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NCERT Solutions for Class 12 Chemistry chapter 15 Polymers

NCERT Solutions for Class 12 Chemistry chapter 15 Polymers

Edited By Sumit Saini | Updated on Aug 16, 2022 01:40 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers - Hey are you stuck while solving your homework problems? Now you can resolve all your doubts here. Just scroll down to get NCERT solutions for class 12 chemistry chapter 15 Polymers. This Class 12 Chemistry Chapter 15 NCERT solutions contains a total of 6 intext questions and 20 questions in the exercise at chapter end. The NCERT Solutions for Class 12 Chemistry Chapter 15 are prepared in a comprehensive manner so that you will also learn how to write answers in your exams. These NCERT solutions help you in your preparation of CBSE board exams as well as in the various competitive exams like JEE, NEET, etc. By referring to the NCERT solutions for class 12, students can understand all the important concepts and practice questions well enough before their examination.

Topics and Sub-topics of NCERT Textbook Class 12 Chemistry Chapter 15 Polymers-

15.1 Classification of Polymers

15.2 Types of Polymerisation Reactions

15.3 Molecular Mass of Polymers

15.4 Biodegradable Polymers

15.5 Polymers of Commercial Importance

Solutions to In-Text Questions Ex 15.1 to 15.6

Question 15.1 What are polymers?

Answer :

Polymers- Poly means many and mer means unit or parts. Polymers are high molecular masses macromolecules (10^3-10^7u) . These are formed by joining of the repeated units of monomers.
15965635536001596563551370

Question 15.3 (1) Write the names of monomers of the following polymers

Screenshot%20-%202022-05-26T180605

Answer :
So, the monomer is adipic acid and hexamethylene diamine

Question 15.3 (2) Write the names of monomers of the following polymers

1649970185488

Answer :

1649970208524 It is a polymer of Nylon6 . So, the monomeric unit is Caprolactum.
1596563535781


Question 15.3 (3) Write the names of monomers of the following polymers

1596563529301

Answer :

1596563519883 the above polymer is a Teflon (PTFE)
the monomeric unit is tetrafluroethene
1596563512703

Question 15.4 Classify the following as addition and condensation polymers: Terylene, Bakelite, Polythene, Teflon.

Answer :

  • Addition Polymers are formed by the direct addition of repeated monomers. Example- polyethene and Teflon
  • Condensation polymers are formed by condensation of two or more than two monomers by eliminating by-product like water and HCl. Example- terylene and bakelite


Question 15.5 Explain the difference between Buna-N and Buna-S.

Answer :

Buna-N
It is a copolymer of 1,3-Butadiene and acrylonitrile. It is resistant to the action of petrol, lubricating oil and organic solvents. It is used in making oil seals and tank lining etc.
1596563501645

Buna-S
It is formed by copolymerisation of 1,3-Butadiene and Styrene. It is used for making automobiles tyres and rubber soles etc.
1596563492635

Question 15.6 Arrange the following polymers in increasing order of their intermolecular forces.
Nylon 6,6, Buna-S, Polythene

Answer :

Increasing order in their intermolecular forces-
Buna-S(elastomers)<Polyethene<Nylon6, 6(fibres

  • (Thermoplastics, intermediate forces between elastomers and fibres)
  • ( strong H-bond or dipole-dipole interaction)
  • elastomers weakest force of attraction

NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers- Exercise Questions

Question 15.1 Explain the terms polymer and monomer.

Answer :

Polymers- Poly means many and mer means unit or parts. Polymers are high molecular masses macromolecules. These are formed by joining of repeating unit of monomers.
1596563481823 Monomers- These are simple reactive units, which combine together to form large molecules through covalent bond.examples- ethene and hexamethylene diamine, adipic acid.


Question 15.2 What are natural and synthetic polymers? Give two examples of each type.

Answer :

Natural Polymers-
Polymers that formed naturally like formed by animals and plants. these are found in nature. Example- protein, starch, cellulose etc.

1596563472006

Synthetic Polymers-
Polymers made by human beings are called synthetic or man-made polymers. Examples- plastic(polyethene), nylon6,6 and nylon 6 etc.
1596563464587

Question 15.3 Distinguish between the terms homopolymer and copolymer and give an example of each.

Answer :

HOMO-POLYMER
These types of polymers are formed by polymerisation of one type of monomers. -[A-A-A-A]_{n}-
examples - polyethene is the homopolymer of ethene monomers.

1596563456683

CO-POLYMER-
These types of polymers are formed by the polymerisation of two different monomers. -[A-B-A-B]_{n}-
example- Nylon6, 6 is the copolymer of adipic acid and hexamethylene diamine.

1596563448705

Question 15.4 How do you explain the functionality of a monomer?

Answer :

The functionality of a monomer is the number of binding sites present in it. For example, for propene and ethene functionality is one but for adipic acid and 1,3- butadiene is two.

Question 15.5 Define the term polymerisation.

Answer :

The process of formation of polymers or high molecular masses (10^3-10^7u) from its respective monomers is known as polymerisation. In polymers monomers are held by cobalent bonds.

Question 15.6 Is ( NH-CHR-CO )_n , a homopolymer or copolymer?

Answer :

( NH-CHR-CO )_n , is a homopolymer because it is obtained from a single monomer of NH_{2}-CHR-COOH

Question 15 . 7 Why do elastomers possess elastic properties?

Answer :

In elastomers, the polymeric chains are held by weak intermolecular forces of attraction. These weak binding forces allow them to stretch and a few cross-links are there in between the chains, which helps them to retract after stretching or releasing forces. Due to this elastomers are elastic in nature.
Example- Buna-S, Buna-N and Neoprene etc.

Question 15.8 How can you differentiate between addition and condensation polymerisation?

Answer :

Addition Polymerisation- The process of repeated addition of monomer, having a double or triple bond to form polymers. For example, polyethene is formed by the addition polymerisation of ethene.

1596563434919 Condensation Polymerisation- Process of formation of polymers by repeated condensation reaction between two different monomers, having bi-functionality or tri-functionality. A small molecule is eliminated like water and HCl in each condensation step. For example Nylon 6, 6 is a condensation polymerisation of adipic acid and hexamethylene diamine.

1596563427472

Question 15.9 Explain the term copolymerisation and give two examples.

Answer :

The process of formation of polymers of two or more different monomeric units is known as copolymerisation. For example, Buna-S is formed by the copolymerisation.
1596563419402

Question 15.10 Write the free radical mechanism for the polymerisation of ethene.

Answer :

In free radical mechanism, there are three main steps-

  1. Chain initiation
  2. Chain propagation
  3. Termination

1. Chain Initiation - the polymerisation of ethene to polythene consists of heating or exposing to light a mixture of ethene with a small amount of benzoyl peroxide initiator. Generating new and larger free radicals.

1596563410560

2. Chain Propagation Step - As the radical reacts with another molecule of ethene. So, another bigger sized radical is formed. The repetition of this step is chain propagation.

1596563402098

3. Chain Termination step - At some time the product radical reacts with another radical to form the polymerised product and this step is called the chain terminating step.

1596563394376

Question 15.11 Define thermoplastics and thermosetting polymers with two examples of each.

Answer :

Thermoplastic polymers are linear or slightly branched chained molecules. It can be repeatedly softened and hardened on heating. Thus they can be modified again and again. Examples- polyethene and polystyrene. These polymers have intermolecular forces of attraction intermediate between elastomers and fibres. Some examples of common thermoplastics are polyethene, polystyrene, polyvinyls, etc.

Thermosetting Plastics are cross-linked and heavily branched molecules, which get hardened during the moulding process. These polymers cannot be reused. For examples bakelite and urea-formaldehyde resin etc.

Question 15.12(i) Write the monomers used for getting the following polymers.
(i) Polyvinyl chloride

Answer :

For PVC (polyvinyl chloride) we use vinyl chloride as a monomer.

(CH_{2}=CH-Cl)

Question 15.12(ii) Write the monomers used for getting the following polymers.
(ii) Teflon

Answer :

The monomeric unit of Teflon (PTFE) is tetraflouroethene ( CF_{2}=CF_{2} ). It is resistant to heat and chemical attack.

Question 15.12(iii) Write the monomers used for getting the following polymers.
Bakelite

Answer :

The monomeric unit of abkelite is phenol and formaldehyde.
(a) phenol- C_{6}H_{5}OH
(b) formaldehude- HCHO

Question 15.14 How does the presence of double bonds in rubber molecules influence their structure and reactivity?

Answer :

Natural rubber is a linear polymer of isoprene (2-methyl-1, 3-butadiene) and is also called as cis - 1, 4 polyisoprenes. Due to the cis configuration about the double bond, it is difficult to come closer for effective compactness due to the weak intermolecular attraction (van der Waals). Thus natural rubber has a coiled structure and it can be stretched like spring(show elastic nature).
1596563377186

Question 15.15 Discuss the main purpose of vulcanisation of rubber.

Answer :

The natural rubber has many flaws in following ways like-

  • it becomes soft at high temperature and brittle at low temperature (<283K). And
  • show very high water absorption capacity
  • soluble in a non-polar solvent and
  • poor resistant to the attack of oxidising agents.

To improve all these physical properties we do vulcanisation of rubber. During this process, sulphur cross-links are formed, which makes it hard, tough with high tensile strength.

Question 15.16 What are the monomeric repeating units of Nylon-6 and Nylon-6,6?

Answer :

The monomeric repeating unit of Nylon 6 is caprolactam and the monomers of Nylon6, 6 is adipic acid and hexamethylene diamine.

1596563367398

1596563360228

Question 15.17(i) Write the names and structures of the monomers of the following polymers:
Buna-S

Answer :

Name of monomer of Buna-S -(a) (1,3 butadiene)
(b) Styrene

159656335157415965633469301596563344806

Question 15.17(ii) Write the names and structures of the monomers of the following polymers:
Buna-N

Answer :

The momomeric unit of buna-N is

  • 1, 3-Butadiene
  • Acrylonitrile

The structure of the monomer are-

1596563335349 1596563325303

Question 15. 17(iii) Write the names and structures of the monomers of the following polymers:
Dacron

Answer :

The monomer of Dacron is-

1596563315838

-Ethylene glycol
-Terephthalic acid

Structures are-
15965632993321596563306910

Question 15.17(iv) Write the names and structures of the monomers of the following polymers:
(iv) Neoprene

Answer :

Monomer of Neoprene is Chloroprene.(2-chloro-1, 3-butadiene)
Structure-
1596563613739

1596563608065

Question 15. 18 (i) Identify the monomer in the following polymeric structures.

1649970265227

Answer :

1649970283846 It is Nylon10, 6
So, the monomer is Decanoic acid [ COOH-(CH_{2})_{8}-COOH ]and hexamethylene diamine [ NH_{2}-(CH_{2})_{6}-NH_{2} ]


Question 15.18 (ii) Identify the monomer in the following polymeric structures.

1649970316563

Answer :

1649970341171 The above polymer is melamine
So, the monomer is melamine and fomaldehyde( HCHO )
1596563268381

Question 15.19 How is dacron obtained from ethylene glycol and terephthalic acid ?

Answer :

Dacron(terelyne) is obtained from polymerisation of ethylene glycol and terepthalic acid at around 420 to 460 K. Also used in blending of cotton and wool fibres.

1596563257006

Question 15.20 What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester.

Answer :

Polymers which are decomposed by bacteria are called bio-degradable polymers. Their degraded product does not cause any serious effect on the environment.
Bio-degradable aliphatic polyesters are -
1. poly \beta -hydroxybutyrates-co- \beta -hydroxy valerate(PHBV)
1596563247904 2. Nylon-2-Nylon-6
15965632430931596563240769

More About NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers

In this chapter, you will deal with the science of polymers this chapter covers important concepts such as polymers, monomers, polymerization, types of polymers, classification of polymers based on their source and structure, cross-linked or linear polymer with their properties, and importance of polymers in daily life. This chapter holds 3 marks in the CBSE boards exams and after completing the NCERT solutions for Class 12 Chemistry chapter 15 Polymers students will be able to explain the terms like polymer, monomer, and polymerization and realize their importance, able to distinguish between different types of polymerization processes and different classes of polymers. This NCERT syllabus chapter also explains the formation of some important synthetic polymers and their uses and properties.

Important terms and points of NCERT Class 12 Chemistry solutions chapter 15-

1. Polymers- They are very high molecular mass macromolecules, which composed of repeating structural units derived from the monomers. Polymers have a high molecular mass (10^3-10^7U). Rubber, polythene, and nylon 6, 6 are examples of polymers.

2. Monomers- Monomers are simple and reactive molecules that combine in large numbers through covalent bonds to give rise to repeating structural units or polymers. For example propane, vinyl chloride, styrene, etc.

NCERT Solutions Class 12 Chemistry

NCERT Solutions for Class 12 Subject wise

  • The solutions are written in a comprehensive manner in the NCERT Class 12 Chemistry solutions chapter 15 will help you in understanding chapter easily.
  • It will be easier to revise because the detailed solutions will help you to remember the concepts and get you good marks.
  • Homework problems will be easier for you, all you need to do is check the detailed NCERT Solutions for Class 12 Chemistry Chapter 15 PDF download and you are ready to sail.
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Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

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  3. Explore Alternative Options:

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  4. Focus on NEET 2025 Preparation:

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Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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