Polymer Chemistry MCQ Quiz in मराठी - Objective Question with Answer for Polymer Chemistry - मोफत PDF डाउनलोड करा

The correct answer is  27,800

Concept:

• Polymers are macromolecules that are formed by the covalent linkage between many repeating small units. These units are called monomers. There are different types of polymers:

1. Homo polymers: contains only one monomer as the repeating unit (........A-A-A-A.........)
2. Co-polymers: consists of two or more different monomers as the repeating unit (-A-A-B-B-B-A-A-B-A-B-B-B-)
3. Block polymers: It contains two or more different monomers as the repeating unit in alternate arrangement (-A-A-B-B-A-A-B-B-A-A-)
4. Grafted polymers: Branches of a different monomeric can be added or grafted onto a linear chain of polymer.

The exact mass of a polymer can not be determined therefore we use the average molar masses. The table given below shows the different types of average molar masses:

a = any variable that depends upon the system or polymer under study called Mark Houwink exponent, and its value is \(0.5 < a < 1\)

• Polydispersity Index (PDI): The heterogeneity of a polymer sample is represented as PDI. Mathematically it is defined as the rate of weight average molar mass (\(\overline{M_{w}}\)) to the number average molar mass (\(\overline{M_{n}}\)); 
• It is represented by,

​\(PDI = \frac{\overline{M_{w}}}{\overline{M_{n}}} \)

\(\rm \text { P.D.I. }=\frac{M_\omega}{M_n}=1 \) (for monodisperse)

\(\rm \therefore \quad \overline{M}_w=\overline{M}_n\)

Explanation:-

• In this case, the formula for the number average molecular weight is provided above is used. Here, the percentages given (50%, 40% and 9%) are the proportional representation of each molecule type in the total protein sample and can be used as the ni values.
• It's important to note that the percentages don't add up to 100%, they add up to 99%. I'll assume it's a typo and proceed with the available data.
• The formula for the Number Average Molecular Weight (Mn) is:

\(Mn = Σ(ni * Mi) / Σni\)

• Substitute in the provided information:

Mn = [(0.50 * 20,000) + (0.40 * 30,000) + (0.09 * 60,000)] / (0.50 + 0.40 + 0.09)

• Now do the math:

Mn = [10,000 + 12,000 + 5,400] / 0.99

= 27,767.67 Da nearly equal to 27,800 Da

Conclusion:-

• So, the Number Average Molecular Weight of the protein sample, based on the data you provided, is 27,800 Da (daltons, the unit used for molecular weight).

Concept:-

• In stepwise polymerization any two monomers present in the reaction mixture can link together at any time and growth of the polymer is not confined to chains that are already forming.
• As a result, monomers are removed early in the reaction and, as we shall see, the average molar mass of the product grows with time.
• In chain polymerization an activated monomer, M, attacks another monomer, links to it, then that unit attacks another monomer, and so on. The monomer is used up as it becomes linked to the growing chains.
• High polymers are formed rapidly and only the yield, not the average molar mass, of the polymer is increased by allowing long reaction times.

Explanation:-

• The fraction (p) condensed(extent of reaction) of a polyesterification reaction time t is given by,

\(p=\frac{kt[A]_o}{1+kt[A]_o}\).....(1)

where, p is fraction condensed,

[A]_o is the initial reactant concentration,

k is the rate constant.

• Now, given time t=1.0 hour, of a polymer formed with kr = 1.80 × 10-2 dm3 mol-1 s-1 and initial monomer concentration of 3.00 × 10-2 mol dm-3

​t = 1 hr = 3600 sec

kr = 1.80 × 10-2 dm3 mol-1 s-1

[A]o = 3.00 × 10-2 mol dm-3

• Thus, from equation (1) we get,

\(p=\frac{1.80 × 10^{-2} dm^3 mol^{-1} s^{-1}\times 3600 sec\times 3.00 × 10^{-2} mol dm^{-3}}{1+1.80 × 10^{-2} dm^3 mol^{-1} s^{-1}\times 3600 sec\times 3.00 × 10^{-2} mol dm^{-3}}\)

p = 0.66

• The degree of polymerization, which is defined as the average number of monomer residues per polymer molecule.
• This quantity is the ratio of the initial concentration of A, [A]₀, to the concentration of end groups, [A], at the time of
  interest, because there is one -A group per polymer molecule.
• The degree of polymerization () of a polyesterification reaction time t is given by,

= 1+ kt[A]o,,,,,(2)

• Now, given time t=1.0 hour, of a polymer formed with kr = 1.80 × 10-2 dm3 mol-1 s-1 and initial monomer concentration of 3.00 × 10-2 mol dm-3

​t = 1 hr = 3600 sec

kr = 1.80 × 10-2 dm3 mol-1 s-1

[A]o = 3.00 × 10-2 mol dm-3

• From equation (2) we got,

= 1 + 1.80 × 10-2 dm3 mol-1 s-1 × 3600 sec ×3.00 × 10-2 mol dm-3

 = 2.94

Conclusion:-

Hence, the fraction condensed (extent of reaction) and the degree of polymerization at time t=1.0 hour, of a polymer formed with kr = 1.80 × 10-2 dm3 mol-1 s-1 and initial monomer concentration of 3.00 × 10-2 mol dm-3, are respectively 0.66 and 2.94

Concept:

• Polymers are macromolecules that are formed by the covalent linkage between many repeating small units. These units are called monomers. There are different types of polymers:

1. Homo polymers: contains only one monomer as the repeating unit (........A-A-A-A.........)
2. Co-polymers: consists of two or more different monomers as the repeating unit (-A-A-B-B-B-A-A-B-A-B-B-B-)
3. Block polymers: It contains two or more different monomers as the repeating unit in alternate arrangement (-A-A-B-B-A-A-B-B-A-A-)
4. Grafted polymers: Branches of a different monomeric can be added or grafted onto a linear chain of polymer

Explanation:

• The exact mass of a polymer can not be determined therefore we use the average molar masses. The table given below shows the different types of average molar masses:

a = any variable that depends upon the system or polymer under study called Mark Houwink exponent, and its value is \(0.5 < a < 1\)

• Polydispersity Index (PDI): The heterogeneity of a polymer sample is represented as PDI. Mathematically it is defined as the rate of weight average molar mass (\(\overline{M_{w}}\)) to the number average molar mass (\(\overline{M_{n}}\)); 
• It is represented by,

​\(PDI = \frac{\overline{M_{w}}}{\overline{M_{n}}} \)

\(\rm \text { P.D.I. }=\frac{M_\omega}{M_n}=1 \) (for monodisperse)

\(\rm \therefore \quad \overline{M}_w=\overline{M}_n\)

Conclusion:-

• Hence, option 2 is the correct answer.

Concept:-

• Polymers are macromolecules that are formed by the covalent linkage between many repeating small units. These units are called monomers. There are different types of polymers:
• Homo polymers: contains only one monomer as the repeating unit (........A-A-A-A.........)
• Co-polymers: consists of two or more different monomers as the repeating unit (-A-A-B-B-B-A-A-B-A-B-B-B-)
• Block polymers: It contains two or more different monomers as the repeating unit in alternate arrangement (-A-A-B-B-A-A-B-B-A-A-)
• Grafted polymers: Branches of a different monomeric can be added or grafted onto a linear chain of polymer.

Explanation:-

• The exact mass of a polymer can not be determined therefore we use the average molar masses. The table given below shows the different types of average molar masses:

a = any variable that depends upon the system or polymer under study called Mark Houwink exponent, and its value is \(0.5 < a < 1\)

• The number average molar mass \((\overline{M}_n)\) of the polymer is

\(\rm \overline{\mathrm{M} }n=\frac{\mathrm{N}_1 \mathrm{M}_1+\mathrm{N}_2 \mathrm{M}_2}{\mathrm{~N}_1+\mathrm{N}_2}=\frac{50 \times 500+75 \times 6000}{50+75}=5600\)

Conclusion:-

Hence, the calculated number average molar mass \((\overline{M}_n)\) of the polymer is 5600.

Correct option is (b)

Concept:

→ The RMS separation between the two ends of a polymer chain is a measure of the average distance between the two ends, taking into account the statistical fluctuations due to the thermal motion of the chain. It is calculated by finding the square root of the sum of the squared distances between each pair of monomer units in the chain.

→ For a polymer chain consisting of N monomer units, the number of possible configurations or conformations of the chain is extremely large, and it is not possible to calculate the RMS separation analytically in most cases. However, it is possible to use statistical mechanics to estimate the RMS separation for a polymer chain in a given state.

Explanation:

→ The root mean square (RMS) separation between the two ends of a polymer chain of N monomer units can be expressed as:

RMS = \(\sqrt{N}\times a\)

where "a" is the average bond length between the monomer units.

This equation shows that the RMS separation is proportional to the square root of N.

This relationship arises because the polymer chain is essentially a random walk, where each step has a certain length (the bond length "a") and direction. The more steps there are (i.e., the larger the value of N), the more likely it is that the polymer chain will have taken a long, meandering path, resulting in a larger RMS separation.

Conclusion:
Therefore, we can conclude that the RMS separation between the two ends of a polymer chain is proportional to the square root of the number of monomer units in the chain.

Explanation:

→ First, we need to calculate the number of moles of NaOH used in the titration. We can use the formula:

moles of NaOH = concentration of NaOH x volume of NaOH

→ moles of NaOH = 0.02 mol/L x 0.011 L

moles of NaOH = 0.00022 mol

→ Since the polymer has carboxylic acid end group, we can assume that all the NaOH reacts with the carboxylic acid group in the polymer.

Therefore, the number of moles of carboxylic acid groups in the polymer is also 0.00022 mol.

→ The molar mass of the polymer can be calculated using the formula:

molar mass = mass of polymer / moles of carboxylic acid groups

molar mass = 4.4 g / 0.00022 mol

= 20,000 g/mol

Since the answer is required in kg/mol, we need to convert the molar mass to kg/mol:

molar mass = 20,000 g/mol / 1000

= 20 kg/mol

Conclusion:
Therefore, the average molar mass of the polymer is 20 kg/mol, which corresponds to option 2.

CONCEPT:

Mark–Houwink Equation for Polymer Molecular Weight

[η] = K × M^a

• The Mark–Houwink equation relates the intrinsic viscosity [η] of a polymer solution to its molar mass (M):
• Where:
  • [η] = intrinsic viscosity (in dL g⁻¹)
  • K = 1.6 × 10⁻⁴ dL g⁻¹
  • a = 0.60 (empirical constant)
  • M = molar mass (g mol⁻¹)

EXPLANATION:

• Given:
  • [η] = 0.04 dL g⁻¹
  • K = 1.6 × 10⁻⁴ dL g⁻¹
  • a = 0.60
• Use the Mark–Houwink equation:

  0.04 = 1.6 × 10⁻⁴ × M^0.60

• M^0.60 = 0.04 / (1.6 × 10⁻⁴) = 250
• M = (250)^1/0.60 = (250)^5/3
• M ≈ 10,000 g/mol

Therefore, the molar mass of the polymer is closest to 10,000 g mol⁻¹.

CONCEPT:

Chain Polymerization

• In chain polymerization, polymer growth occurs through a reactive center (radical, cation, or anion) that is generated during initiation.
• The process involves three main steps:
  • Initiation: Formation of a reactive species.
  • Propagation: Successive addition of monomer units to the growing chain.
  • Termination: Deactivation of the reactive center.
• The average molecular weight (molar mass) of the polymer depends on how many chains are initiated versus how many monomer units are available.

EXPLANATION:

• If the initiation is slow, then fewer chains begin growing.
• Each of these chains has access to more monomer units, so they grow longer before termination.
• This is because the rate of chain growth (propagation) is often much faster than the rate of chain initiation. If initiation is slow, fewer polymer chains will be formed, but those that do form will have more time to grow before termination, resulting in longer chains and a higher average molar mass. 
• This leads to a higher average molar mass of the polymer product.
• This is a key concept in controlled/living polymerization, where limiting initiation increases chain length.

Therefore, the correct answer is The slower the initiation of the chain, the higher the average molar mass of the polymer in chain polymerization.

The correct answer is 0.993

Explanation:-

Extent of Reaction Calculation:

Hydroxyacid HO-(CH₂)₅-COOH, and it is polymerized. The number-average molar mass (M_n) of the resulting polymer is 20,000 g mol^-1.

Using the expression for number-average molar mass in terms of the extent of reaction (p) given by:

M_n = M₁ / (1 - p)

Where:

• M_n = 20,000 g mol^-1 (Given)
• M₁ is the molar mass of the monomer unit
• p is the extent of the reaction

First, calculate the molar mass of the monomer unit:

M₁ = molar mass of HO-(CH₂)₅-COOH

M₁ = (Hydroxyl group) + 5 × (CH₂) + (Carboxyl group)

• Molar mass of hydroxyl group (OH) = 17 g mol^-1
• Molar mass of CH₂ group = 14 g mol^-1
• Molar mass of carboxyl group (COOH) = 45 g mol^-1

Thus, M₁ = 17 + 5 × 14 + 45 = 17 + 70 + 45 = 132 g mol^-1

We now substitute the values into the given expression:

20,000 = 132 / (1 - p)

Solving for p:

1 - p = 132 / 20,000

1 - p = 0.0066

p = 1 - 0.0066

p = 0.9934

Conclusion:-

Thus, the extent of the reaction is approximately 0.993, making the correct answer option 4.

Concept:

The number average molar mass (\( \overline{M}_n \)) of a polymer is calculated by taking the sum of the products of the number of molecules (N_i) of each type and their corresponding molar masses (Mi), and then dividing by the total number of molecules (ΣN_i). The formula is given by:

\( \overline{M}_n = \frac{∑ (N_i \cdot M_i)}{∑ N_i} \)

Explanation:

Given the molar mass distribution:

• First, calculate the total number of molecules:

  • ∑Ni = 15 + 20 + 25 = 60 

• Next, calculate the sum of the products of the number of molecules and their corresponding molar masses:

  • ∑ (Ni.Mi) = (15× 4000) + (20× 5000) + (25× 2000)

  • = 60000 + 100000 + 50000 = 210000

• Finally, calculate the number average molar mass:

  • \( \overline{M}_n = \frac{∑ (N_i \cdot M_i)}{∑ N_i} = \frac{210000}{60} = 3500 \text{ g/mol} \)

​Conclusion:

Therefore, the correct answer is 3500.