R: [Histonet] equation problem PLEASE help

[Massimo]

Your final solution:

1ml. will contain 10^-8 mole
Its molarità will be:
M = (10^-8 * 10^3)/1 = 10^-5 (mole / litre)

Starting solution:
1ml. contains 10^-3/1064 = 9.40*10^-7 mole

So you have to take:

1 : (9.40*10^-7) = Xml : 10^-5

Xml = (10^-5) / (9.40*10^-7) = 10,64 ml. of starting solution and bring them to 1 litre by adding 989.36 ml. of distilled water.

Best Regards,

Massimo Tosi

--- Mer 7/1/09, eca9 <@t> georgetown.edu <eca9 <@t> georgetown.edu> ha scritto:
Da: eca9 <@t> georgetown.edu <eca9 <@t> georgetown.edu>
Oggetto: [Histonet] equation problem PLEASE help
A: histonet <@t> lists.utsouthwestern.edu
Data: Mercoledì 7 gennaio 2009, 22:46

Good afternoon,
I am hoping someone out there will take pity on a "mathematically"
challenged individual such as myself. I have been trying for hours to wrap my
head around this equation and am now getting to the point where I am more
confused than ever. Please help me.
The protocol calls for 10nmol of a substance per 1ml needed. It comes in a
1mg/ml solution and has a molecular weight of 1064g/mol. How do I do this? If
for example I needed 2ml of the solution...
The clearer the explanation the better. I really want to understand the
calculation not just have an answer. PLEASE HELP ME.
Thank you,
Eva

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