Polymer Characterization I: Solutions, Molecular Weight Averages and Their Determination, GPC and Mooney Viscosity

This chapter opens Section XE5.2 “Polymer Characterization” of XE5 Polymer Science and Engineering, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam. It answers the first part of the heading: “Solubility and swelling; Concept of molecular weight distribution and its significance; Concept of average molecular weight; Determination of number average, weight average, viscosity average and Z-average molecular weights; GPC; Mooney viscosity.” It teaches the thermodynamics of polymer solutions (Flory–Huggins, solubility parameter), the molecular-weight distribution and the Mn, Mw, Mv and Mz averages with the polydispersity index, how each average is measured (end-group analysis, osmometry, light scattering, ultracentrifuge, viscometry with the Mark–Houwink equation), the principle of GPC, and Mooney viscosity. The numericals are Mn, Mw and PDI from a distribution table, Mv from Mark–Houwink and Mn from end-group titration.

1. Solubility and swelling: Flory–Huggins and the solubility parameter

A polymer dissolves only if the Gibbs energy of mixing is negative, ΔG_m = ΔH_m − TΔS_m < 0. Long chains gain far less entropy than small molecules, so the entropy term is small, and dissolution depends mainly on a small or negative ΔH_m. The Flory–Huggins lattice theory gives, per mole of lattice sites, for a solvent (volume fraction φ₁, size 1) and a polymer of N segments (φ₂):

ΔG_m/RT = φ₁ ln φ₁ + (φ₂/N) ln φ₂ + χφ₁φ₂

The first two terms are the combinatorial entropy (always favourable, small because of the 1/N), and χ is the Flory–Huggins interaction parameter, the enthalpic term: χ small or negative means a good solvent. The osmotic second virial coefficient is A₂ ∝ (½ − χ), so χ = ½ is the theta (Flory) condition, where A₂ = 0 and the chain behaves ideally, with unperturbed dimensions. In the limit of very long chains the critical value for phase separation is χ_c = ½(1 + 1/√N)² → 0.5; hence a polymer of high molecular weight dissolves only if χ < about 0.5, and the higher the molecular weight, the narrower the window.

Solubility parameter. The Hildebrand parameter δ = (ΔE_vap/V)^½ = (cohesive energy density)^½ has units of (MPa)^½. For a regular solution ΔH_m/V = φ₁φ₂(δ₁ − δ₂)², which gives χ ≈ 0.34 + V₁(δ₁ − δ₂)²/RT, the 0.34 being an empirical entropic contribution. The rule of thumb is that a polymer dissolves in solvents with δ close to its own (“like dissolves like”), and polar or hydrogen-bonded polymers need these interactions matched, for which one splits δ into dispersion, polar and hydrogen-bonding parts. The parameter is useful only for non-specific interactions, and it can never predict a negative ΔH_m.

Swelling. A crosslinked network cannot dissolve; it absorbs solvent until the entropic elastic retraction of the network balances the mixing drive (Flory–Rehner). Equilibrium swelling is lower for higher crosslink density, so swelling measurements give crosslink density. A linear polymer dissolves slowly: solvent first diffuses in and swells the polymer to a gel layer, and only then do chains disentangle and dissolve; higher molecular weight and crystallinity slow this and crystalline polymers (polyethylene, nylon) often dissolve only when hot.

2. Molecular weight distribution and the averages Mn, Mw, Mv, Mz

A synthetic polymer is a mixture of chains of different length, so a single molecular weight does not exist. With N_i molecules of molar mass M_i (weight w_i = N_iM_i), the averages are defined as follows.

Molecular-weight averages
AverageFormulaMeasured by
Mn, numberΣN_iM_i / ΣN_i = Σw_i / Σ(w_i/M_i)End-group analysis, osmometry, cryoscopy
Mw, weightΣN_iM_i² / ΣN_iM_i = Σw_iM_i / Σw_iLight scattering, sedimentation equilibrium
Mz, zΣN_iM_i³ / ΣN_iM_i²Sedimentation equilibrium
Mv, viscosity[ΣN_iM_i^(1+a) / ΣN_iM_i]^(1/a)Dilute-solution viscometry

For any distribution Mn ≤ Mv ≤ Mw ≤ Mz, with equality only for a monodisperse sample; since 0.5 ≤ a ≤ 0.8 usually, Mv lies just below Mw, and Mv = Mw if a = 1. The polydispersity index PDI = Mw/Mn ≥ 1 measures breadth: about 1.0x for living anionic polymer, ≈ 1.5 (coupling) to 2 (disproportionation) for radical polymerization at low conversion, ≈ 2 for step-growth and metallocene polyolefins, and 4–10 or more for Ziegler–Natta polyethylene or highly branched LDPE.

Worked example. A sample has 4 chains of 10,000 g/mol, 4 chains of 20,000 g/mol and 2 chains of 50,000 g/mol (10 chains in all). ΣN_iM_i = 4×10⁴ + 4×2×10⁴ + 2×5×10⁴ = 2.2×10⁵ g/mol, so Mn = 2.2×10⁵/10 = 22,000. ΣN_iM_i² = 4×10⁸ + 4×4×10⁸ + 2×25×10⁸ = 7.0×10⁹, so Mw = 7.0×10⁹/2.2×10⁵ = 31,818, and PDI = 31,818/22,000 = 1.45.

⚠️ Which table do you have?
If the table gives weight fractions w_i, use Mn = 1/Σ(w_i/M_i) and Mw = Σw_iM_i; averaging M_i with weight fractions directly gives Mw, not Mn. If it gives mole fractions or numbers, use Mn = ΣN_iM_i/ΣN_i. The same table read in the two ways gives different Mn.

Why the distribution matters. Mn is dominated by the many short chains, Mw by the heavy ones. Properties that depend on the number of chains or on end groups follow Mn; properties that depend on chain entanglement follow Mw. Above the critical entanglement molar mass, zero-shear melt viscosity scales as η₀ ∝ Mw^3.4, so a small high-molar-mass tail raises viscosity and melt strength sharply. Toughness and tensile strength rise with Mn and level off, low-molar-mass fractions act as plasticiser (lower Tg, tackiness, extractables), and a narrow distribution gives sharp melting and predictable processing, while a broad one eases flow in injection moulding.

3. Number-average methods: end groups and osmometry

Mn is a colligative quantity: it counts molecules. End-group analysis titrates or spectroscopically counts the chain-end groups (carboxyl or hydroxyl ends of a polyester by titration with KOH; amine ends of nylon; NMR or UV for others). If each chain carries ν end groups of the type counted, Mn = (mass of sample × ν)/(moles of end groups). Example: 2.00 g of a linear polyester with one COOH at each end (ν = 2) consumes 0.80 mL of 0.100 M KOH: moles COOH = 0.080 mmol, chains = 0.040 mmol, Mn = 2.00 g/0.040×10⁻³ mol = 50,000 g/mol. The method works only if the number of ends per chain is known, so branching or loss of ends gives error, and it is insensitive above about 2×10⁴ g/mol because the end-group count becomes too small.

Membrane osmometry measures the osmotic pressure π of a dilute solution across a membrane that passes solvent but not polymer. The van’t Hoff expansion is π/c = RT(1/Mn + A₂c + …), so a plot of π/c against c extrapolated to c → 0 has intercept RT/Mn (c in g per unit volume) and slope RTA₂. Example: with R = 8.314 J mol⁻¹ K⁻¹, T = 300 K and intercept π/c = 0.0500 Pa m³ g⁻¹, Mn = 8.314 × 300/0.0500 = 49,884 g/mol. Because the membrane leaks low-molar-mass chains, osmometry reads slightly high for broad samples, and the useful range is roughly 10⁴ to 10⁶ g/mol. Vapour-pressure osmometry, ebulliometry and cryoscopy extend the same colligative idea to Mn below about 10⁴ g/mol.

🎯 Why colligative methods give Mn
Osmotic pressure depends on the number of solute particles per unit volume and not on their size. A 1 g sample of 2000-g/mol chains has 100 times as many particles as 1 g of 200,000-g/mol chains, which is exactly the weighting that makes Mn small whenever low-molar-mass chains are present.

4. Weight-average methods: light scattering and the ultracentrifuge

Rayleigh scattering from a dilute solution increases with the mass of the scatterer, so large chains scatter strongly: the intensity is weighted by M², which makes the technique an absolute measurement of Mw (no calibration against standards). The Zimm equation is Kc/R_θ = 1/(Mw P(θ)) + 2A₂c, where R_θ is the excess Rayleigh ratio at angle θ, c the concentration, K a constant containing the refractive-index increment (dn/dc)² and the wavelength, and P(θ) the particle scattering factor. The Zimm plot extrapolates to c → 0 and θ → 0 together: the common intercept is 1/Mw, the slope in c gives A₂, and the slope in sin²(θ/2) gives the radius of gyration Rg. Dust, which scatters strongly, must be filtered out, and dn/dc must be measured.

Ultracentrifugation. In sedimentation velocity a centrifugal field drives the polymer to sediment at a rate expressed by the sedimentation coefficient s; combined with the diffusion coefficient D, the Svedberg equation gives M = sRT/[D(1 − v̄ρ)], with v̄ the partial specific volume of the polymer and ρ the solvent density. In sedimentation equilibrium the centrifugal and diffusive fluxes balance and the concentration profile across the cell gives Mw and Mz (and the distribution), without needing s or D. Both are accurate but slow, which is why GPC has largely replaced them for routine work.

5. Dilute-solution viscometry and the Mark–Houwink equation

A capillary (Ubbelohde) viscometer gives the efflux time t of a polymer solution and t₀ of the pure solvent. Define: relative viscosity η_r = t/t₀ (density correction neglected in dilute solution); specific viscosity η_sp = η_r − 1; reduced viscosity η_sp/c; inherent viscosity (ln η_r)/c. The intrinsic viscosity [η] = lim(c→0) η_sp/c = lim(c→0) (ln η_r)/c, with the units of reciprocal concentration (dL/g or mL/g), is found by extrapolating the two lines to c = 0: Huggins η_sp/c = [η] + k_H[η]²c and Kraemer (ln η_r)/c = [η] − k_K[η]²c.

The Mark–Houwink equation [η] = K·M^a relates intrinsic viscosity to molar mass for a given polymer–solvent–temperature set, K and a being empirical constants found from fractions of known M. The exponent reports the chain shape: a = 0.5 at the theta condition, 0.5 to 0.8 for a flexible coil in a good solvent (the more expanded the coil, the larger a), near 1 for stiff chains and up to about 2 for rigid rods; a = 0 for compact spheres. Because a polydisperse sample contains different chains, the M obtained is the viscosity-average Mv, with Mn < Mv < Mw and Mv = Mw only when a = 1. In practice Mv lies just below Mw.

Worked example: for a theta solvent, a = 0.5, K = 0.10 mL/g (in mL g⁻¹ (g/mol)^−0.5) and a measured [η] = 30 mL/g: M = ([η]/K)^(1/a) = (30/0.10)² = 300² = 90,000 g/mol. Doubling M changes [η] by 2^0.5 = 1.41 only. Viscometry is simple and cheap but relative: K and a must be known, and the result is valid only for the conditions of their calibration.

⚠️ Units of K and the exponent
K carries units that depend on a and on the units of [η]; mixing dL/g with mL/g shifts M by a power of ten when a = 1, and by 10^(1/a) in general. Always check that the K you use matches the [η] units and the solvent and temperature.

6. Gel permeation chromatography and Mooney viscosity

GPC (SEC, size-exclusion chromatography) is the one routine method that returns the whole distribution. A dilute polymer solution is pumped through columns packed with porous cross-linked beads (polystyrene–divinylbenzene for organic solvents). Separation is by hydrodynamic volume, not by chemical interaction: large coils cannot enter the pores and run through the interstitial volume, so they elute first; smaller coils sample more pore volume and elute later. A concentration detector (refractive index, UV) records the signal against retention volume V_e, and a calibration of log M against V_e (a nearly linear curve in the working range) from narrow standards, usually polystyrene, converts it to a distribution from which Mn, Mw and PDI follow by summation.

The calibration is valid only for polymers with the same size-to-mass relation as the standards; for a different polymer the result is a polystyrene-equivalent mass. Because coils of different polymers with equal [η]M have equal hydrodynamic volume, the universal calibration plots log([η]M) against V_e and works for all polymers; with an on-line viscometer or a light-scattering detector, absolute masses and branching can be measured. Chains of equal size but different chemistry (or branching) co-elute, and sample–column adsorption distorts the result.

Mooney viscosity is the plant test for raw rubber and rubber compounds. A rotor of standard dimensions is turned at a fixed speed (2 rpm) in a closed, heated cavity filled with the sample, and the torque is read as an arbitrary scale of Mooney units. The usual designation is ML(1+4) at 100 °C: the large rotor (L), 1 minute of preheat, then reading after 4 minutes of rotation, at 100 °C. A higher Mooney number means a higher viscosity, which reflects a higher molecular weight, more long-chain branching or gel, or filler. It is an empirical single-point measure of processability, not an absolute viscosity or a molecular weight, and cannot be converted to Mn or Mw without a calibration for the specific polymer.

Key takeaways

  • Dissolution needs ΔG_m < 0; Flory–Huggins ΔG_m/RT = φ₁lnφ₁ + (φ₂/N)lnφ₂ + χφ₁φ₂, theta at χ = 0.5, and δ matching (χ ≈ 0.34 + V₁(δ₁ − δ₂)²/RT) predicts solvents; networks only swell.
  • Mn = ΣN_iM_i/ΣN_i, Mw = ΣN_iM_i²/ΣN_iM_i, Mz = ΣN_iM_i³/ΣN_iM_i²; Mn ≤ Mv ≤ Mw ≤ Mz; PDI = Mw/Mn; with weight fractions Mn = 1/Σ(w_i/M_i).
  • Mn: end-group titration (Mn = mass × ν/moles of ends) and osmometry (π/c → RT/Mn); Mw: light scattering (Zimm, absolute); Mz and Mw: sedimentation equilibrium.
  • Viscometry: [η] = lim η_sp/c, [η] = K·M^a gives Mv, relative method; a = 0.5 theta, 0.5–0.8 good solvent, up to 2 for rods.
  • GPC separates by hydrodynamic volume, large chains elute first, calibrated with narrow standards (universal calibration log[η]M); it alone gives the whole distribution.
  • Mooney viscosity ML(1+4) at 100 °C is an empirical single-point torque measure of rubber processability, rising with molar mass, branching and gel.

Practice questions (16)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In Flory–Huggins theory, the theta condition for a polymer–solvent pair corresponds to

    1. χ = 0, where the enthalpy of mixing is zero and the polymer is infinitely soluble
    2. χ = 1, where the polymer just fails to dissolve at any molecular weight
    3. χ = 0.5, where the second virial coefficient is zero and the chain has unperturbed dimensions
    4. χ < 0, where the chain collapses into a compact globule
    Show answer

    Answer: C — χ = 0.5, where the second virial coefficient is zero and the chain has unperturbed dimensions

    A₂ ∝ (½ − χ), so χ = ½ makes polymer–solvent and polymer–polymer contacts energetically balanced, giving A₂ = 0 and the ideal (unperturbed) chain size. Negative χ is a very good solvent that expands the coil, not a collapsed globule.

  2. Two samples of the same elastomer, one lightly and one heavily crosslinked, are immersed in a good solvent until equilibrium. The heavily crosslinked sample

    1. swells less, because the stiffer network retracts more strongly against the mixing drive
    2. swells more, because it has more segments able to interact with the solvent molecules
    3. dissolves completely, because the good solvent overcomes the crosslinks given enough time
    4. swells to the same degree, because equilibrium swelling depends only on χ
    Show answer

    Answer: A — swells less, because the stiffer network retracts more strongly against the mixing drive

    Equilibrium swelling is where the entropic retraction of the network chains balances the mixing drive; more crosslinks mean shorter network chains, stronger retraction and less swelling (Flory–Rehner), and covalent crosslinks never dissolve. Swelling depends on both χ and crosslink density.

  3. For a polydisperse polymer sample, which ordering of the averages is correct?

    1. Mn > Mv > Mw > Mz
    2. Mw < Mn < Mz < Mv
    3. Mz < Mw < Mv < Mn
    4. Mn < Mv < Mw < Mz
    Show answer

    Answer: D — Mn < Mv < Mw < Mz

    Each higher moment weights the heavy chains more, so Mn < Mv < Mw < Mz, with Mv just below Mw for typical exponents. Equality holds only for a monodisperse sample.

  4. Which technique determines the weight-average molar mass of a polymer in dilute solution without calibration against polymer standards?

    1. End-group titration, which counts the chain ends of each molecule
    2. Static light scattering, via the Zimm plot
    3. Capillary viscometry, using the Mark–Houwink constants of the system
    4. Membrane osmometry, from the osmotic pressure of dilute solutions
    Show answer

    Answer: B — Static light scattering, via the Zimm plot

    Light scattering intensity is theoretically related to Mw, so the Zimm plot gives an absolute Mw (plus A₂ and Rg). End-group titration and osmometry give Mn; viscometry gives Mv and needs K and a calibrated against other methods.

  5. In membrane osmometry of a broad-distribution polymer, chains of very low molar mass pass through the membrane. The Mn calculated from the intercept RT/(π/c)c→0 is therefore

    1. lower than the true Mn, because the leaked chains reduce the osmotic pressure
    2. equal to Mw, since the remaining chains are the heavy ones
    3. higher than the true Mn, because the chains that leak no longer contribute to the pressure
    4. unaffected, as osmotic pressure depends only on the mass concentration
    Show answer

    Answer: C — higher than the true Mn, because the chains that leak no longer contribute to the pressure

    Only retained chains generate π; the leaked, light chains are missed, so the apparent number of particles per gram is too small and Mn reads high. The pressure depends on the number concentration, not the mass concentration alone, which rules out the claim that it is unaffected.

  6. A monodisperse polymer of molar mass 100,000 g/mol is contaminated with 10 wt% of a monodisperse oligomer of 10,000 g/mol. Compared with the pure polymer, the blend has

    1. Mn ≈ 53,000 and Mw ≈ 91,000, so Mn falls much more than Mw
    2. Mn ≈ 91,000 and Mw ≈ 53,000, so Mw falls much more than Mn
    3. Mn ≈ 91,000 and Mw ≈ 91,000, since the weight average of the two components applies to both
    4. Mn ≈ 55,000 and Mw ≈ 55,000, the simple mean of the two molar masses
    Show answer

    Answer: A — Mn ≈ 53,000 and Mw ≈ 91,000, so Mn falls much more than Mw

    With weight fractions 0.9 and 0.1: Mn = 1/(0.9/10⁵ + 0.1/10⁴) = 1/(1.9×10⁻⁵) = 52,600 g/mol and Mw = 0.9 × 10⁵ + 0.1 × 10⁴ = 91,000 g/mol. Mn counts molecules, and the oligomer supplies many, so Mn is far more sensitive; a simple mean of 55,000 is the tempting error.

  7. For the same polymer, the Mark–Houwink exponent measured in a good solvent is 0.78 and in a theta solvent is 0.50. This indicates that in the good solvent the chain

    1. is a rigid rod, since a is close to 1
    2. is more expanded than its unperturbed random coil, so [η] rises faster with M
    3. has collapsed into a compact sphere, so [η] is nearly independent of M
    4. has a lower Mv than in the theta solvent for the same [η], because a is larger
    Show answer

    Answer: B — is more expanded than its unperturbed random coil, so [η] rises faster with M

    a = 0.5 is the unperturbed coil; a rise to 0.5–0.8 means polymer–solvent contacts expand the coil. A rigid rod needs a near 2 and a compact sphere a ≈ 0. The claim about a lower Mv confuses K and a; M also depends on K, which differs between the solvents.

  8. In gel permeation chromatography of a polystyrene sample dissolved in tetrahydrofuran on a porous polystyrene–divinylbenzene gel, which species elute first?

    1. The smallest chains, because they diffuse fastest through the column and reach the outlet first
    2. The largest coils, because they are excluded from the pores and travel only through the interstitial volume
    3. The most polar chains, because they adsorb only weakly on the gel and are carried along by the solvent
    4. All chains together, because separation depends only on the solvent and not on chain size
    Show answer

    Answer: B — The largest coils, because they are excluded from the pores and travel only through the interstitial volume

    GPC separates by hydrodynamic volume, not interaction: big coils cannot enter the pores, take the shortest path and elute first, while small coils explore pore volume and elute later. Mass diffusion speed is not the criterion, and adsorption is a nuisance, not the mechanism.

  9. A natural-rubber sample is reported as having ML(1+4) at 100 °C equal to 60. This number is best described as

    1. its number-average molar mass in kg/mol, read from a calibration curve
    2. its glass transition temperature in °C, measured under shear in the cavity
    3. an empirical viscosity torque reading under standard test conditions
    4. its degree of crystallinity as a percentage of the rubber mass
    Show answer

    Answer: C — an empirical viscosity torque reading under standard test conditions

    ML(1+4) at 100 °C is the Mooney reading with the large rotor after 1 min of preheat and 4 min of running; it is an arbitrary-unit viscosity measure that correlates with molar mass and branching but is not Mn, Tg or crystallinity.

  10. Which of the following techniques give a molar-mass average without needing a calibration curve made from standards of the same polymer?
    P: Membrane osmometry (Mn)
    Q: Static light scattering (Mw)
    R: Dilute-solution viscometry (Mv)
    S: Conventional GPC with narrow polystyrene standards

    1. P
    2. Q
    3. R
    4. S
    Show answer

    Answer: A — P; B — Q

    Osmotic pressure and light scattering follow from theory (RT/Mn intercept; Zimm intercept 1/Mw), so they are absolute. Viscometry needs K and a, and conventional GPC needs the calibration curve of log M against elution volume.

  11. Which of the following statements about polymer solubility in Flory–Huggins theory are correct?
    P: A small or negative χ favours dissolution.
    Q: The second virial coefficient A₂ is positive when χ < 0.5.
    R: A very high molar mass polymer dissolves more readily than a low molar mass one of the same type, because its entropy of mixing is larger.
    S: A covalently crosslinked polymer dissolves completely in a solvent with χ = 0.

    1. P
    2. Q
    3. R
    4. S
    Show answer

    Answer: A — P; B — Q

    A₂ ∝ (½ − χ), so a small χ gives a positive A₂ and favours mixing. The entropy of mixing per unit mass falls as chain length grows (the (φ₂/N)lnφ₂ term), so long chains dissolve less readily, and a network can only swell, never dissolve.

  12. Which of the following statements about molar-mass averages and the polydispersity index PDI = Mw/Mn are correct?
    P: PDI is never less than 1.
    Q: PDI = 1 only for a sample in which all chains have the same molar mass.
    R: Mw is more sensitive to the low-molar-mass chains than Mn is.
    S: Mv equals Mw when the Mark–Houwink exponent a = 1.

    1. P
    2. Q
    3. R
    4. S
    Show answer

    Answer: A — P; B — Q; D — S

    Mw ≥ Mn always, with equality only for a uniform sample, and Mv = [Σw_iM_i^a]1/a reduces to Σw_iM_i = Mw at a = 1. Mn is the average that is more sensitive to short chains; Mw is weighted toward the heavy ones.

  13. A polymer sample contains 4 chains of molar mass 10,000 g/mol, 4 chains of 20,000 g/mol and 2 chains of 50,000 g/mol. Using Mn = ΣN_iM_i/ΣN_i, calculate Mn in g/mol. Give the answer as an integer.

    Numerical answer — type the value.

    Show answer

    Answer: 22000

    ΣN_iM_i = 4×10,000 + 4×20,000 + 2×50,000 = 40,000 + 80,000 + 100,000 = 220,000 and ΣN_i = 10, so Mn = 22,000 g/mol.

  14. A sample consists of three fractions with weight fractions 0.2, 0.5 and 0.3 and molar masses 10,000, 30,000 and 100,000 g/mol respectively. Using Mn = 1/Σ(w_i/M_i) and Mw = Σw_iM_i, calculate the polydispersity index Mw/Mn to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 1.86

    Σ(w_i/M_i) = 0.2/10⁴ + 0.5/(3×10⁴) + 0.3/10⁵ = 2.0×10⁻⁵ + 1.667×10⁻⁵ + 0.3×10⁻⁵ = 3.967×10⁻⁵, so Mn = 25,210. Mw = 0.2×10⁴ + 0.5×3×10⁴ + 0.3×10⁵ = 47,000. PDI = 47,000/25,210 = 1.864, i.e. 1.86.

  15. In a theta solvent a polymer obeys the Mark–Houwink equation [η] = K·M^a with a = 0.5 and K = 0.10 mL g⁻¹ (g/mol)^−0.5. The measured intrinsic viscosity is [η] = 30 mL/g. Calculate the viscosity-average molar mass in g/mol. Give the answer as an integer.

    Numerical answer — type the value.

    Show answer

    Answer: 90000

    M = ([η]/K)^(1/a) = (30/0.10)^(1/0.5) = 300² = 90,000 g/mol; this is Mv, the viscosity average.

  16. A linear polyester has one carboxyl end group on each chain end (two per chain). A 2.00 g sample requires 0.80 mL of 0.100 mol/L KOH for titration of the end groups. Calculate Mn in g/mol, giving the answer as an integer.

    Numerical answer — type the value.

    Show answer

    Answer: 50000

    Moles COOH = 0.80 mL × 0.100 mol/L = 0.080 mmol. Two COOH per chain gives 0.040 mmol of chains, so Mn = 2.00 g/(0.040×10⁻³ mol) = 50,000 g/mol.