dilutions again!please check - molar dilution (Nov/29/2008 )

Hi there, I am new to calculations and need help! I am trying to make PAF (PHORBOL 12-MYRISTATE 13-ACETATE) solution whose MW is 616.8. PAF is supplies as 1mg powder. How can I make 0.001M solution? Is the following calculation correct?

616830mg(616.8gr)=1M in 1000ml
0.61683 =0.001M in 1ml

1:0.61683= 1.62 so I add 1.62 ml to 1mg of PAF. Then this final conc. is 0.001M in 1ml.

If I take out 50microlitres of final conc. does it become 5x10^-5M. As I saw someone elses calculation like this 5microlitres of 10^-3M stock +495microlitres would make 10^-5 M. So this final 10^-5 is in 500microlitres isnt it?

Please expalin this to me thoroughly as it is ruining my lab life!!

cheers,

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Hi Elizabeth,

The first part of your calculation is correct - you need 0.61683mg powder in 1ml in order to achieve a final C of 0.001M.
As for the rest of the calculation - I didn't completely understand... what I would do, is simply weigh 0.61683mg, and add 1ml of fluid. True - this method considers the powder V as negligable.

Good Luck...

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First welcome, second don't cry by a big font it's irritating

If I take out 50microlitres of final conc. does it become 5x10^-5M ?
I don't know because you did not give the total volume

5 microlitres of 10^-3 M stock +495 microlitres is 500/5 = 10^-2 dillution therefore the concentration will 10^-3 . 10^-2 = 10^-5 M

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[quote name='Gerard' date='Nov 30 2008, 12:56 PM' post='158979']

Thanks Gerard,yes I am new and thats why I made a mistake of using big font! It does look irritating!! I was trying to understand how to post.

Thnaks for your answer. What I am confused about is when you have a say 5M solution (which is in 1000ml). Does 5 microlitre of this solution is 5M as well? Thats what I read somewhere online. I will find that quote and post it.

best wishes,

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[quote name='Gat' date='Nov 30 2008, 09:03 AM' post='158976']
Hi Elizabeth,

The first part of your calculation is correct - you need 0.61683mg powder in 1ml in order to achieve a final C of 0.001M.
As for the rest of the calculation - I didn't completely understand... what I would do, is simply weigh 0.61683mg, and add 1ml of fluid. True - this method considers the powder V as negligable.

Good Luck...



Thank you...I am kind of understanding. cheers

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Hello

Solutions, solutions, all your calculations are correct (I had someone check over my shoulder just to be sure).

You have to keep two basic formulas in your head to get through this:
First one: rule of three
Second: C1*V1=C2*V2 (where C is concentration and V is volume).

SO:

616.8 g/Mol --> 1M--> 1000 ml
(Divide two of them by 1000, and bring to mg and 1 ml)

616.8 mg/Mol --> 1M--> 1 ml

(reduce your concentration by 1000, you need less compound)

0.616.8 mg/Mol --> 0.001M--> 1 ml

(you are now in range)
0.6168 mg-->1 ml
1.0 mg -----> X
X=1.62

Now, if you take 5 ul of this solution (0.001M) and bring it to 500 ul,

C1*V1=C2*V2
(0.001 M)*(5ul)= (X M)* (500ul)

(0.001 M)*(5ul)= (X M)
(500ul)

X= 1*10(-5)M

(1 mM)*(5ul)= (X M)
(500ul)

X= 1*10(-2)mM= 10 uM


Good luck!

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Dear Elizabeth

Do not try to weigh 0.6168 mg of PMA. Jut add the solvent to it.
Aliquot it, hopefully in glass containers, but if not, plastic microtubes will have to do.
Close the tubes in a nitrogen enriched atmosphere. If you do not have a nitrogen gas tank, you can do this by placing a rack inside a cooler and adding a small amount of Liquid Nitrogen to the bottom. Place a second rack on top of the first one and dispense your solution into your tubes (in the second rack). Close them and freeze them. If you unsure and/or are by yourself in the lab, this can wait till tomorrow.

good luck

Alejandro

PS
I assume you know your way around liquid nitrogen, if not, wait for help

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For any solution preparation go to: http://www.graphpad.com/quickcalcs/Molarityform.cfm
I added to my favorites!!!

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Hi, it´s a wonderful page, but it might be better to leave it for double checking.
I jus think that learning how to prepare solutions is part of the basic training. And asking someone to check on your calculations actually leads to learning new ways to do stuff. Examples? Did you know that it is a LOT easier to dissolve high concentrations of BSA (Albumin) by putting your solution in the fridge? You can go to 20% in minutes ( and I learned by asking someone to check my calculations because I could not dissolve).

A.

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You are right all person should know first how to prepare solutions and how make calculations by hand. Thats the first thing that I teach to my students. But also give the confidence to them to ask me if any question or doubt instead of running to ask to other person. I notice that people don't ask to their lab mates and they should be the first to been ask, if they don't know then turn to here. Don't take me wrong...I love this forum and is good to ask in here, but the calculations of concentrations is something that is seen very often here and people should know that there is a section of Media and Solutions that could clarify doubts with out waiting for somebody to reply to a post.

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