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NCERT Exemplar Class 11 Chemistry Solutions 2021

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Q: 4. What are the 3 parts of chemistry?

Class 11 chemistry is divided into 3 parts:1) Physical2) Organic3) Inorganic

NCERT Exemplar Class 11 Chemistry Solutions 2021 - NCERT Exemplar Class 11 Chemistry Solutions 2021 is given here. Students who are looking for NCERT Solutions for Class 11 Chemistry here. NCERT Solutions for Class 1 includes detailed answers to all the questions of all chapters of Chemistry which is given in the NCERT book. Download the NCERT Exemplar Class 11 Chemistry Solutions 2021 PDF for free here. We provide detailed NCERT Exemplar Class 11 Chemistry Solutions 2021 for the students. Read the NCERT Exemplar Class 11 Chemistry Solutions 2021 for more details on NCERT Class 11 Chemistry.

B Hari Aditya | Updated: Jun 09, 2021, 16:30 IST

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New Update on 09-06-2021: FreshersLive acts as a great provider of NCERT Exemplar Class 11 Chemistry Solutions 2021. Students who are looking for Class 11 Chemistry notes can get them here. NCERT Exemplar Class 11 Chemistry Solutions 2021 for all chapters is provided here for student's reference. Download the complete solution for Class 11 Chemistry Question and Answers for free. Besides, students can acquire the NCERT Exemplar Class 11 Chemistry Solutions 2021 in PDF format soon. CBSE students of Class 11 can use the solutions given here for all chapters to prepare for the exams. Prepare for your exam by referring to the NCERT Exemplar Class 11 Chemistry Solutions 2021 to score good marks.

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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)

(ii)
(iii)

Answer:

(i)

The molecular mass of water,

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

(ii)

The molecular mass of carbon dioxide,

(1
Atomic mass of carbon) + (2
Atomic mass of oxygen)

[1(12.011 u) + 2 (16.00 u)]

12.011 u + 32.00 u

44.01 u

(iii)

The molecular mass of methane,

(1
Atomic mass of carbon) + (4
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 (

Answer:

The molecular formula of sodium sulphate is

Molar mass of

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

Mass percent of an element

Mass percent of sodium:

Mass percent of sulphur:

Mass percent of oxygen:

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:

Relative moles of oxygen in iron oxide:

Simplest molar ratio of iron to oxygen:

1.25: 1.88

1: 1.5

=2: 3

The empirical formula of the iron oxide is
.

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:

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

required to make 500 mL of

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

Answer:

0.375 M aqueous solution of sodium acetate

1000 mL of a solution containing 0.375 moles of sodium acetate

∴Number of moles of sodium acetate in 500 mL

Given

Molar mass of sodium acetate = 82.0245 g

Required mass of sodium acetate = (82.0245 g

) * (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

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

1 + 14 + 48

63 g

Number of moles in 69 g of HNO₃

Volume of 100g of nitric acid solution

Concentration of nitric acid

∴Concentration of nitric acid = 15.44 mol/L

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

)? 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.

⇒ 100 g of CuSO₄ will contain

of copper.

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

Answer:

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

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

Number of moles of iron present in the oxide

= 1.25

Number of moles of oxygen present in the oxide

= 1.88

Ratio of iron to oxygen in the oxide,

1: 1.5

2: 3

The empirical formula of the oxide is

Empirical formula mass of

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

Molar mass of = 159.69 g

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

Thus, the empirical formula of the given oxide is

and n is 1.

Hence, the molecular formula of the oxide is

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

Answer:

The average atomic mass of chlorine

26.4959 + 8.9568

35.4527 u

The average atomic mass of chlorine = 35.4527 u

10. In three moles of ethane

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

contains 2 moles of carbon atoms.

Number of moles of carbon atoms in 3 moles of

= 2 × 3 = 6

ii. 1 mole of

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

1 mole of

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

= 3 × 6.023 × 1023 =

11. What is the concentration of sugar (

) in mol
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,

Molar concentration of sugar = 0.02925

12. If the density of methanol is 0.793 kg

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

Answer:

Molar mass of methanol (

) = (1 × 12) + (4 × 1) + (1 × 16)

32 g

0.032 kg

Molarity of methanol solution

= 24.78

(Since density is mass per unit volume)

Applying,

(Given solution) (Solution to be prepared)

(24.78

)

= (2.5 L) (0.25

)

- 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 =

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

Answer:

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

= 1.01332 × 105 kg

We know, 1 N = 1 kg

Then,

1 Pa = 1

= 1 kg

1 Pa = 1 kg

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:

Answer:

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NCERT Solutions for Class 11 Ncert solutions 2021 - 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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