Polymers Test

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Polymers Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

Terylene (Dacron) is the polyester of

A
Hexamethylenediamine and adipic acid

B
Vinyl chloride and formaldehyde

C
Melamine and formaldehyde

D
Ethylene glycol and terephthalic acid

Solution
Question 2
1 / -0

GRA is a copolymer of __________________.

A
butadiene and acrylonitrile

B
butadiene and adipic acid

C
chloroprene and acrylonitrile

D
chloroprene and adipic acid
Question 3
1 / -0

Free radical polymerisation requires a free radical initiator. The most commonly used free radical initiator is:

A
$$Ph-CO-O-O-COPh$$ (benzoylperoxide)

B
$$(CH_3)_3C-O-O-C(CH_3)_3$$ (tert-butyl peroxide)

C
$$Ph-N-(NO)-Ph$$ (azoxybenzene)

D
$$CH_2N_2$$ (diazomethane)

Solution

In a free radical addition polymerization, the choice of polymerization initiator depends mainly on two factors:
a) Solubility and b) Decomposition temperature.
If the polymerization is performed in an organic solvent, then the initiator should be soluble in that solvent and the decomposition temperature of the initiator must be at or below the boiling point of the solvent.
Commonly, Benzoyl peroxide suit these requirements.
For e.g., Benzoyl peroxide initiates the methacrylate (MMA) monomer to form a radical that can attack the double bound of another MMA monomer, to form PMMA.
Question 4
1 / -0

Saran is a copolymer of ___________.

A
vinyl chloride and vinyl acetate

B
vinylidene chloride and vinyl chloride

C
ethylene chloride and vinyl chloride

D
vinyl acctate and methyl acetate

Solution

Saran is a copolymer of vinylidene chloride ($$1,\:1$$-dichloroethene) and vinyl chloride.
Question 5
1 / -0

If $$N_1,\ N_2, \ N_3,....$$ are the number of molecules with molecular masses $$M_1,\ M_2,\ M_3, ...$$ respectively, then mass average molar mass is expressed as :

A
$$\dfrac{\sum NiMi^2}{\sum NiMi}$$

B
$$\dfrac{\sum NiMi}{\sum Ni}$$

C
$$\dfrac{\sum Mi^2}{\sum Ni}$$

D
$$\dfrac{\sum NiMi}{\sum Mi}$$

Solution

Polymer molecules come in different sizes (chain lengths for linear polymers) even ones of the same type.
So, the average molecular mass will depend on the method of averaging. The number average molecular mass is the ordinary arithmetic mean or average of the molecular masses of the individual macromolecules.
Thus, the right equation for mass average molar mass measurement is $$\dfrac{\sum NiMi^2}{\sum NiMi}$$.
Question 6
1 / -0

Molecular weights of macromolecules are determined by :

A
elevation of boiling point

B
depression in freezing point

C
osmotic pressure

D
None of these

Solution

The number average molar mass can be determined by osmometry. The osmotic pressure of a solution is dependent only on the relative number of particles in the solution (in case of dilute solutions), irrespective of the nature (called colligative property).
Therefore, by taking certain known weight of the macromolecule, one wants to calculate the molecular mass of a solution is prepared and the osmotic pressure is measured. Then, this is compared to the known osmotic pressure of standard solution which gives the relative number of particles required to create such a pressure. From this and the weight of macromolecule used to prepare the solution, one can obtain the required molecular weight (weight of Avogadro's number of particles).
Question 7
1 / -0

Identify $$X$$ in the reaction.

$$CHCl_3 \xrightarrow[HF]{SbF_3} CHF_2Cl \overset{1070\, K}{\rightarrow} (X)+2HCl$$

A
$$F_2C=CF_2$$

B
$$ClFC=CFCl$$

C
$$F_2C=CFCl$$

D
$$F_2C=CCl_2$$

Solution

When chloroform is treated with antimony trifluoride and $$HF$$, 2 chlorine atoms are replaced with 2$$ F$$ atoms.

When $$\displaystyle CHF_2Cl $$ is heated to 1070 K, terafluoroethylene (compound X) is obtained.

$$\displaystyle 2CHF_2Cl \xrightarrow {1070K} F_2C=CF_2 $$
Question 8
1 / -0

Average molecular mass of a polymer is determined by:

A
light scattering and ultracentrifuge method.

B
osmotic pressure.

C
depression of freezing point.

D
elevation in boiling point.
Question 9
1 / -0

The PDI (polydisersity index) is the ratio of weight to number-average molecular masses, $$M \bar{w} / M \bar{n}$$. In natural polymers, which are generally monodispersed, PDI is ............. and in synthetic polymers which are always polydispersed, PDI is ............. because $$M \bar {w}$$ is always ............. than $$M \bar {n}$$. Fill in the blanks.

A
greater than $$1$$, $$1$$, higher

B
$$1$$, greater than $$1$$, higher

C
less than $$1$$, $$1$$, lower

D
$$1$$, less than $$1$$, lower

Solution

The PDI (polydisersity index) is the ratio of weight to number-average molecular masses, $$M \bar{w} / M \bar{n}$$. In natural polymers, which are generally monodispersed, PDI is 1 and in synthetic polymers which are always polydispersed, PDI is greater than 1 because $$M \bar {w}$$ is always higher than $$M \bar {n}$$.
Question 10
1 / -0

A polymeric sample in which $$30\%$$ molecules have a molecular mass $$20,000$$, $$40\%$$ have $$30,000$$ and the rest $$30\%$$ have $$60,000$$. The $$( \bar{M}_{n})$$ and $$( \bar{M}_{w})$$ of this sample are, respectively:

A
$$36,000,\: 43,333$$

B
$$43,000,\: 36,000$$

C
$$72,000,\: 86,000$$

D
$$86,000, \:72,000$$

Solution

Number average molecular weight
$$( \bar{M}_{n}) = \dfrac {\Sigma N_iM_i}{\Sigma N_i} = \dfrac {30(20000)+40(30000)+30(60000)}{30+40+30}=36000$$

Weight average molecular weight

$$( \bar{M}_{w}) = \dfrac {\Sigma N_iM_i^2}{\Sigma N_iM_i}=
\dfrac {30(20000)^2+40(30000)^2+30(60000)^2}{30(20000)+40(30000)+30(60000)}=43333$$

Hence, the $$( \bar{M}_{n})$$ and $$( \bar{M}_{w})$$ of this sample are, respectively $$36,000$$ and $$43,333$$

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Polymers Test - 57

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Polymers Test - 57

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Question 1

Hexamethylenediamine and adipic acid

Vinyl chloride and formaldehyde

Melamine and formaldehyde

Ethylene glycol and terephthalic acid

Solution

Question 2

butadiene and acrylonitrile

butadiene and adipic acid

chloroprene and acrylonitrile

chloroprene and adipic acid

Question 3

$$Ph-CO-O-O-COPh$$ (benzoylperoxide)

$$(CH_3)_3C-O-O-C(CH_3)_3$$ (tert-butyl peroxide)

$$Ph-N-(NO)-Ph$$ (azoxybenzene)

$$CH_2N_2$$ (diazomethane)

Solution

Question 4

vinyl chloride and vinyl acetate

vinylidene chloride and vinyl chloride

ethylene chloride and vinyl chloride

vinyl acctate and methyl acetate

Solution

Question 5

$$\dfrac{\sum NiMi^2}{\sum NiMi}$$

$$\dfrac{\sum NiMi}{\sum Ni}$$

$$\dfrac{\sum Mi^2}{\sum Ni}$$

$$\dfrac{\sum NiMi}{\sum Mi}$$

Solution

Question 6

elevation of boiling point

depression in freezing point

osmotic pressure

None of these

Solution

Question 7

$$F_2C=CF_2$$

$$ClFC=CFCl$$

$$F_2C=CFCl$$

$$F_2C=CCl_2$$

Solution

Question 8

light scattering and ultracentrifuge method.

osmotic pressure.

depression of freezing point.

elevation in boiling point.

Question 9

greater than $$1$$, $$1$$, higher

$$1$$, greater than $$1$$, higher

less than $$1$$, $$1$$, lower

$$1$$, less than $$1$$, lower

Solution

Question 10

$$36,000,\: 43,333$$

$$43,000,\: 36,000$$

$$72,000,\: 86,000$$

$$86,000, \:72,000$$

Solution

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