An aqueous solution of barium nitrate has a density of 1.09 g/cm3. The solution is 18.7% barium nitrate by mass. How many nitrate ions are present in 87.6 dL of this solution?

Answer 1

18.7% by mass means 18.7 g Ba(NO3)2 in 100 g solution. Use density to calculate the volume. mass = volume x density or volume = mass/density = 100 g/1.09 = approx 91.7 mL. Therefore, you have 18.7 g Ba(NO3)2 in 91.7 mL or 18.7 g Ba(NO3)2 in 9.17 dL. In 87.6 dL you will have
18.7 g x 87.6/9.17 = approx 179 g Ba(NO3)2/87.6 dL. How many mols is that?
mols = grams/molar mass = approx 179/261 = approx 0.7 mols Ba(NO3).
There are 6.02E23 molecules in 1 mol Ba(NO3)2, convert that to molecules in approx 0.7 mol. Then multiply by 2 since there are two nitrate ions/molecule Ba(NO3)2.

Answer 2

Thank you for such a quick response DrBob222.

Answer 3

It took 20 minutes to type all of that stuff. :-)