Answer to Question #136319 in Chemistry for rolly

Question

How much NAD+ (mw=663g/mol) would I need to weigh to make up a 10nM solution of 1ml?

Answer 1

The number of the moles $n$ of NAD+ contained in $V =$ 1 mL of $c=$ 10nM solution is:

$n=cV = 10·10^{-9}\text{ (M)}·1·10^{-3}\text{(L)} =1·10^{-11}$ mol.

Therefore, its mass will be:

$m = nM = 10^{-11}\text{(mol)}·663\text{(g/mol)} = 6.63·10^{-9}$ g.

As this mass is very small, one can advice to prepare a more concentrated solution from the available amount of NAD+ and then dilute it.