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NCERT Solutions for Class 11
NCERT Solutions for Class 11 2023NCERT 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.
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)
Answer:
(i)
The molecular mass of water,
- (2 Atomic mass of hydrogen) + (1Atomic 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) + (2Atomic 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) + (4Atomic 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
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
0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g
Answer:
0.375 M aqueous solution 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
6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g
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 (
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.
= 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
Empirical formula mass of
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
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
10. In three moles of ethane
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
Number of moles of carbon atoms in 3 moles of
= 2 × 3 = 6
ii. 1 mole of
1 mole of
= 3 × 6.023 × 1023 =
11. What is the concentration of sugar (
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
Answer:
Molar mass of methanol (
- 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
-
- 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 =
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 Pa = 1 kg
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 2023 - FAQ
Students can download the class 11 chemistry chapter wise pdf on Fresherslive for free.
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.
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.
Class 11 chemistry is divided into 3 parts:
1) Physical
2) Organic
3) Inorganic
To download the pdf of Chemistry class 11 open this article and click on the chapters links.
There are a total of 14 chapters in the Class 11 chemistry book