creating a 100 nanomolar stock solution

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creating a 100 nanomolar stock solution

Postby sarbot » Fri Sep 05, 2008 9:19 am

It's been more than 30 years since I took college chemistry. Appreciate it if you can confirm my calcs.

Molecular weight of my compound is 319.9, so I'll say 320.

If I remember correctly, then 1 mole of this compound is 320 grams.

So, through four successive dilutions using my pipet I come up with:

.0032 mg. in 1 liter of water is a 100 nanomolar stock solution.

Is that correct ?
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