Chemistry- finding the milliliters of an aqueous solution

Question

How many milliliters of an aqueous solution of 0.199 M aluminum nitrate is needed to obtain 14.6 grams of the salt? 

Answer 1

This is a simple derivative equation of molarity where M=n/v. M represents the molarity which in this case is given in the problem as .199M. N represents the number of moles present in solution. You are not given the number of moles, but you are given a mass of 14.6 grams. We know that the compund is aluminum nitrate so it will have a formula of Al(NO3)3. We can use stoichiometry:

Start with what you are given-

14.6g Al(NO3)3 x 1 mole Al(NO3)3/212.996 g Al(NO3)3=this will give you n for your equation. Since we are looking for V, we need to arrange the formula to v=n/M. Keep in mind that the units for M are moles of solute per L so your answer will ne in L. Simply convert to milliliters.