CBSE Class 11 Chemistry Chapter 2: Structure of Atom NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 11 PDF

CBSE Class 11 Chemistry Chapter 2: Structure of Atom NCERT Solutions offers a thorough exploration of atomic structure. This resource breaks down complex concepts, starting with the fundamental properties of subatomic particles like electrons, protons, and neutrons, including their mass and charge. It guides students through calculating the number of these particles in various chemical species, such as methane and ammonia, and determining their total mass. The solutions emphasize a clear, step-by-step problem-solving approach, making quantitative aspects of atomic structure more accessible. This makes it an invaluable tool for students aiming to solidify their understanding and excel in their examinations by mastering the calculations involved in atomic composition.

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Board	CBSE
Class	Class 11
Subject	Chemiry
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 2: Structure of Atom

Chapter summary

Chapter 2 of the NCERT Class 11 Chemistry textbook focuses on the Structure of the Atom. These NCERT Solutions break down complex calculations involving electrons, protons, and neutrons. Students will learn to calculate the number of electrons, their total mass and charge, and the quantities of neutrons and protons in given amounts of substances like methane and ammonia. The solutions emphasize understanding the relationship between moles, Avogadro's number, and the properties of subatomic particles.

Learning outcomes

Calculate the number of electrons that weigh a specific mass.
Determine the mass and charge of one mole of electrons.
Calculate the total number of electrons in a given quantity of a molecule.
Find the total number and mass of neutrons in a given mass of an isotope.
Calculate the total number and mass of protons in a given mass of a compound.
Understand the role of Avogadro's number in calculations involving moles of particles.

Topics covered

Paper topics

Mass and charge of electrons
Number of electrons
Mass of one mole of electrons
Charge of one mole of electrons
Number of electrons in molecules
Number of neutrons
Mass of neutrons
Number of protons
Mass of protons
Atomic structure calculations
Stoichiometry of subatomic particles
Avogadro's number applications

Important topics

Calculating mass and charge of electrons
Determining total electrons in molecules
Calculating number and mass of neutrons
Calculating number and mass of protons
Application of Avogadro's number

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Questions and Solutions

Question 2.1

Calculate the number of electrons which will together weigh one gram.

Calculate the mass and charge of one mole of electrons.

Solution:

To calculate the number of electrons that weigh one gram, we first need the mass of a single electron. The mass of one electron is given as $9.10939 \times 10^{-31}$ kg. We want to find the number of electrons in 1 gram, which is equal to $1 \times 10^{-3}$ kg. The number of electrons can be found by dividing the total mass (1 g) by the mass of a single electron: Number of electrons = $\frac{\text{Total mass}}{\text{Mass of one electron}}$ Number of electrons = $\frac{1 \times 10^{-3} \text{ kg}}{9.10939 \times 10^{-31} \text{ kg}}$ Number of electrons = $\frac{1}{9.10939} \times 10^{-3 - (-31)}$ Number of electrons = $0.10977 \times 10^{28}$ Number of electrons = $1.0977 \times 10^{27}$ Therefore, approximately $1.098 \times 10^{27}$ electrons will together weigh one gram.
To calculate the mass and charge of one mole of electrons: First, let's find the mass of one mole of electrons. One mole contains Avogadro's number of particles, which is $6.022 \times 10^{23}$ . Mass of one mole of electrons = (Number of electrons in one mole) $\times$ (Mass of one electron) Mass of one mole of electrons = $(6.022 \times 10^{23}) \times (9.10939 \times 10^{-31} \text{ kg})$ Mass of one mole of electrons = $5.4819 \times 10^{-7} \text{ kg}$ Next, let's find the charge of one mole of electrons. The charge on a single electron is $1.6022 \times 10^{-19}$ C. Charge on one mole of electrons = (Number of electrons in one mole) $\times$ (Charge of one electron) Charge on one mole of electrons = $(6.022 \times 10^{23}) \times (1.6022 \times 10^{-19} \text{ C})$ Charge on one mole of electrons = $9.648 \times 10^{4} \text{ C}$ So, the mass of one mole of electrons is approximately $5.48 \times 10^{-7}$ kg, and the charge is approximately $9.65 \times 10^{4}$ C.

Question 2.2

Calculate the total number of electrons present in one mole of methane.

Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C. (Assume that mass of a neutron = $1.675 \times 10^{-27}$ kg).

Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP. Will the answer change if the temperature and pressure are changed?

Solution:

To find the total number of electrons in one mole of methane (CH₄): First, determine the number of electrons in one molecule of methane. Carbon (C) has 6 electrons, and Hydrogen (H) has 1 electron. So, in CH₄, the total number of electrons is $1 \times 6 + 4 \times 1 = 10$ electrons per molecule. One mole of any substance contains Avogadro's number of particles, which is $6.022 \times 10^{23}$ molecules. Total number of electrons in one mole of methane = (Number of electrons per molecule) $\times$ (Avogadro's number) Total number of electrons = $10 \times 6.022 \times 10^{23}$ Total number of electrons = $6.022 \times 10^{24}$ electrons.
To find the total number and mass of neutrons in 7 mg of ¹⁴C: (a) Total number of neutrons: The atomic mass of ¹⁴C is 14, and its atomic number (number of protons) is 6. Therefore, the number of neutrons in one atom of ¹⁴C is $14 - 6 = 8$ neutrons. One mole of ¹⁴C weighs 14 grams and contains Avogadro's number ( $6.022 \times 10^{23}$ ) of atoms. The total number of neutrons in 14 g of ¹⁴C is $8 \times 6.022 \times 10^{23}$ . We have 7 mg of ¹⁴C, which is $7 \times 10^{-3}$ g. Number of neutrons in 7 mg = $\frac{\text{Number of neutrons in 14 g}}{14 \text{ g}} \times 7 \times 10^{-3} \text{ g}$ Number of neutrons = $\frac{8 \times 6.022 \times 10^{23}}{14 \text{ g}} \times 7 \times 10^{-3} \text{ g}$ Number of neutrons = $8 \times 6.022 \times 10^{23} \times \frac{7 \times 10^{-3}}{14}$ Number of neutrons = $8 \times 6.022 \times 10^{23} \times 0.5 \times 10^{-3}$ Number of neutrons = $4 \times 6.022 \times 10^{20}$ Number of neutrons = $2.4088 \times 10^{21}$ neutrons. (b) Total mass of neutrons: The mass of one neutron is given as $1.675 \times 10^{-27}$ kg. Total mass of neutrons = (Number of neutrons) $\times$ (Mass of one neutron) Total mass of neutrons = $(2.4088 \times 10^{21}) \times (1.675 \times 10^{-27} \text{ kg})$ Total mass of neutrons = $4.0345 \times 10^{-6} \text{ kg}$ . So, in 7 mg of ¹⁴C, there are approximately $2.409 \times 10^{21}$ neutrons, and their total mass is approximately $4.035 \times 10^{-6}$ kg.
To find the total number and mass of protons in 34 mg of NH₃: (a) Total number of protons: First, consider the molar mass of ammonia (NH₃). Nitrogen (N) has an atomic mass of approximately 14, and Hydrogen (H) has an atomic mass of approximately 1. So, the molar mass of NH₃ is $14 + (3 \times 1) = 17$ g/mol. This means 17 g of NH₃ contains Avogadro's number ( $6.022 \times 10^{23}$ ) of molecules. Now, let's find the number of protons in one molecule of NH₃. Nitrogen (N) has an atomic number of 7 (meaning 7 protons), and Hydrogen (H) has an atomic number of 1 (meaning 1 proton). Number of protons per molecule of NH₃ = (Protons in N) + 3 $\times$ (Protons in H) Number of protons per molecule = $7 + (3 \times 1) = 10$ protons. So, 17 g of NH₃ contains $10 \times 6.022 \times 10^{23}$ protons. We have 34 mg of NH₃, which is $34 \times 10^{-3}$ g. Number of protons in 34 mg = $\frac{\text{Number of protons in 17 g}}{17 \text{ g}} \times 34 \times 10^{-3} \text{ g}$ Number of protons = $\frac{10 \times 6.022 \times 10^{23}}{17 \text{ g}} \times 34 \times 10^{-3} \text{ g}$ Number of protons = $10 \times 6.022 \times 10^{23} \times \frac{34 \times 10^{-3}}{17}$ Number of protons = $10 \times 6.022 \times 10^{23} \times 2 \times 10^{-3}$ Number of protons = $20 \times 6.022 \times 10^{20}$ Number of protons = $1.2044 \times 10^{22}$ protons. (b) Total mass of protons: The mass of one proton is approximately $1.67493 \times 10^{-27}$ kg. Total mass of protons = (Number of protons) $\times$ (Mass of one proton) Total mass of protons = $(1.2044 \times 10^{22}) \times (1.67493 \times 10^{-27} \text{ kg})$ Total mass of protons = $2.0173 \times 10^{-5} \text{ kg}$ . The answer will not change if the temperature and pressure are changed because the number of protons and their mass are intrinsic properties of the NH₃ molecules and do not depend on the physical state (like temperature and pressure) of the gas. The number of molecules in 34 mg of NH₃ remains constant regardless of T and P.

Common mistakes

Incorrectly converting units (e.g., grams to kilograms).
Errors in applying Avogadro's number in calculations.
Miscalculating the number of neutrons or protons based on atomic mass and atomic number.
Confusing mass of a single particle with the mass of a mole of particles.
Arithmetic errors in scientific notation calculations.

Revision tips

Review the mass and charge of fundamental particles (electron, proton, neutron).
Practice unit conversions carefully, especially between grams and kilograms.
Understand the relationship between moles, Avogadro's number, and the number of particles.
Break down complex molecules to find the total number of protons and neutrons per molecule.
Use the provided formulas and constants consistently for accurate calculations.

Practice MCQs

Q1. What is the mass of a single electron?

Q2. How many electrons are present in one mole of methane (CH4)?

Q3. What is the approximate charge on one mole of electrons?

Q4. How many neutrons are in 7 mg of Carbon-14 (^14C)?

Q5. The mass of protons in 34 mg of NH3 is approximately:

Frequently asked questions

What are the key concepts covered in these NCERT Solutions for Class 11 Chemistry Chapter 2?

These solutions cover calculations related to the number, mass, and charge of electrons, protons, and neutrons. They also include determining these quantities for moles of substances and specific masses of elements and compounds.

How do these solutions help in understanding the structure of an atom?

By working through these problems, students gain a quantitative understanding of the subatomic particles that constitute an atom and how to relate their properties to macroscopic quantities like grams and moles.

Are the calculations for mass and charge of electrons explained step-by-step?

Yes, each solution provides a clear, step-by-step breakdown of the calculations, making it easier for students to follow the logic and arrive at the correct answer.

What is the significance of Avogadro's number in these problems?

Avogadro's number (6.022 x 10^23) is crucial for converting between the number of particles (atoms, molecules, electrons) and the mole concept, which is used extensively in these calculations.

Can these solutions be used for exam revision?

Absolutely. They provide practice on common calculation-based questions found in exams, helping students reinforce their understanding and improve their problem-solving speed and accuracy.

Does the answer change if temperature and pressure change for proton calculations in NH3?

No, the number and mass of protons in a given mass of NH3 do not depend on temperature and pressure, as these factors affect the volume and density of the gas, not the number of protons within the molecules themselves.

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