Answer
(a) $t=\frac{L^{2}}{2D}$
(b) 31.25 s
Work Step by Step
(a) We can write,
$C_{Avg}=\frac{C_{1}+C_{2}}{2}=\frac{C_{2}}{2}=\frac{m}{V}=\frac{m}{AL}$
$C_{2}=\frac{2m}{AL}-(1)$
Let's apply Fick's law as follows.
$m=\frac{DAC_{2}t}{L}$
(1)=>
$m=\frac{DA(\frac{2m}{AL})t}{L}=>t=\frac{L^{2}}{2D}$
(b) Let's plug known values into the above equation.
$t=(2.5\times10^{-2}m)^{2}\div [2(1\times10^{-5}m^{2}/s)]=31.25\space s$
