Calculate the volume (mL) of 2.06 M magnesium phosphate solution require to prepare 46.20 mL of 2.3998 %w/v solution of Mg3(PO4)2

Answer 1

I would convert 2.3998 w/v% to M first.
2.l3998 g Mg3(PO4)2/100 mL soln
mols Mg3(PO4)2 = grams/molar mass = 2.13998/molar mass = approximately but you need to do it more accurately = 0.00814 (which is far short of the number of significant figures allowed based on the % in the problem).
0.00814 moles/100 mL = 0.0814 mols/L = 0.0814M( approx).
Then use the dilution formula of
c1v1 = c2v2

Answer 2

If I make 1 Liter of a 2M magnesium phosphate solution, how many grams of magnesium phosphate do I need