[SamanthaMcOw]

Hi Folks!

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

[Aldebaran]

I haven’t checked your Mr but assuming it’s correct at 63.03 then you’d need 63.03g per litre for a molar solution. Thus for 0.08 molar you would need ….?

[Babcock_Hall]

Think about units cancellation.

[SamanthaMcOw]

I haven’t checked your Mr but assuming it’s correct at 63.03 then you’d need 63.03g per litre for a molar solution. Thus for 0.08 molar you would need ….?

then I will need 5,04 g/L

Is that correct ?

[Borek]

Looks OK now.

[Aldebaran]

Correct!