I would like to know if I am thinking correctly when solving this problem. I have to prepare 80mM ammonium formate (1L volume), MW 63.03 g / mol.
I calculated it this way:
(0,08 M* 1L)/63,03 g/mol= 0,13 g
So I need to weight 0,13 g into 1 L of mili water....is that correct, weight seems a bit low