I wrote the answer for problem two, I just need help figuring out how to get 3 a
ID: 511703 • Letter: I
Question
I wrote the answer for problem two, I just need help figuring out how to get 3 and 4.
2. If a portable glucose meter wicked in 1 µL out of 100 µL drop of blood and operated with 1% efficiency, i.e., 1% of the glucose reacted to give current, how many coulombs are exchanged when the glucose concentration is 120 mg/dL? The molecular weight of glucose is 180 g/mol. The answer is: 6.4x10-6 Coulombs
3. For problem 2, if this number of coulombs exchanged through the circuit over 0.1 s, what is the average current in amps?
4. For problem 2, how many mAh of the battery were used for the amperometry? Mention why this meter allows for tiny batteries to be used.
Explanation / Answer
Concentration of Glucose = 120 mg/dl
=> 120 mg of glucose present in 0.1 L of Blood
=> Amount of glucose present in 10-4 L of Blood
Amount of glucose = 12 x 10-2 mg
But portable glucose meter wicked 1 micro litre for every 100 micro litre blood.
Therefore mass of glucose wicked = 1/100 x 12 x 10-2 mg = 12x10-4 mg = 12x10-7g
Since efficiency is 1% therefore mass of glucose reacted = 12x10-9g
Number of moles of glucose present = 12x10-9g/180 = 6.67 x 10-11moles
Using Faraday Law
Q = (ne)F
= 6.67 x 10-11 x 96500
= 6.44 x 10-6 C
3) q = 6.44 x 10-6 / 0.1
q = 6.44 x 10-5 amps
4) battery power = (.44 x 10-5 /60)*10-3 = (1.0666 x 10-6) *10-3 = 1.0666 x 10-9 in mAh
meter allows the tiny batteries beacause batteries are used as a power source for various functions. In electricity meters, this is typically to provide a backup power supply for the real-time clock in the event of a power cut. Smart meters for gas, water and heating on the other hand are often powered by the battery alone.