# Question d3ac2

Jan 26, 2015

The answer is $\text{0.237 mM}$.

So, you start with $\text{11.2 mg}$ of glucose and $\text{7.45 mL}$ of water. In order to determine the molarity of this solution, you must find the number of moles of glucose you have

$11.2 \cdot {10}^{- 3} \text{g" * ("1 mole")/("180 g") = 0.0622 * 10^(-3)"moles}$

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

$C = 0.237 \text{mmol"/"L" = "0.237 mM}$