Water and Solute Permeability of Plant Cuticles: Measurement and ...

204 6 Diffusion of Non-Electrolytes
10. Diffusion coefficients in the barrier must have been similar for all solutes.
11. If the ratio volume/area decreased by a factor of 0.5 and the half time by the
same factor, this implies that P was constant.
12. The wetting agent improved the contact between donor solutions and cuticle.
This increased contact area and uniformity of rates of penetration. As Tween 20
is not an accelerator adjuvant (Chap. 7) and Alar is an ionic solute, this is the
best explanation.
13. In steady state penetration, diffusion in the waxy limiting skin is rate-controlling,
while in simultaneous bilateral desorption the limiting skin is not involved
because most solutes are desorbed from the inner surface of the CM.
14. They differ in the nature of the donor. With a liquid donor, (2.27) k is determined.
If solutes dissolved in cutin and wax as in UDOS, the rate constant (k ∗ )
is marked with an asterisk. k ∗ is independent of the partition coefficient, while
k depends on K.
15. Diffusion of lipophilic solutes proceeds in the amorphous wax fraction, and
fluidity of this fraction is apparently the same in waxes of all species.
16. Differences among species in solute mobility observed with the same solute are
proportional to k ∗ 0
, which reflects differences in the tortuosity factors.
17. These ratios vary between 0.87 and 1.03 and the average of these values is 0.95,
which is not bad since different populations were used for these determinations.
18. According to (6.21), the product β ′ ×Vx must be added to log k ∗ .
19. If the same solute is used for determining k ∗ , differences in k ∗ among species
are proportional to the ratio A/Vdonor (6.14) and (6.16).
20. Prior to droplet drying P was 1.22 ×10 −7 m s −1 , and after droplet drying it was
1.28 × 10 −7 m s −1 . Hence, permeance was independent on urea concentration.
21. A thickness of 1.33µm can be calculated by dividing the amount of reconstituted
wax of 120µg cm −2 by wax density of 0.9g cm −3 . Using (2.35), D is
3.47 × 10 −17 m 2 min −1 or 5.79 × 10 −19 m 2 s −1 . Increasing the thickness of the
wax layer by a factor of 2 leads to a four-fold smaller slope, because at constant
D the product of ℓ 2 × slope 2 must be constant.
22. Using (6.23), a D of 5.25 × 10 −18 m 2 s −1 can be calculated.
23. Rearranging (6.24) or (2.18), a thickness ℓ of 5.95µm can be calculated.