[jennipaige]

If I have 2 mM of pyrene in decane, how do I make 10 mL each of 10 μM, 0.5 mM, 1.0 mM, and 1.5 mM solutions using the density of decane (0.73 g/mL)?

[Dan]

M = mol/L, you do not need density.

[jennipaige]

Well, don't I need the density to calculate the molarity of decane? Then I would get 5.13 mol/L. And do I just use decane to find V₁ from C₁V₁=C₂V₂? Do I take the original concentration of pyrene (2mM) into account at all? And is this a situation that would lend itself to serial dilution nicely?

[Dan]

Well, don't I need the density to calculate the molarity of decane?

No, if I understand your first post correctly decane is the solvent.

When you calculate how much NaCl needed to make 10 mL of 1 M NaCl(aq), do you calculate the molarity of the solvent (water)?

A 1.5 mM solution of pyrene in decane means that there is 1.5 mmol of pyrene per litre of solution and the solvent is decane.

[EpicWinston]

Okay, remember that 2mM is the original concentration (Molarity (M) = mols/Liters). The question is asking you to make many 10mL solutions with many different concentrations (10 μM, 0.5 mM, 1.0 mM, and 1.5 mM).

This question requires that you know the definition of several prefixes (Which I am sure you already know), but I will write them here to make the answer more clear. The following pre-fixes  you should know (for this question):

mili (m) = 10^-3
Micro (μ) = 10^-6

Knowing these conversion units, the problem should be easier to read/answer. {If I have 0.002 M of pyrene in decane, how do I make 10 mL each of 0.000010 M, 0.0005 M, 0.0010 M, and 0.0015 M solutions using the density of decane (0.73 g/mL)?}

Recall that the equation for dilution is M₁V₁ = M₂V₂
So,  0.002 M * V₁ = 10mL * 0.000010 M
0.002 M * V₁ = 10mL * 0.0005 M
0.002 M * V₁ = 10mL * 0.0010 M
0.002 M * V₁ = 10mL * 0.0015 M

How will you do it? Just take out what ever volume you find and add water until it reaches 10mL.