Answer :
Answer:
L = 0.0464 m
Explanation:
Given;
The relation between the time 't' and the distance 'L' as:
[tex]t=\frac{L^2}{2D}[/tex]
and,
Diffusion constant, D = 1.8 x 10⁻⁵ m²/s
time, t = 60 s
on substituting the values in the above relation, we get
[tex]60=\frac{L^2}{2\times1.8\times10^{-5}}[/tex]
or
[tex]L=\sqrt{2\times1.8\times10^{-5}\times60}[/tex]
or
L = 0.0464 m