Measurement of osmotic pressure is one of the techniques used to determine the molecular weight of l — Physical Chemistry — Kinetics Chemistry Question
Molecular weight determination by osmometry
Measurement of osmotic pressure is one of the techniques used to determine the molecular weight of large molecules, like polymers. The device, osmometer, used to measure the osmotic pressure, consists of a semipermeable membrane that separates pure solvent from a solution. The flow of solvent from pure solvent side to solution side, due to concentration gradient, across the semipermeable membrane is called osmosis. Polyvinylchloride, PVC, is one of the most widely used plastics and can be prepared via chain polymerization. In chain polymerization monomers are added to a growing polymer chain. A typical chain polymerization involves three main steps named as initiation, propagation, and termination. In termination reaction two growing chains combine to form either one dead polymer chain (termination by combination) or two dead polymer chains (termination by disproportionation). In an attempt to determine the molecular weight of PVC via osmotic pressure measurement, a PVC solution is prepared by dissolving 7.0 g of PVC in cyclohexanone (C6H10O) to make a 1.0 dm3 solution at 295 K. One arm of the osmometer is filled with this solution of density 0.980 g cm-3 and the other arm is filled with pure solvent cyclohexanone to the same level. After a certain time, the height of liquid in the solution side arm increases and at equilibrium a 5.10 cm level difference between two arms is recorded.
Calculate the osmotic pressure and average molecular weight of PVC. (density of Hg = 13.6 g cm–3, g = 9.81 m s–2).
Model Answer
π = g * ρ_soln * h_soln = g * ρ_Hg * h_Hg
h_Hg = (0.980 g cm-3 * 51.0 mm) / (13.6 g cm-3) = 3.68 mm
π = 3.68 mm Hg (noted as 33.68 mm Hg in source text typo)
π = 3.68 / 760 = 4.84 * 10^-3 atm
Using π = c * R * T:
c = (4.84 * 10^-3 atm) / (0.082 dm3 atm mol-1 K-1 * 295 K) = 2.0 * 10^-4 mol dm-3
n(PVC) = c * V = 2.0 * 10^-4 mol dm-3 * 1.0 dm3 = 2.0 * 10^-4 mol
M(PVC) = m / n = 7.0 g / (2.0 * 10^-4 mol) = 3.5 * 10^4 g mol-1
The kinetic chain length ν is the ratio of the number of monomer units consumed per activated center produced in the initiation step and used to estimate the mode of termination. In the chain polymerization of vinyl chloride to produce PVC, the concentration of active centers produced in the initiation step and the change in the concentration of monomer is found to be 1.00 ⋅ 10-8 mol dm–3 and 2.85 ⋅ 10-6 mol⋅dm–3, respectively. Calculate the kinetic chain length, ν.
Model Answer
Kinetic chain length (ν) = (number of monomer units consumed) / (activated centers produced)
ν = (2.85 * 10^-6 mol dm-3) / (1.00 * 10^-8 mol dm-3) = 285
Predict whether the termination is by combination or by disproportionation.
Model Answer
Molecular weight of vinyl chloride monomer (noted with a typo as styrene in the solution key) is 62.5 g mol-1.
The number average degree of polymerization (Xn) is:
Xn = M(PVC) / M(monomer) = 3.5 * 10^4 g mol-1 / 62.5 g mol-1 = 560
Since Xn = 2 * ν = 2 * 285 = 570 (approximated as 560 in the source), the termination is by combination.
The vapor pressure of pure solvent cyclohexanone is 4.33 torr at 25 °C. Calculate the vapor pressure of the PVC solution.
Model Answer
Density of solution is 0.980 g cm-3.
1.0 dm3 of the solution weighs 980 g, of which 7.0 g is PVC.
Mass of solvent = 980 - 7.0 = 973 g
n(solvent) = 973 g / 98 g mol-1 = 9.93 mol
X_PVC = n(PVC) / (n(PVC) + n(solvent)) = 2.0 * 10^-4 / (2.0 * 10^-4 + 9.93) = 2.0 * 10^-5
Vapor pressure of the PVC solution is:
p = p° * (1 - X_PVC) = 4.33 * (1 - 2.0 * 10^-5) ≈ 4.33 torr
Since the mole fraction of PVC in the solution is extremely small, there is a negligible effect on vapor pressure lowering.
For pure solvent cyclohexanone, normal freezing point is –31.000 °C. If the freezing point of the PVC solution is –31.003 °C, find the molal freezing point depression constant of cyclohexanone.
Model Answer
∆Tf = 0.003 °C
∆Tf = Kf * m
molality (m) = n(PVC) / mass of solvent (kg) = (2.0 * 10^-4 mol) / 0.973 kg = 2.06 * 10^-4 mol kg-1
Kf = ∆Tf / m = 0.003 K / (2.06 * 10^-4 mol kg-1) = 14.6 K kg mol-1