How would you prepare 100 ml of 0.4 M MgSO4 from a stock solution of 2 M MgSO4?

My attempt at answering this:

Using the formula (Vi) (Mi) = (Vf) (Mf)

(Vi) (2M) = (100) (0.4) = 200-100 (final volume) = 100 ml

I am assuming you would add 100 ml of water to the 2 M MgSO4 200 ml to get the 0.4 M MgSO4 100 ml

OK, so to answer this question, you will simply use the molality equation which is as follows:

M1V1 = M2V2

In the givens you have:

M1 = 2M

V1 is the unknown

M2 = 0.4M

V2 = 100 ml

plug in the givens in the above equation:

2 x V1 = 0.4 x 100

therefore:

V1 = 20 ml

Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.