Hello Eva
Put simply
1nmol is 10-9 moles
Therefore 10nmol is 10-8 moles
1 mole = 1,064g
Therefore 10-8 moles = 0.00001064g or 0.01064mg
Therefore your working solution of 10nmol/ml = 0.01064mg/ml
Your stock solution is 1mg/ml
Therefore your dilution factor is 1.0 =
94 (93.984962 exactly)
0.01064
Therefore you can add 1ml stock to 93ml of diluent to get your desired
concentration..
I hope this is useful
regards
Tony Reilly
Chief Scientist
Anatomical Pathology
Pathology Queensland
Level 1, Building 15
Princess Alexandra Hospital
Ipswich Rd,
Woolloongabba Q 4102
Australia
Ph: 07 32402412
Fax:07 32402930
[email protected]
>>> <[email protected]> 8/01/2009 7:46 am >>>
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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