The answer is "0.237 mM".
So, you start with "11.2 mg" of glucose and "7.45 mL" of water. In order to determine the molarity of this solution, you must find the number of moles of glucose you have
11.2 * 10^(-3)"g" * ("1 mole")/("180 g") = 0.0622 * 10^(-3)"moles"
This makes the concentration of your parent solution
C = n/V = (0.0622 * 10^(-3)"moles")/(7.45 * 10^(-3)L) = "0.00835 M"
The next step is to determine how many moles of glucose the aliquot will contain
n = C * V = 0.00835("mol")/L * 17.0 * 10^(-6)"L" = 0.142 * 10^(-6)
The volume of the new solution will be
V_("solution") = 17.0mu"L" + 583mu"L" = 600.0mu"L"
Therefore, the concentration will be
C = n/V = (0.142 * 10^(-6)"moles")/(600.0 * 10^(-6)"L") = 0.000237"moles"/"L", or
C = 0.237"mmol"/"L" = "0.237 mM"
