How Do You Prepare a 13.3 mM Trypan Blue Solution in 30% Ethanol?

[plutolover]

Concentrations and Molarity??

Homework Statement

This problem was given in class; the teacher left it to us to solve it on our own. He wrote this problem himself:

you have 150 millimolar of Trypan Blue. you need to get 13.3 millimolar Trypan Blue in a 30% ethanol solution in EXACTLY 10 milliliters. How do you do it?

Homework Equations

The Attempt at a Solution

at first, i thought it was a simple solution. I thought that because you need a 30% solution in 10 milliliters, that means that you need 3 mL of Ethanol, right? Then I thought that the other 7 mL would be Trypan Blue. I determined the formula weight of Trypan to be 960.81. But I don't know how much Trypan is needed in correlation to the Ethanol. Is this right? Is there a better way of doing this problem??

Thank you! This problem has been bothering me.

 

[Borek]

30% ethanol means ethanol/water mixture in which there are 30% of ethanol. Technically, as there will be 13.3 milimolar TB in that solution, final concentration of ethanol will be slightly lower (but you can safely ignore it).

Is the stock solution made using water, or ethanol as a solvent? You will need to take some of this solution (enough to prepare 10 mL of 13.3 mM TB), and then dilute it to 10 mL adding correct amounts of water and ethanol.

It is not clear if the solution should be 30% w/w, w/v or v/v.

I don't like the question, it tries to be clever, but is IMHO poorly defined and ambiguous.

 

[epenguin]

I don't like the question, it tries to be clever, but is IMHO poorly defined and ambiguous.

I seem to have seen a number of questions like that here of late, which are probably explained by desire of teachers to vary the same usual limited number of standard type questions, maybe their job is not easy.

Because of the several indefinitions pointed out by Borek, you should clearly state your assumptions when writing your calculation for presentation.

I think the most reasonable assumptions are that the stock solution is in 30% v/v ethanol. Reason: why would anyone want to dilute into part ethanol solution? Because I guess the dye has limited solubility in water. But if that's the problem then for the same reason the original solution would have had ethanol. You'd prefer 30% ethanol to 100% because evaporation would make the concentration less defined when it is 100%. (get points for looking up the solubility of this dye which I don't know.)

Then I said v/v because the others are not hugely different and the exact ethanol concentration of little importance, anyone in the lab would just do the easiest thing.

Even v/v is ambiguous offhand, because the /v could be water or mixture. 3 in 10 or 3 in 13? Top of head I'll say 3 in 10.

So if I had to do this in the lab (I think I have) I would make 10 ml of 30% ethanol water mixture, and then calculate x ml of the dye solution, and (10 - x) ml of the solvent mixture.

Of course I could always make larger volumes of the diluted solution and take 10ml of it.

Anyway, repeat just state in your setting out exactly what you are doing and assuming.

You are confusing yourself by considering the molecular mass of the dye - be clear that this does not enter into the problem; you are just diluting a solution of known molarity to make one of less molarity.