Asked by alex
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)?
No idea where to start for this question.
No idea where to start for this question.
Answers
Answered by
DrBob222
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.
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.
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