Ncert Solutions for Class 12 Chemistry Chapter 15 Polymers

NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 12 exams must go through NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 12 Chemistry Chapter 15 Polymers NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time. Students can also download NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers PDF to access them even in offline mode.

NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers

NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

INTEXT Questions

Question 1.
What are polymers ?
Solution:
Polymers are high molecular mass substances consisting of large number of repeating structural units. They are also called as macromolecules. Some examples of polymers are polythene, bakelite, rubber, nylon 6, 6 etc.

Question 2.
How are polymers classified on the basis of structure?
Solution:
On the basis of structure, the polymers are classified as below :

1. Linear polymers such as polythene, polyvinyl chloride, etc.
2. Branched chain polymers such as low density polythene.
3. Cross linked polymers such as bakelite, melamine, etc.

Question 3.
Write the names of monomers of the following polymers :

Solution:

1. Hexamethylene diamine and adipic acid
2. Caprolactam
3. Tetrafluoroethene

Question 4.
Classify the following as addition and condensation polymers : Terylene, Bakelite, Polyvinyl chloride, Polythene.
Solution:
Addition polymers : Polyvinyl chloride, Polythene.
Condensation polymer : Terylene, Bakelite.

Question 5.
Explain the difference between Buna-N and Buna-S.
Solution:
Buna-N is a copolymer of 1, 3-butadiene and acrylonitrile and Buna-S is a copolymer of 1, 3-butadiene and styrene.

Question 6.
Arrange the following polymers in increasing order of their intermolecular forces.

1. Nylon 6,6, Buna-S, Polythene
2. Nylon 6, Neoprene, Polyvinyl chloride.

Solution:
The increasing order of intermolecular forces is :

1. Buna-S, Polythene, Nylon 6, 6
2. Neoprene, Polyvinyl chloride, Nylon 6.

NCERT Exercises

Question 1.
Explain the terms polymer and monomer.
Solution:
Polymers are very large molecules having high molecular mass, which are formed by joining of repeating structural units on a large scale derived from monomers, e.g., Polythene, PVC, Nylon-6, 6 etc.

Monomer is a simple molecule capable of undergoing polymerisation and leading to the formation of the corresponding polymer e.g., Ethene, Vinyl chloride, etc.

Question 2.
What are natural and synthetic polymers? Give two examples of each type.
Solution:
Natural Polymers – These are substances of natural origin, i.e., these are found in nature mainly in plants and animals. The well known natural polymers are proteins (polymers of amino acids), polysaccharides (polymers of monosaccharides), etc.

Synthetic Polymers – These are man made polymers i.e., polymers synthesised in laboratory. These include synthetic plastics, fibres and synthetic rubber. Specific examples are polythene and dacron.

Question 3.
Distinguish between the terms homopolymer and copolymer and give an example of each.
Solution:
A polymer which is obtained from only one type of monomer molecules is known as homopolymer e.g., Polythene, Polyvinyl chloride, etc.

Question 4.
How do you explain the functionality of a monomer?
Solution:
Functionality of a monomer is defined as the number of bonding sites in the monomers.

Question 5.
Define the term polymerisation.
Solution:
The process of joining together of a large number of simple small molecules (monomers) to make very large molecules (polymer) is termed polymerisation.

Question 6.

Solution:

Question 7.
In which classes, the polymers are classified on the basis of molecular forces?
Solution:
On the basis of molecular forces present between the chains of various polymers, the polymers are classified into the following four groups

1. Elastomers
2. Fibres
3. Thermoplastic polymers and
4. Thermosetting polymers.

Question 8.
How can you differentiate between addition and condensation polymerisation?
Solution:
In addition polymerisation, the molecules of the same or different monomers add together to form a large polymer molecule without the elimination of simple molecules like H₂O, HCl etc. Condensation polymerisation is a process in which two or more bifunctional molecules undergo a series of condensation reactions with the elimination of some simple molecules like H₂O, HCl, alcohol leading to the formation of polymers.

Question 9.
Explain the term copolymerisation and give two examples.
Solution:
Copolymerisation is a process in which a mixture of more than one monomeric species is allowed to polymerise. The copolymer contains multiple units of each monomer in the chain. The examples are copolymer of 1, 3-butadiene and styrene. Another example is the copolymer of 1, 3-butadiene and acrylonitrile.

Question 10.
Write the free radical mechanism for the polymerisation of ethene.
Solution:
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. The process starts with the addition of phenyl free radical formed by the peroxide to the ethene double bond thus regenerating a new and larger free radical. This step is called chain initiating step. As this radical reacts with another molecule of ethene, another bigger sized radical is formed. The repetition of this sequence with new and bigger radicals carries the reaction forward and the step is termed as chain propagating step. Ultimately, at some stage the product radical thus formed reacts with another radical to form the polymerised product. This step is called the chain terminating step. The sequence of steps may be depicted as follows :
Chain initiating steps :

Chain terminating step :
For termination of the long chain, these radicals can combine in different ways to form polythene. One mode of termination of chain is shown as under :

Question 11.
Define thermoplastics and thermosetting polymers with two examples of each.
Solution:
Thermoplastic polymers : These are the linear or slightly branched long chain molecules capable of repeatedly softening on heating and hardening on cooling. These polymers possess intermolecular forces of attraction intermediate between elastomers and fibres. Some common thermoplastics are polythene, polystyrene, polyvinyls, etc.

Thermosetting polymers : These polymers are cross linked or heavily branched molecules, which on heating undergo extensive cross linking in moulds and become infusible. These cannot be reused. Some common examples are bakelite, urea-formaldelyde resins, etc.

Question 12.
Write the monomers used for getting the following polymers.

1. Polyvinyl chloride
2. Teflon
3. Bakelite

Solution:

Question 13.
Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Solution:
Benzoyl peroxide

Question 14.
How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Solution:
Natural rubber is a linear polymer of isoprene i.e., 2-methy l-1, 3-butadiene.

In the polymer double bonds are located between C₂ and C₃ isoprene units. The cis- polyisoprene molecule consists of various chains held together by weak van der Waals interactions and has coiled structure. This cis-configuration about double bonds does not allow the chains to come closer for effective attraction due to weak van der Waals interactions. Thus, it can be stretched like a spring and exhibits elastic properties.

Question 15.
Discuss the main purpose of vulcanisation of rubber.
Solution:
Natural rubber becomes soft at high temperature (>335 K) and brittle at low temperature (<283 K) and shows high water absorption capacity. It is soluble in non¬polar solvents and is non-resistant to attack by oxidising agents. To improve upon these physical properties, a process of vulcanisation is carried out. This process consists of heating a mixture of raw rubber with sulphur and an appropriate additive at a temperature range between 373 K to 415 K. On vulcanisation, sulphur forms cross links at the reactive sites of double bonds and thus the rubber gets stiffened.

Question 16.
What are the monomeric repeating units of Nylon-6 and Nylon-6,6?
Solution:

Question 17.
Write the names and structures of the monomers of the following polymers :

1. Buna-S
2. Buna-N
3. Dacron
4. Neoprene.

Solution:

Question 18.
Identify the monomers in the following polymeric structures.

Solution:

Question 19.
How is dacron obtained from ethylene glycol and terephthalic acid?
Solution:
Dacron is obtained from ethylene glycol and terephthalic acid by condensation polymerisation reaction.

Question 20.
What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester.
Solution:
A large number of polymers are quite resistant to the environmental degradation processes and are thus responsible for the accumulation of polymeric solid waste materials. These solid wastes cause acute environmental problems and remain undegraded for quite a long time.

Aliphatic polyesters are one of the important classes of biodegradable polymers. One such example is of PHBV.

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NCERT Solutions for Class 12 Chemistry

Ncert Solutions for Class 12 Chemistry Chapter 15 Polymers

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