[sammyjo06]

Hi everyone .

In my Physio Psych class, my group will be making intracranial injections of L-Dopa into the nucleus accuments of obese rats.

We need to know how much L-Dopa in grams per microliter to use to make a solution with a concentration of 10 mM.

The chemical formula of the compound we will be using is:

(HO)₂2C₆H₃CH₂CH(NH₂)CO₂H

The molecular weight is: 197.19 g/mol

I tried to solve the problem myself, but I am not sure if the amount I came up with is correct.

First, I converted mM to moles.

10 mM = .01 mol

.01 mol L-Dopa x (197.19g L-Dopa/1 mol L-Dopa) = 1.97 g L-Dopa

After that, I set up a proportion.

(1.97 g L-Dopa/1 L L-Dopa) = (x g L-Dopa/1 x 10^-6 L L-Dopa)

x = 1.97 x 10^-6 g / µL

Did I do this correctly? I haven't taken chemistry in a little while...

[enahs]

Yes, you did it correctly.

I am not really sure why in the end you put it in micrograms/ microliter. But whatever makes you happy.

[sammyjo06]

Yes, you did it correctly.

I am not really sure why in the end you put it in micrograms/ microliter. But whatever makes you happy.

Are those the correct units?

Microliters because we will be injecting one microliter of the solution into each hemisphere daily.

Grams because it's in a powdered form.

[Arkcon]

The question is, can you accurately mass such a tiny amount?  Can you then add a tiny amount of solvent, and stir to dissolve?  Also, check the potency on the bottle.  Biological reagents tend to have a purity factor you may need to compensate for.  Or you may be provided with an L-Dopa salt, and you will have to adjust for the correct quantity of L-Dopa.