STP, 1 mol of O2 = 22.4L

Then volume of O2 = 22.4 × 0.05 = 1.12L

V1 = 1.12L

V2 =?

P1 = 1atm

P2 = ½ = 0.5atm

According to Boyle’s law, p1V1 = p2V2

Substituting the values

V2 = 1 × 1.12/0.5 = 2.24L

(ii) No of molecules in 1.6g or 0.005mol = 6.022 × 1023 × 0.05 = 3.011 ×  1022

43. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according

to the reaction given below:

CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)

What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with

1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles

of CaCl2 formed in the reaction.

Solution:

No: of moles of HCl taken = MV/1000 = 0.76*250/1000 = 0.19

No: of moles of CaCO3 = Mass/Molar mass = 1000/100 = 10

1. When CaCO3 is completely consumed

1 mol of CaCO3 = 1 mol CaCl2

10 mol CaCO3 = 10mol CaCl2

2. When HCl is completely consumed.

2 mol HCl = 1 mol CaCl2

0.19mol HCl = ½  × 0.19mol CaCl2 = 0.095 mol CaCl2

HCl will be the limiting reagent and the number of moles of CaCl2 formed will be 0.095mol

44. Define the law of multiple proportions. Explain it with two examples. How

does this law point to the existence of atoms?

Solution:

When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other bear a simple ratio to one another is the law of multiple proportions.

For example, carbon combines with oxygen to form two compounds they are carbon dioxide and carbon monoxide

The masses of oxygen which combine with a fixed mass of carbon in carbon dioxide and carbon monoxide are 32 and 16. Therefore oxygen bear: 32:16 ratio or 2:1

Example 2: Sulphur combines with oxygen to form sulphur trioxide and sulphur dioxide

The masses of oxygen which combine with a fixed mass of sulphur in SO3 and SO2 are 48 and 32. Therefore oxygen bear a ratio of 48:32 or 3:2

45. 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, AB2, A2B and A2B3

and show that the law of multiple proportions is applicable.

Solution:

 ABab2A,BA2B3
Mass of A (in g)224415
Mass of B (in g)5105 

Masses of B which combine with a fixed mass of A are

10g, 20g, 5g, 15g

2 : 4 : 1 : 3

This is the simple whole-number ratio.


Important Topics of Chemistry Chapter Some Basic Concepts of Chemistry

  • Importance of Chemistry
  • Nature of Matter
  • Properties of Matter and Their Measurement
    • The International System of Units (Si)
    • Mass and Weight
  • Uncertainty in Measurement
    • Scientific Notation
    • Significant Figures
    • Dimensional Analysis
  • Laws of Chemical Combinations
    • Law of Conservation of Mass
    • Law of Definite Proportions
    • Law of Multiple Proportions
    • Gay Lussac’s Law of Gaseous Volumes
    • Avogadro Law
  • Dalton’s Atomic Theory
  • Atomic and Molecular Masses
    • Atomic Mass
    • Average Atomic Mass
    • Molecular Mass
    • Formula Mass
  • Mole Concept and Molar Masses
  • Percentage Composition
  • Empirical Formula for Molecular Formula
  • Stoichiometry and Stoichiometric Calculations
    • Limiting Reagent
    • Reactions in Solutions

At ANAND CLASSES (A School Of Competitions), students are provided with sample papers, previous years’ question papers, notes, exemplars, study materials, exercises, worksheets, tips, and tricks to help them prepare for their Class 11 exam and entrance exams in a more effective way.

Frequently Asked Questions on NCERT Exemplar Solutions for Class 11 Chemistry Chapter Some Basic Concepts of Chemistry

Q1

What concepts can I learn using NCERT Exemplar Solutions for Class 11 Chemistry Chapter 1?

By using NCERT Exemplar Solutions for Class 11 Chemistry Chapter 1, you can learn about concepts like
1. Importance of Chemistry
2. Nature of Matter
3. Properties of Matter and Their Measurement
4. Uncertainty in Measurement
5. Laws of Chemical Combinations
6. Dalton’s Atomic Theory
7. Atomic and Molecular Masses
8. Mole Concept and Molar Masses
9. Percentage Composition
10. Empirical Formula for Molecular Formula
11. Stoichiometry and Stoichiometric Calculations

Q2

Explain the concept of the law of definite proportions of NCERT Exemplar Solutions for Class 11 Chemistry.

The law of constant proportions states that chemical compounds are made up of elements that are present in a fixed ratio by mass. This implies that any pure sample of a compound, no matter the source, will always consist of the same elements that are present in the same ratio by mass. The law of constant proportions is often referred to as Proust’s law or as the law of definite proportions. NCERT Exemplar Solutions act as a major source of reference material for the Class 11 students from the exam perspective. The solutions are student-friendly and help them to get a grip on the important concepts.

Q3

List out the postulates of Dalton’s Atomic Theory discussed in Chapter 1 of NCERT Exemplar Solutions for Class 11 Chemistry.

Postulates of Dalton’s Atomic Theory are as follows:
1. All matter is made up of tiny, indivisible particles called atoms.
2. All atoms of a specific element are identical in mass, size and other properties. However, atoms of different elements exhibit different properties and vary in mass and size.
3. Atoms can neither be created nor destroyed. Furthermore, atoms cannot be divided into smaller particles.
4. Atoms of different elements can combine with each other in fixed whole-number ratios in order to form compounds.
5. Atoms can be rearranged, combined or separated in chemical reactions.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

CBSE Class 11 Chemistry Syllabus

CBSE Class 11 Chemistry Syllabus is a vast which needs a clear understanding of the concepts and topics. Knowing CBSE Class 11 Chemistry syllabus helps students to understand the course structure of Chemistry.

Unit-wise CBSE Class 11 Syllabus for Chemistry

Below is a list of detailed information on each unit for Class 11 Students.

UNIT I – Some Basic Concepts of Chemistry

General Introduction: Importance and scope of Chemistry.

Nature of matter, laws of chemical combination, Dalton’s atomic theory: concept of elements,
atoms and molecules.

Atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry.

UNIT II – Structure of Atom

Discovery of Electron, Proton and Neutron, atomic number, isotopes and isobars. Thomson’s model and its limitations. Rutherford’s model and its limitations, Bohr’s model and its limitations, concept of shells and subshells, dual nature of matter and light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbitals, quantum numbers, shapes of s, p and d orbitals, rules for filling electrons in orbitals – Aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of atoms, stability of half-filled and completely filled orbitals.

UNIT III – Classification of Elements and Periodicity in Properties

Significance of classification, brief history of the development of periodic table, modern periodic law and the present form of periodic table, periodic trends in properties of elements -atomic radii, ionic radii, inert gas radii, Ionization enthalpy, electron gain enthalpy, electronegativity, valency. Nomenclature of elements with atomic number greater than 100.

UNIT IV – Chemical Bonding and Molecular Structure

Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization, involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules(qualitative idea only), Hydrogen bond.

UNIT V – Chemical Thermodynamics

Concepts of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics – internal energy and enthalpy, measurement of U and H, Hess’s law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution. Second law of Thermodynamics (brief introduction)
Introduction of entropy as a state function, Gibb’s energy change for spontaneous and nonspontaneous processes.
Third law of thermodynamics (brief introduction).

UNIT VI – Equilibrium

Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium – Le Chatelier’s principle, ionic equilibrium- ionization of acids and bases, strong and weak electrolytes, degree of ionization,
ionization of poly basic acids, acid strength, concept of pH, hydrolysis of salts (elementary idea), buffer solution, Henderson Equation, solubility product, common ion effect (with illustrative examples).

UNIT VII – Redox Reactions

Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions, in terms of loss and gain of electrons and change in oxidation number, applications of redox reactions.

UNIT VIII – Organic Chemistry: Some basic Principles and Techniques

General introduction, classification and IUPAC nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyper conjugation. Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions.

UNIT IX – Hydrocarbons

Classification of Hydrocarbons
Aliphatic Hydrocarbons:
Alkanes – Nomenclature, isomerism, conformation (ethane only), physical properties, chemical reactions.
Alkenes – Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markovnikov’s addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition.
Alkynes – Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of – hydrogen, halogens, hydrogen halides and water.

Aromatic Hydrocarbons:

Introduction, IUPAC nomenclature, benzene: resonance, aromaticity, chemical properties: mechanism of electrophilic substitution. Nitration, sulphonation, halogenation, Friedel Craft’s alkylation and acylation, directive influence of functional group in monosubstituted benzene. Carcinogenicity and toxicity.

To know the CBSE Syllabus for all the classes from 1 to 12, visit the Syllabus page of CBSE. Meanwhile, to get the Practical Syllabus of Class 11 Chemistry, read on to find out more about the syllabus and related information in this page.

CBSE Class 11 Chemistry Practical Syllabus with Marking Scheme

In Chemistry subject, practical also plays a vital role in improving their academic scores in the subject. The overall weightage of Chemistry practical mentioned in the CBSE Class 11 Chemistry syllabus is 30 marks. So, students must try their best to score well in practicals along with theory. It will help in increasing their overall academic score.

CBSE Class 11 Chemistry Practical Syllabus

The experiments will be conducted under the supervision of subject teacher. CBSE Chemistry Practicals is for 30 marks. This contribute to the overall practical marks for the subject.

The table below consists of evaluation scheme of practical exams.

Evaluation SchemeMarks
Volumetric Analysis08
Salt Analysis08
Content Based Experiment06
Project Work04
Class record and viva04
Total30

CBSE Syllabus for Class 11 Chemistry Practical

Micro-chemical methods are available for several of the practical experiments. Wherever possible such techniques should be used.

A. Basic Laboratory Techniques
1. Cutting glass tube and glass rod
2. Bending a glass tube
3. Drawing out a glass jet
4. Boring a cork

B. Characterization and Purification of Chemical Substances
1. Determination of melting point of an organic compound.
2. Determination of boiling point of an organic compound.
3. Crystallization of impure sample of any one of the following: Alum, Copper Sulphate, Benzoic Acid.

C. Experiments based on pH

1. Any one of the following experiments:

  • Determination of pH of some solutions obtained from fruit juices, solution of known and varied concentrations of acids, bases and salts using pH paper or universal indicator.
  • Comparing the pH of solutions of strong and weak acids of same concentration.
  • Study the pH change in the titration of a strong base using universal indicator.

2. Study the pH change by common-ion in case of weak acids and weak bases.

D. Chemical Equilibrium
One of the following experiments:

1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing the concentration of either of the ions.
2. Study the shift in equilibrium between [Co(H2O)6] 2+ and chloride ions by changing the concentration of either of the ions.

E. Quantitative Estimation
i. Using a mechanical balance/electronic balance.
ii. Preparation of standard solution of Oxalic acid.
iii. Determination of strength of a given solution of Sodium hydroxide by titrating it against standard solution of Oxalic acid.
iv. Preparation of standard solution of Sodium carbonate.
v. Determination of strength of a given solution of hydrochloric acid by titrating it against standard Sodium Carbonatesolution.

F. Qualitative Analysis
1) Determination of one anion and one cation in a given salt
Cations‐ Pb2+, Cu2+, As3+, Al3+, Fe3+, Mn2+, Ni2+, Zn2+, Co2+, Ca2+, Sr2+, Ba2+, Mg2+, NH4 +
Anions – (CO3)2‐ , S2‐, NO2 , SO32‐, SO2‐ , NO , Cl , Br, I‐, PO43‐ , C2O2‐ ,CH3COO
(Note: Insoluble salts excluded)

2) Detection of ‐ Nitrogen, Sulphur, Chlorine in organic compounds.

G) PROJECTS
Scientific investigations involving laboratory testing and collecting information from other sources.

A few suggested projects are as follows:

  • Checking the bacterial contamination in drinking water by testing sulphide ion
  • Study of the methods of purification of water.
  • Testing the hardness, presence of Iron, Fluoride, Chloride, etc., depending upon the regional
    variation in drinking water and study of causes of presence of these ions above permissible
    limit (if any).
  • Investigation of the foaming capacity of different washing soaps and the effect of addition of
    Sodium carbonate on it.
  • Study the acidity of different samples of tea leaves.
  • Determination of the rate of evaporation of different liquids Study the effect of acids and
    bases on the tensile strength of fibres.
  • Study of acidity of fruit and vegetable juices.

Note: Any other investigatory project, which involves about 10 periods of work, can be chosen with the approval of the teacher.

Practical Examination for Visually Impaired Students of Class 11

Below is a list of practicals for the visually impaired students.

A. List of apparatus for identification for assessment in practicals (All experiments)
Beaker, tripod stand, wire gauze, glass rod, funnel, filter paper, Bunsen burner, test tube, test tube stand,
dropper, test tube holder, ignition tube, china dish, tongs, standard flask, pipette, burette, conical flask, clamp
stand, dropper, wash bottle
• Odour detection in qualitative analysis
• Procedure/Setup of the apparatus

B. List of Experiments A. Characterization and Purification of Chemical Substances
1. Crystallization of an impure sample of any one of the following: copper sulphate, benzoic acid
B. Experiments based on pH
1. Determination of pH of some solutions obtained from fruit juices, solutions of known and varied
concentrations of acids, bases and salts using pH paper
2. Comparing the pH of solutions of strong and weak acids of same concentration.

C. Chemical Equilibrium
1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing
the concentration of eitherions.
2. Study the shift in equilibrium between [Co(H2O)6]2+ and chloride ions by changing the
concentration of either of the ions.

D. Quantitative estimation
1. Preparation of standard solution of oxalic acid.
2. Determination of molarity of a given solution of sodium hydroxide by titrating it against standard
solution of oxalic acid.

E. Qualitative Analysis
1. Determination of one anion and one cation in a given salt
2. Cations – NH+4
Anions – (CO3)2-, S2-, (SO3)2-, Cl-, CH3COO-
(Note: insoluble salts excluded)
3. Detection of Nitrogen in the given organic compound.
4. Detection of Halogen in the given organic compound.

Note: The above practicals may be carried out in an experiential manner rather than recording observations.

We hope students must have found this information on CBSE Syllabus useful for their studying Chemistry. Learn Maths & Science in interactive and fun loving ways with ANAND CLASSES (A School Of Competitions) App/Tablet.

Frequently Asked Questions on CBSE Class 11 Chemistry Syllabus

Q1

How many units are in the CBSE Class 11 Chemistry Syllabus?

There are 9 units in the CBSE Class 11 Chemistry Syllabus. Students can access various study materials for the chapters mentioned in this article for free at ANAND CLASSES (A School Of Competitions).

Q2

What is the total marks for practicals examination as per the CBSE Class 11 Chemistry Syllabus?

The total marks for the practicals as per the CBSE Class 11 Chemistry Syllabus is 30. It includes volumetric analysis, content-based experiment, salt analysis, class record, project work and viva.

Q3

Which chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry?

The organic chemistry chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry.