Answer
(a) $2.08\space s$
(b) $1.6\times10^{-5}s$
(c) The answer is given below
Work Step by Step
(a) Time (t) required for the first solute to = $\frac{L^{2}}{2D}$
traverse a channel of length 0.01 m
$$t=\frac{(0.01\space m)^{2}}{2(2.4\times10^{-5}m^{2}/s)}=2.08\space s$$
(b) Here we use equation 14.6 $KE_{Avg}=\frac{1}{2}mv_{rms}^{2}$ to find the time.
$KE_{Avg}=\frac{1}{2}mv_{rms}^{2}=>v_{rms}=\sqrt {\frac{2(KE_{Avg})}{m}}-(1)$
We can write,
mass of a single = $\frac{[2(1.00794)+15.9994]\times10^{-3}kg/mol}{6.022\times10^{23}mol^{-1}}=2.99\times10^{-26}kg-(2)$
molecule (m)
$KE_{Avg}=\frac{3}{2}kT=\frac{3(1.38\times10^{-23}J/K)(293\space K)}{2}=6.07\times10^{-21}J-(3)$
(2),(3)=>(1),
$v_{rms}=\sqrt {\frac{2(6.07\times10^{-21}J)}{2.99\times10^{-26}kg}}=637\space m/s$
So, the time required for a water molecule to travel the distance 0.01 m is,
$t=\frac{0.01\space m}{637\space m/s}=1.6\times10^{-5}s$
(c) When a water molecule diffuses through air, it makes a very large amount of collisions with air molecules. So, speed and direction change abruptly as a result of each collision. Due to this case, the water molecules move slowly in a zigzag path from one end of the channel to the other. But in part (b) there are no obstacles and will have a larger displacement over a much shorter time. Therefore, the answer to part (a) is much larger than the answer to part (b)
