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NCERT Exemplar Class 11 Chemistry Solutions 2022 - NCERT Books for Class 11 Chemistry are the bible for students studying for 11th grade exams as well as competitive exams such as NEET, JEE Mains, BITSAT, and others. NCERT Chemistry books for class 11 make sophisticated processes simple for kids to learn. NCERT Chemistry Books for Class 11 contain the most comprehensive explanations of each concept, as well as solved solutions. All CBSE students' curriculum and syllabus are governed by the NCERT (National Council of Educational Research and Training). For more information on NCERT Exemplar Class 11 Chemistry Solutions 2022, read this article.

by Keerthika | Last updated: Feb 02, 2022

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NCERT Solutions for Class 11

NCERT Solutions for Class 11 2023

NCERT Exemplar Class 11 Chemistry Solutions 2021 and Read NCERT Class 11 Chemistry Notes

New Update on 02-02-2022: NCERT Exemplar Class 11 Chemistry Solutions 2022 - NCERT Books for Class 11 Chemistry are the bible for students studying for their 11th grade exams as well as competitive exams such as NEET, JEE Mains, BITSAT, and other exams. NCERT Chemistry books for class 11 assist students in comprehending complex concepts in a straightforward manner. NCERT Chemistry Books for Class 11 contain the finest possible explanations of each concept, as well as solved solutions. The curriculum and syllabus for all CBSE students are regulated by the NCERT (National Council of Educational Research and Training). Around 15 to 20% of KVPY, JEE Mains and Advanced, and NEET questions are based on principles covered in these texts.

Check - NCERT Exemplar Class 11 Chemistry Solutions 2022


NCERT Class 11 Chemistry Chapter Wise Solutions

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 Reactions 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 13 – Hydrocarbons Chapter 14 – Environmental Chemistry

Chapter 1 - SOME BASIC CONCEPTS OF CHEMISTRY

1. Calculate the molecular mass of the following:

(i)

image
(ii)
image
(iii)
image

Answer:

(i)

image
:

The molecular mass of water,

image

  • (2
    image
    Atomic mass of hydrogen) + (1
    image
    Atomic mass of oxygen)
  • [2(1.0084) + 1(16.00 u)]
  • 2.016 u + 16.00 u
  • 18.016
  • 18.02 u

(ii)

image
:

The molecular mass of carbon dioxide,

image

  • (1
    image
    Atomic mass of carbon) + (2
    image
    Atomic mass of oxygen)
  • [1(12.011 u) + 2 (16.00 u)]
  • 12.011 u + 32.00 u
  • 44.01 u

(iii)

image
:

The molecular mass of methane,

image

  • (1
    image
    Atomic mass of carbon) + (4
    image
    Atomic mass of hydrogen)
  • [1(12.011 u) + 4 (1.008 u)]
  • 12.011 u + 4.032 u
  • 16.043 u

2. Calculate the mass percent of different elements present in sodium sulphate (

image
).

Answer:

The molecular formula of sodium sulphate is

image
.

Molar mass of

image
= [(2 × 23.0) + (32.066) + 4 (16.00)] = 142.066 g

image

Mass percent of an element

  • Mass percent of sodium:

image

Mass percent of sulphur:

image

Mass percent of oxygen:

image

3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

Answer:

% of iron by mass = 69.9 % [Given]

  • of oxygen by mass = 30.1 % [Given] Relative moles of iron in iron oxide:

image

Relative moles of oxygen in iron oxide:

image

Simplest molar ratio of iron to oxygen:

  • 1.25: 1.88
  • 1: 1.5

=2: 3

  • The empirical formula of the iron oxide is
    image
    .

4. Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

(ii) a mole of carbon is burnt in 16 g of dioxygen.

(iii) 2 moles of carbon is burnt in 16 g of dioxygen.

Answer:

The balanced reaction of combustion of carbon can be written as:

image

i. As per the balanced equation, 1 mole of carbon burns in1 mole of dioxygen (air) to produce1 mole of carbon dioxide.

ii. According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant.

iii. According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

5. Calculate the mass of sodium acetate

image
required to make 500 mL of

0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g

image

Answer:

0.375 M aqueous solution of sodium acetate

image
1000 mL of a solution containing 0.375 moles of sodium acetate

∴Number of moles of sodium acetate in 500 mL

image

Given

Molar mass of sodium acetate = 82.0245 g

image

Required mass of sodium acetate = (82.0245 g

image
) * (0.1875 mole) = 15.38 g

6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g

image
and the mass per cent of nitric acid in it being 69%.

Answer: Mass percent of nitric acid in the sample = 69 % [Given]

Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.

Molar mass of nitric acid (HNO3 )

  • {1 + 14 + 3(16)} g
    image
  • 1 + 14 + 48
  • 63 g
    image
  • Number of moles in 69 g of HNO3

image

Volume of 100g of nitric acid solution

image

Concentration of nitric acid

image

∴Concentration of nitric acid = 15.44 mol/L

7. How much copper can be obtained from 100 g of copper sulphate (

image
)? Ans. 1 mole of CuSO4 contains 1 mole of copper.

Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)

  • 63.5 + 32.00 + 64.00
  • 159.5 g

159.5 g of CuSO4 contains 63.5 g of copper.

image

⇒ 100 g of CuSO4 will contain

of copper.

image

image
Amount of copper that can be obtained from 100 g CuSO4

= 39.81 g

8. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide

is 159.69 g

image
.

Answer:

Mass percent of iron (Fe) = 69.9% (Given)

Mass percent of oxygen (O) = 30.1% (Given)

image

Number of moles of iron present in the oxide

= 1.25

image

Number of moles of oxygen present in the oxide

= 1.88

Ratio of iron to oxygen in the oxide,

image

1: 1.5

2: 3

image
The empirical formula of the oxide is
image
.

Empirical formula mass of

image
= [2(55.85) + 3(16.00)] g

image

Molar mass of = 159.69 g

image

Molecular formula of a compound is obtained by multiplying the empirical formula with n.

Thus, the empirical formula of the given oxide is

image
and n is 1.

Hence, the molecular formula of the oxide is

image
.

9. Calculate the atomic mass (average) of chlorine using the following data:

Answer:

The average atomic mass of chlorine

image
image

  • 26.4959 + 8.9568
  • 35.4527 u

image
The average atomic mass of chlorine = 35.4527 u

10. In three moles of ethane

image
calculate the following:

i. Number of moles of carbon atoms.

ii. Number of moles of hydrogen atoms.

iii. Number of molecules of ethane.

Answer:

(i) 1 mole of

image
contains 2 moles of carbon atoms.

Number of moles of carbon atoms in 3 moles of

= 2 × 3 = 6

ii. 1 mole of

image
contains 6 moles of hydrogen atoms.
image
Number of moles of carbon atoms in 3 moles of
image
= 3 × 6 = 18

1 mole of

image
contains
image
molecules of ethane. Number of molecules in 3 moles of
image

= 3 × 6.023 × 1023 =

image

11. What is the concentration of sugar (

image
) in mol
image
if its 20 g
are dissolved in enough water to make a final volume up to 2 L?

Answer:

Molarity (M) of a solution is given by,

image

Molar concentration of sugar = 0.02925

image

12. If the density of methanol is 0.793 kg

image
, what is its volume needed for making 2.5 L of its 0.25 M solution?

Answer:

Molar mass of methanol (

image
) = (1 × 12) + (4 × 1) + (1 × 16)

  • 32 g
    image
  • 0.032 kg
    image

image

Molarity of methanol solution

= 24.78

image

(Since density is mass per unit volume)

Applying,

(Given solution) (Solution to be prepared)

(24.78

image
)
image
= (2.5 L) (0.25
image
)

image

    • 0.0252 L
    • 25.22 mL

13. Pressure is determined as force per unit area of the surface. The SI unit of pressure,

Pascal is as shown below: 1Pa =

image
If mass of air at sea level is 1034 g
image
, calculate the pressure in Pascal.

image

Answer:

Pressure is defined as force acting per unit area of the surface.

image

= 1.01332 × 105 kg

image

We know, 1 N = 1 kg

Then,

1 Pa = 1

image
= 1 kg
image

1 Pa = 1 kg

image

image
Pressure = 1.01332 x 105 Pa

14. What is the SI unit of mass? How is it defined?

Answer. The SI unit of mass is kilogram (kg). 1 Kilogram is defined as the mass equal to the mass of the international prototype of kilogram.

15. Match the following prefixes with their multiples:

image

Answer:

image

Disclaimer: The above information is for general informational purposes only. All information on the Site is provided in good faith, however we make no representation or warranty of any kind, express or implied, regarding the accuracy, adequacy, validity, reliability, availability or completeness of any information on the Site.
Also Learn

NCERT Solutions for Class 11 Ncert solutions 2023 - FAQ

1. Where can I get NCERT Solutions for Class 11 Chemistry pdf?

Students can download the class 11 chemistry chapter wise pdf on Fresherslive for free.

2. How NCERT Solutions for Class 11 Chemistry can help you perform well in the Exam?

These solutions act as a reference guide on which you can rely. These will help you revise the syllabus quickly. Being available in both online and offline mode you can use them whenever you wish to. The steps available will help you understand the numerical with great ease.

3. Is class 11 Chemistry tough?

That purely depends on you. If you stay smartly, use the resources well, take mock tests then you will enjoy the learning process. However, considering the course the subject is lengthy compared to other subjects.

4. What are the 3 parts of chemistry?

Class 11 chemistry is divided into 3 parts:
1) Physical
2) Organic
3) Inorganic

5. How can I download NCERT Chemistry Class 11 part 1 pdf?

To download the pdf of Chemistry class 11 open this article and click on the chapters links.

6. How many chapters are there in Class 11 chemistry?

There are a total of 14 chapters in the Class 11 chemistry book

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