NCERT Exemplar Class 11 Chemistry solutions chapter 1 Some Basic Concepts of Chemistry provides detailed explanations on the importance of some of the basic concepts and laws of Chemistry. These laws include Avogadro’s Law, Law of Conservation of Mass, and many more. NCERT Exemplar Class 11 Chemistry chapter 1 solutions have been drafted by the experts in such a way that anyone who goes through it gets a deep insight into the topics related to the basic terms used in Chemistry. The experts have provided solutions for numerous questions in a detailed step-by-step process to ensure understanding of the students on the various topics covered.
What is the mass percent of carbon in carbon dioxide?
The answer is the option (ii) 27.27%Explanation:
Molar mass of
Therefore, 1 g of
contains 1 g of carbon
44 g of
therefore contains 12 g of carbon.
% of C in
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.
The answer is the option (iii) A compound retains the physical properties of its constituent elements.
The compound can be completely indifferent and in contrast to the physical properties of its constituent elements.
Which of the following statements is correct about the reaction given below:
(i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows Law of conservation of mass.
(ii) Total mass of reactants = total mass of product; therefore, Law of multiple proportions is followed.
(iii) Amount of can be increased by taking any one of the reactants (iron or oxygen) in excess.
(iv) Amount of produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.
The answer is the option (i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows Law of conservation of mass.Explanation:
Keeping the Law of conservation in mind, we evaluate both the sides of every equation. The mass of the reactants should be equal to the of the product, and this is only followed in equation (i), therefore it is correct.
Which of the following statements indicates that Law of multiple proportion is being followed?
(i) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the simple ratio 1:2,
(ii) Carbon forms two oxides namely and . where masses of oxygen which combine with 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.
The answer is the option (ii) Carbon forms two oxides namely
. where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1.Explanation:
According to the Law of multiple proportions,
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 2)
One mole of oxygen gas at STP is equal to ________
(i) molecules of oxygen
(ii) atoms of oxygen
(iii) 16 g of oxygen
(iv) 32 g of oxygen
The answer are the options (i) and (iv)
1 mole of oxygen gas at STP =
molecules of oxygen (which is the Avogadro number)
Thus, one mole of
is equal to the molecular weight of oxygen and the Avogadro number.
Sulphuric acid reacts with sodium hydroxide as follows :
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 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
The answer are the options (ii) and (iii)
Given, (0.1 M), Moles of H2
= 0.1 moles
Given, (0.1M), Moles of NaOH = 0.1 moles
The ratio of H2
and NaOH is 1:2, using the ratio, we figure for 0.1 mole of H2
, 0.2 moles of NaOH is used, which is not available.
Therefore if 0.1 moles of NaOH reacts, then only
moles of H2
is used. NaOH is the limiting reactant, and the reaction and products are evaluated according to the limiting reactant.
Therefore the number of moles formed of Na2
will be 0.05 as well.
This is because 2 moles of NaOH produces 1 mole of H2
, therefore 1 mole of NaOH will logically produce 0.05 moles of Na2
Therefore, mass of Na2
= 0.05 × (molar mass)
0.05 × (23 ×2 + 32 + 16× 4)
0.05 × 142 = 7.1 g
Molarity = Number of moles / Volume in L
Molarity = 0.05/ 2 L = 0.023 mol/L
Which of the following pairs have the same number of atoms ?
(i) 16 g of (g) and 4 g of (g)
(ii) 16 g of and 44 g of
(iii) 28 g of and 32 g of
(iv) 12 g of C (s) and 23 g of Na(s)
The answer are the options (iii) and (iv)
One mole is measured as the amount of a substance that equal number of atoms as contained in exactly 12 g of Carbon.
1 mole of C = 12g C
While 23 grams of Na (mass) = 1 mole of Na
= 1 mole of nitrogen = 2 × atoms of nitrogen
atoms of nitrogen
= 1 mole of oxygen =
atoms of oxygen
Which of the following solutions have the same concentration?
(i) 20 g of NaOH in 200 mL of solution
(ii) 0.5 mole 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
The answer are the options (i) and (ii)Explanation:
(i) 20 g NaOH in 200 mL solution.
mol of NaOH
(ii) Molar concentration Of NaOH
as 200 mL = 0.2 L
Molar concentration of KCl =
16 g of oxygen has same of molecules as in
(i) 16 g of CO
(ii) 28 g of
(iii) 14 g of
(iv)) 1.0 g of
The answer are the options (iii) and (iv)Explanation:
1 mole or the number of molecules of oxygen in 16g Of oxygen is
In 14g or N2
and 1 g of H2
contain 0.5 moles. Therefore they will also contain
Number of molecules of N2
Number of molecules of H2
Which of the following terms are unitless?
(iii) Mole fraction
(iv) Mass percent
The answer are the options (iii) and (iv)Explanation:
Mole Fraction: It is the particular ratio of the substance in the solution in comparison to the total components of the solution. It is the ratio of the number of moles of the particular component to the total number of moles in the solution.
While, Mass percent=
What is the difference between molality and molarity?
Molarity: It is the number of moles of solute present in 1 litre of the solution. It is denoted by M. It is dependent on temperature.
Molarity (M) =
Molality: It is the number of moles of solute present in 1 kg of solvent. It is denoted by m. It is independent on temperature.
Thus, Molality (m)=
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Matching Type
Match the following:
A. 88 g of CO2
a. 0.25 mol
B. 6.022 x 1023 molecules of H2O
b. 2 mol
C. 5.6 litres of O2 at STP
c. 1 mol
D. 96 g of O2
d. 6.022 x 1023 molecules
E. 1 mole of any gas
e. 3 mol
i) – b
ii) – c
iii) – a
iv) - e
v) – d
Match the following physical quantities with units
Column I (Physical)
Column II (Unit)
a. g mL-1
(ii) Mole fraction
e. mol L-1
(vi) Luminous intensity
g. mol kg-1
h. N m-1
(i) – e
(ii) – d
(iii) – b
(iv) – g
(v) – c
(vi) – f
(vii) – a
(viii) - i
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Long Answer Type
A vessel contains 1.6 g of dioxygen at STP (273.15K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate
(i) volume of the new vessel.
(ii) number of molecules of dioxygen.
We will be calculating V1
We know that 32 g of oxygen occupies 22.4 L of volume at STP+
Hence, 1.6 g Of O2
We know that,
Now we must calculate
As per Boyle's law,
(iii) Number of molecules of O2
that are present in the vessel,
Define the Law of multiple proportions. Explain it with two examples. How does this law point to the existence of atoms?
The Law of multiple proportions states that "If two elements can combine to form more than one compound, the mass of element combines with the fixed mass of the other element and is in the ratio of small whole numbers." For example, hydrogen reacts with oxygen to form water and
Hydrogen + Oxygen → Water
2g 16g 18g
Hydrogen + Oxygen → Hydrogen Peroxide
2g 32g 34g
Here, the masses of oxygen combine with a fixed mass of hydrogen and have a simple ratio of 16:32 or 1:2. This Law reflects that reactants always combine in fixed proportions.
Students preparing for their examinations can make use of the NCERT Exemplar Class 11 Chemistry solutions chapter 1 PDF download feature to download a PDF to prepare for their exams, as it covers many important topics which are asked in the competitive exams. NCERT Exemplar solutions for Class 11 Chemistry chapter 1 includes the following topics:
Main subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 1
Importance of Chemistry
Nature of matter
Properties of matter and their measurement
The International System of Units (S.I)
Mass and weight
Uncertainty in measurement
Laws of Chemical Combinations
Law of conservation of mass
Law of definite proportions
Law of Multiple Proportions
Dalton Atomic Theory
Atomic and Molecular masses
Average atomic Mass
Mole Concept and Molar Mass
Stoichiometry and stoichiometric calculations
Reactions in solutions
What Will Students Learn?
Through NCERT Exemplar Class 11 Chemistry solutions chapter 1 Some Basic Concepts of Chemistry students will learn the basic properties of matter and also about the various S.I units that can be used. All the laws and theories such as Dalton’s Atomic Theory and Avogadro Law among others have been explained in a detailed way to form a good base for the students. The students will also get to learn and solve numerical based on topics such as molecular weight, mass concentration, and mole concept among others which will be helpful if they are preparing for competitive exams and also for their class 12.
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 - Chapter-Wise
Important Topics To Cover For Exams From NCERT Exemplar Class 11 Chemistry Solutions Chapter 1
Here are the important topics that the students must specifically learn in this chapter:
The students will get to learn about the various parameters to denote the strength of the solution such as Molarity, Normality, Molality and their important relations. The related numerical and their solutions are given in Class 11 Chemistry NCERT Exemplar solutions chapter 1.
NCERT Exemplar Class 11 Chemistry solutions chapter 1 also introduces important concepts such as Dilution law, Equivalence concept, and Stoichiometry and has exercises for the revision of students.
Students will study various laws of chemical combination like Law of Multiple Proportion, Law of Definite proportion.
Check Solutions of Textbook Chapters
Chapter-1 - Some Basic Concepts of Chemistry
Chapter-2 - Structure of Atom
Chapter-3 - Classification of Elements and Periodicity in Properties
Chapter-4 - Chemical Bonding and Molecular Structure
Chapter-5 - States of Matter
Chapter-6 - Thermodynamics
Chapter-7 - Equilibrium
Chapter-8 - Redox Reaction
Chapter-9 - Hydrogen
Chapter-10 - The S-Block Elements
Chapter-11 - The P-Block Elements
Chapter-12 - Organic chemistry- some basic principles and techniques
Chapter-14 - Hydrocarbons
Chapter-15 - Environmental Chemistry
NCERT Exemplar Class 11 Solutions
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