Posted by myles on Wednesday, February 15, 2012 at 9:00pm.
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
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