how many grams of solid magnesium nitrate should be added to 5.0 grams of solid sodium nitrate in a 500 mL volumetric flask in order to produce an aqueous 0.25M solution of NO3- ions (once the flask is filled to the 500mL volume line)?

Question

No idea where to start for this question.

Answer 1

How many mols NO3^- do you want.
That's M x L = mols = 0.25 x 0.5L = 0.125

How many mols do you have in the NaNO3? That's mols = grams/molar mass = 5.0/85 = approx 0.06 but you need a more accurate answer that the estimate.

How many more mols do you need to be furnished by Mg(NO3)? That's approx 0.125 - 0.06 = approx 0.06 mols NO3^-
Since Mg(NO3)2 furnishes 2 NO3^- for each 1 mol Mg(NO3)2 you need only 0.03 mols Mg(NO3)2. How many grams is that? That's grams = mols Mg(NO3)2 x molar mass Mg(NO3)2.