Given the polmers : A = Nylon ; B = Buna-S ; C=Polythene. Arrange these in increasing order of their intermolecular forces (lower to higher)

B(Buna-S) < C(polythene) < A(nylon)

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B(Buna-S) > C(polythene) > A(nylon)

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B(Buna-S) > C(polythene) < A(nylon)

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B(Buna-S) < C(polythene) > A(nylon)

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Solution

The correct option is A
B(Buna-S) < C(polythene) < A(nylon)

By definition, elastomers are characterized by weak intermolecular forces. Fibres tend to have the strongest intermolecular forces.

On the basis of intermolecular forces:
Elastomers(Buna-S) < Thermoplastics(polythene) < fibres(Nylon-66) < Thermosetting

Thus, the correct trend is: B(Buna-S) < C(polythene) < A(nylon).

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Similar questions

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

(i) Nylon 6, 6, Buna-S, Polythene.

(ii) Nylon 6, Neoprene, Polyvinyl chloride.

Given the polymers (i) Nylon-6,6 (ii) Buna-S (iii) Polythene.

Arrange these in the increasing order of their intermolecular forces (Lower to higher)

6

6

6

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