Molecular Weight and Distribution Questions and Answers

Q: The z-average molecular weight, \(\overline{M_{z}}\) can be estimated using

The z-average molecular weight is measured by Ultracentrifugation.In sedimentation equilibrium in an ultracentrifuge, the concentration gradient set up in the cell weights the molecules in a way that yields \(\overline{M_{z}}\) (and \(\overline{M_{w}}\)) directly. \(\overline{M_{z}}\) gives extra weight to the largest molecules and is sensitive to the high-mass tail of the distribution.End group analysis and osmometry count molecules, so they give the number average \(\overline{M_{n}}\). Viscometry gives the viscosity average \(\overline{M_{v}}\), which lies between the number and weight averages.

Q: The unit of intrinsic viscosity is

The unit of intrinsic viscosity is dl/g.Intrinsic viscosity is the limit of the reduced viscosity as concentration goes to zero:\[ [\eta] = \lim_{c \to 0} \frac{\eta_{sp}}{c} \]The specific viscosity \(\eta_{sp}\) is dimensionless, so \([\eta]\) has the units of reciprocal concentration, volume per mass, conventionally dl/g (or ml/g). It measures the volume a gram of polymer occupies in solution.g/dl is a concentration, the inverse. Poise is the unit of dynamic viscosity itself, which intrinsic viscosity is not, and g/mol is a molar mass.

Q: The order of average molecular weights of a polymer is

The order is \(M_{z} \gt M_{w} \gt M_{v} \gt M_{n}\).The averages weight the molecules differently. \(M_{n}\) counts every chain equally; \(M_{w}\) weights each chain by its mass, and \(M_{z}\) by its mass squared, so each is pulled further towards the heavy chains. For any sample that is not monodisperse this gives \(M_{z} \gt M_{w} \gt M_{n}\).The viscosity average \(M_{v}\) depends on the Mark-Houwink exponent a, usually 0.5 to 0.8, and lies between \(M_{n}\) and \(M_{w}\) (it equals \(M_{w}\) when a = 1). The other orders put \(M_{n}\) above \(M_{w}\), or \(M_{v}\) outside that range, which cannot happen.

Q: A certain polymer synthesized in the laboratory shows that all the chains have same number of repeat units (i.e., same degree of polymerization). The relationship between weight-average (\(\bar{M}_w\)), number-average (\(\bar{M}_n\)), and z-average (\(\bar{M}_z\)) molecular weights for this polymer can be expressed as

For this polymer \(\bar{M}_z = \bar{M}_w = \bar{M}_n\).If every chain has the same degree of polymerization, every chain has the same molar mass M. Each average is a ratio of moments of the distribution, for example \(\bar{M}_w = \sum n_iM_i^{2}/\sum n_iM_i\), and with only one value of M every such ratio equals M. The polymer is monodisperse, with dispersity \(\bar{M}_w/\bar{M}_n = 1\).The order \(\bar{M}_z \gt \bar{M}_w \gt \bar{M}_n\) holds for any polydisperse sample, not this one. The reverse order and the mixed inequality can never occur.

Q: For a polydisperse polymer, the viscosity-average molecular weight \((\bar{M}_{v})\) becomes equal to the weight-average molecular weight \((\bar{M}_{w})\), when the Mark-Houwink constant a is ___.

\(\bar{M}_{v} = \bar{M}_{w}\) when a = 1.0.The viscosity-average molecular weight is defined through the Mark-Houwink exponent a:\[ \bar{M}_v^{\,a} = \frac{\sum N_i M_i^{1+a}}{\sum N_i M_i} \]With a = 1 this becomes \(\sum N_{i}M_{i}^{2}/\sum N_{i}M_{i}\), which is exactly the weight average. For the usual values of a, between 0.5 (theta solvent) and 0.8 (good solvent), \(\bar{M}_{v}\) lies between the number and weight averages, closer to the weight average, so 0.5 and 0.7 do not make them equal; 0.1 is an unrealistic exponent for flexible coils.

Q: For any polydispersed system, which of the following statements is true ?

For a polydisperse system the number average molar mass is less than the mass average.The two averages are\[ \bar{M}_{n} = \frac{\sum n_{i}M_{i}}{\sum n_{i}} \]\[ \bar{M}_{w} = \frac{\sum n_{i}M_{i}^{2}}{\sum n_{i}M_{i}} \]The mass average weights each molecule by its mass, so the heavier chains count for more, and whenever the molecules differ in size it exceeds the number average. Their ratio, the polydispersity index Mw/Mn, is therefore greater than 1. The two coincide (option C) only for a monodisperse sample, and option A has the inequality the wrong way round.

Q: The number average molecular weight of a polymer can be determined by :

The number average molecular weight can be determined by vapour pressure osmometry.Colligative methods, such as vapour pressure osmometry, membrane osmometry, boiling point elevation and freezing point depression, depend on the number of solute molecules present, not on their size. Dividing the mass of polymer by the number of molecules counted in this way gives the number average molecular weight, Mn.Light scattering responds more strongly to larger molecules and gives the weight average Mw. Sedimentation equilibrium gives Mw or the z-average Mz, and solution viscosity gives the viscosity average Mv.

Q: The weight average molar mass (\(\overline{\mathrm{M}}\)w) of a polymer having 50 and 75 molecules with corresponding molecular mass of 5000 and 6000 g/mol respectively is :

The weight-average molar mass is 5,642 g/mol.\[ \bar{M}_{w} = \frac{\sum N_{i} M_{i}^{2}}{\sum N_{i} M_{i}} \]\[ \begin{aligned} \sum N_{i} M_{i}^{2} &= 50(5000)^{2} + 75(6000)^{2} \\ &= 1.25 \times 10^{9} + 2.70 \times 10^{9} \\ \sum N_{i} M_{i} &= 2.5 \times 10^{5} + 4.5 \times 10^{5} \\ \bar{M}_{w} &= 3.95 \times 10^{9} / 7.0 \times 10^{5} \\ &= 5642.9 \end{aligned} \]The key's 5,642 is this value without rounding. The number average would be 7.0 × 105/125 = 5,600, and 5,800, 6,800 and 10,000 fit neither average.

Direction for [ Question No: 1 To 2 ] :

A polymer mixture contains three monodisperse polystyrene samples A, B and C of molecular weights 10000, 20000 and 30000 g mol^-1, respectively.

If the mixture contains equal number of molecules of A, B and C, the weight average molecular weight, \(\overline{M_{w}}\) will be

A.\(1.93 \times 10^{4}\) g mol\(^{-1}\)
B.\(2.13 \times 10^{4}\) g mol\(^{-1}\)
C.\(2.33 \times 10^{4}\) g mol\(^{-1}\)
D.\(2.53 \times 10^{4}\) g mol\(^{-1}\)

Exam Year(s)

GATE XE - 2012,

Answer: C

The weight average molecular weight is \(2.33 \times 10^{4}\) g mol^-1.

With equal numbers of molecules, take one of each:

\[ \begin{aligned} \overline{M_{w}} &= \frac{\sum N_{i}M_{i}^{2}}{\sum N_{i}M_{i}} = \frac{(1 + 4 + 9) \times 10^{8}}{(1 + 2 + 3) \times 10^{4}} \\ &= \frac{14 \times 10^{8}}{6 \times 10^{4}} = 2.33 \times 10^{4} \end{aligned} \]

The number average for this mixture is simply 20000, and \(\overline{M_{w}}\) is higher because it weights the heavier molecules more. 1.93 and 2.13 fall too close to the number average, and 2.53 overweights the heaviest component.

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If the mixture contains equal masses of A, B and C, then the number average molecular weight, \(\overline{M_{n}}\) will be

A.\(1.64 \times 10^{4}\) g mol\(^{-1}\)
B.\(1.74 \times 10^{4}\) g mol\(^{-1}\)
C.\(1.84 \times 10^{4}\) g mol\(^{-1}\)
D.\(1.94 \times 10^{4}\) g mol\(^{-1}\)

Exam Year(s)

GATE XE - 2012,

Answer: A

The number average molecular weight is \(1.64 \times 10^{4}\) g mol^-1.

With equal masses w of each, the number of moles of each is w/M:

\[ \begin{aligned} \overline{M_{n}} &= \frac{\sum w_{i}}{\sum w_{i}/M_{i}} = \frac{3}{\frac{1}{10000} + \frac{1}{20000} + \frac{1}{30000}} \\ &= \frac{3}{1.833 \times 10^{-4}} = 1.64 \times 10^{4} \end{aligned} \]

This is the harmonic mean of the three masses. It is pulled below the simple average of 20000 because, at equal mass, there are more of the light molecules. The larger options do not account for that.

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The z-average molecular weight, \(\overline{M_{z}}\) can be estimated using

A.End group analysis
B.Osmometry
C.Viscometry
D.Ultracentrifugation

Exam Year(s)

GATE XE - 2012,

Answer: D

The z-average molecular weight is measured by Ultracentrifugation.

In sedimentation equilibrium in an ultracentrifuge, the concentration gradient set up in the cell weights the molecules in a way that yields \(\overline{M_{z}}\) (and \(\overline{M_{w}}\)) directly. \(\overline{M_{z}}\) gives extra weight to the largest molecules and is sensitive to the high-mass tail of the distribution.

End group analysis and osmometry count molecules, so they give the number average \(\overline{M_{n}}\). Viscometry gives the viscosity average \(\overline{M_{v}}\), which lies between the number and weight averages.

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The unit of intrinsic viscosity is

A.dl/g
B.g/dl
C.poise
D.g/mol

Exam Year(s)

GATE XE - 2012,

Answer: A

The unit of intrinsic viscosity is dl/g.

Intrinsic viscosity is the limit of the reduced viscosity as concentration goes to zero:

\[ [\eta] = \lim_{c \to 0} \frac{\eta_{sp}}{c} \]

The specific viscosity \(\eta_{sp}\) is dimensionless, so \([\eta]\) has the units of reciprocal concentration, volume per mass, conventionally dl/g (or ml/g). It measures the volume a gram of polymer occupies in solution.

g/dl is a concentration, the inverse. Poise is the unit of dynamic viscosity itself, which intrinsic viscosity is not, and g/mol is a molar mass.

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The order of average molecular weights of a polymer is

A.\(M_{z} \gt M_{w} \gt M_{v} \gt M_{n}\)
B.\(M_{w} \gt M_{z} \gt M_{n} \gt M_{v}\)
C.\(M_{n} \gt M_{w} \gt M_{v} \gt M_{z}\)
D.\(M_{z} \gt M_{v} \gt M_{n} \gt M_{w}\)

Exam Year(s)

GATE XE - 2018,

Answer: A

The order is \(M_{z} \gt M_{w} \gt M_{v} \gt M_{n}\).

The averages weight the molecules differently. \(M_{n}\) counts every chain equally; \(M_{w}\) weights each chain by its mass, and \(M_{z}\) by its mass squared, so each is pulled further towards the heavy chains. For any sample that is not monodisperse this gives \(M_{z} \gt M_{w} \gt M_{n}\).

The viscosity average \(M_{v}\) depends on the Mark-Houwink exponent a, usually 0.5 to 0.8, and lies between \(M_{n}\) and \(M_{w}\) (it equals \(M_{w}\) when a = 1). The other orders put \(M_{n}\) above \(M_{w}\), or \(M_{v}\) outside that range, which cannot happen.

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A certain polymer synthesized in the laboratory shows that all the chains have same number of repeat units (i.e., same degree of polymerization). The relationship between weight-average (\(\bar{M}_w\)), number-average (\(\bar{M}_n\)), and z-average (\(\bar{M}_z\)) molecular weights for this polymer can be expressed as

A.\(\bar{M}_z \gt \bar{M}_w \gt \bar{M}_n\)
B.\(\bar{M}_z = \bar{M}_w = \bar{M}_n\)
C.\(\bar{M}_z \lt \bar{M}_w \lt \bar{M}_n\)
D.\(\bar{M}_z \gt \bar{M}_w \lt \bar{M}_n\)

Exam Year(s)

GATE XE - 2024,

Answer: B

For this polymer \(\bar{M}_z = \bar{M}_w = \bar{M}_n\).

If every chain has the same degree of polymerization, every chain has the same molar mass M. Each average is a ratio of moments of the distribution, for example \(\bar{M}_w = \sum n_iM_i^{2}/\sum n_iM_i\), and with only one value of M every such ratio equals M. The polymer is monodisperse, with dispersity \(\bar{M}_w/\bar{M}_n = 1\).

The order \(\bar{M}_z \gt \bar{M}_w \gt \bar{M}_n\) holds for any polydisperse sample, not this one. The reverse order and the mixed inequality can never occur.

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For a polydisperse polymer, the viscosity-average molecular weight \((\bar{M}_{v})\) becomes equal to the weight-average molecular weight \((\bar{M}_{w})\), when the Mark-Houwink constant a is ___.

A.0.1
B.0.5
C.0.7
D.1.0

Exam Year(s)

GATE XE - 2026,

Answer: D

\(\bar{M}_{v} = \bar{M}_{w}\) when a = 1.0.

The viscosity-average molecular weight is defined through the Mark-Houwink exponent a:

\[ \bar{M}_v^{\,a} = \frac{\sum N_i M_i^{1+a}}{\sum N_i M_i} \]

With a = 1 this becomes \(\sum N_{i}M_{i}^{2}/\sum N_{i}M_{i}\), which is exactly the weight average. For the usual values of a, between 0.5 (theta solvent) and 0.8 (good solvent), \(\bar{M}_{v}\) lies between the number and weight averages, closer to the weight average, so 0.5 and 0.7 do not make them equal; 0.1 is an unrealistic exponent for flexible coils.

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For any polydispersed system, which of the following statements is true ?

A.Number average relative molar mass is greater than mass average relative molar mass
B.Number average relative molar mass is less than mass average relative molar mass
C.Number average relative molar mass coincide with mass average relative molar mass
D.None of the above

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

For a polydisperse system the number average molar mass is less than the mass average.

The two averages are

\[ \bar{M}_{n} = \frac{\sum n_{i}M_{i}}{\sum n_{i}} \]

\[ \bar{M}_{w} = \frac{\sum n_{i}M_{i}^{2}}{\sum n_{i}M_{i}} \]

The mass average weights each molecule by its mass, so the heavier chains count for more, and whenever the molecules differ in size it exceeds the number average. Their ratio, the polydispersity index M_w/M_n, is therefore greater than 1. The two coincide (option C) only for a monodisperse sample, and option A has the inequality the wrong way round.

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The number average molecular weight of a polymer can be determined by :

A.Vapour pressure osmometry
B.Sedimentation equilibrium method
C.Light scattering method
D.Viscosity method

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The number average molecular weight can be determined by vapour pressure osmometry.

Colligative methods, such as vapour pressure osmometry, membrane osmometry, boiling point elevation and freezing point depression, depend on the number of solute molecules present, not on their size. Dividing the mass of polymer by the number of molecules counted in this way gives the number average molecular weight, M_n.

Light scattering responds more strongly to larger molecules and gives the weight average M_w. Sedimentation equilibrium gives M_w or the z-average M_z, and solution viscosity gives the viscosity average M_v.

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10.

The weight average molar mass (\(\overline{\mathrm{M}}\)w) of a polymer having 50 and 75 molecules with corresponding molecular mass of 5000 and 6000 g/mol respectively is :

A.5,800
B.5,642
C.6,800
D.10,000

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The weight-average molar mass is 5,642 g/mol.

\[ \bar{M}_{w} = \frac{\sum N_{i} M_{i}^{2}}{\sum N_{i} M_{i}} \]

\[ \begin{aligned} \sum N_{i} M_{i}^{2} &= 50(5000)^{2} + 75(6000)^{2} \\ &= 1.25 \times 10^{9} + 2.70 \times 10^{9} \\ \sum N_{i} M_{i} &= 2.5 \times 10^{5} + 4.5 \times 10^{5} \\ \bar{M}_{w} &= 3.95 \times 10^{9} / 7.0 \times 10^{5} \\ &= 5642.9 \end{aligned} \]

The key's 5,642 is this value without rounding. The number average would be 7.0 × 10⁵/125 = 5,600, and 5,800, 6,800 and 10,000 fit neither average.

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Molecular Weight and Distribution Questions and Answers

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