Answer
lons to be transported = (10 mM)(100um³) (N)
= (0.010 mol)(10-13 L)(6.02×1023 ions mol)
=(6.02 x 106 ions)
because 100 ionophores are present, every one of them need to transport value of 6.02× 106 ions.
Hence the time needed would be (6.02 x 106 ions) (1sec/104 ions)
= 602 seconds or 10 minutes.
Work Step by Step
lons to be transported = (10 mM)(100um³) (N)
= (0.010 mol)(10-13 L)(6.02×1023 ions mol)
=(6.02 x 106 ions)
because 100 ionophores are present, every one of them need to transport value of 6.02× 106 ions.
Hence the time needed would be (6.02 x 106 ions) (1sec/104 ions)
= 602 seconds or 10 minutes.
