The diffusion constant for glucose in water is 0.70 105 cm2/s. About how much ti
ID: 61554 • Letter: T
Question
The diffusion constant for glucose in water is 0.70 105 cm2/s. About how much time would it take to get significant diffusion across the following structures? Assume that glucose diffuses through cells and membranes as fast as it does through water.
1. across a cell membrane 8 nm thick
_______seconds
2. from the center of a eukaryotic cell 6µm in radius to the cell membrane
______seconds
3.across the wall of the human heart 2.5 cm in thickness
________seconds
4. from the cells in the center of the cactus to the photosynthetic cells on the surface of a barrel cactus if the cactus has a radius of 12 cm
______seconds
Explanation / Answer
so according to question diffusion cofficient value is 0.7X 10^-5 cm2/sec that means a particle moves 0.7X 10-5 cm2 area during 1 sec
A. so the thickness of cell membrane is 9nm that means it 4.22/7.(9nm)2. and hence to cross such area required time is (4.22/7.(9nm)2)/ 0.7X 10-5 i.e. equal to 1454 X 10 -9 sec.
B. t= 4X3.1732X (6X10-4)^2 divided by .7X10^-5 i.e. equal to 0.646 sec.
C. similarly here time required is 113X10^5 sec.
d. t= 2608X10^5 sec