A solute diffuses through a stagnant edible packaging film of thickness $\Delta z = 2\text{ mm}$. The diffusion coefficient is $D_{AB} = 1.6 \times 10^{-9}\text{ m}^2/\text{s}$. The concentration drop across the film is $\Delta C = 80.0\text{ mol/m}^3$. Under steady-state equimolar counter-diffusion, the mass diffusion flux is ________ $\text{mol/(m}^2\,\text{s)}$ (round off to four decimal places).
Answer Key:
0.0001
Step-by-Step Solution & Explanation
From Fick's first law of diffusion:
$J_A = D_{AB} \frac{\Delta C}{\Delta z}$
Given $D_{AB} = 1.6 \times 10^{-9}\text{ m}^2/\text{s}$, $\Delta C = 80.0\text{ mol/m}^3$, and $\Delta z = 0.002\text{ m}$:
$J_A = (1.6 \times 10^{-9}) \times \frac{80.0}{0.002} = 0.000064\text{ mol/(m}^2\text{ s)}$
Rounding to four decimal places: $0.0001$.