posted by Cass on .
how would you prepare 250 mL of 0.69M Al2(SO4)3 solution from a 2.0M Al2(SO4)3 stock solution?
Ah Ha! So you're posting the problem right. And you see there is a difference in m and M.
How many mols do you want? That's M x L = mols.
Then mols = grams/molar mass. You know molar mass and mols, solve for grams.