Question

When CL was 2.3×10−6 M and CR was zero, the flux across a microporous membrane was found to be 0.234 pmol cm−2 s−1. The free diffusion coefficient of the material being measured was 0.8×10−5 cm2 s−1. A. What is the permeability of the membrane to the material? B. If the thickness of the membrane is 10×10−6 m, what is the equivalent relative area available for diffusion of the material?

Answer 1

Answer:

A) 8.08*10^-10 cm^-1

B)  1.27 * 10^-5

Explanation:

A) PERMEABILITY OF THE MEMBRANE to the material = K * D/(dx) ---- (1)

K = J / (Cl - Cr )

  = 0.234 * 10^-12 / (2.3*10^-6 M - 0 )

  = 10 *10^-6 M

dx = 10*10^-6 m * 100 cm

    = 10 *10^-4 cm

D = 0.8 * 10^-5 cm^2s^-1

input the calculated values into equation 1

permeability of the membrane = 8.08*10^-10 cm^-1

B) equivalent relative area available for the diffusion of the material

at thickness of membrane = 10*10^-6 m

can be calculated using this relation = $\frac{flux * thickness}{diffusion coefficient * ( Cl -Cr)}$ ---------- (2)

flux  = 0.234 * 10^-12 mol.cm^-2.s^-1

thickness = 10*10^-4 cm

diffusion coefficient = 0.8 * 10^-5 cm^2.s

Cl = 2.3*10^-6 m

input the above values into equation 2

equivalent relative area available for the diffusion of the material

= 1.27 * 10^-5