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?

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.

Thank you for such a quick response DrBob222.

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

To find the number of nitrate ions in the solution, we need to determine the mass of barium nitrate in 87.6 dL of the solution and then convert it to moles of nitrate ions.

First, let's calculate the mass of the barium nitrate solution in grams. We know that the density of the solution is 1.09 g/cm³, and we have 87.6 dL of the solution.

To convert dL to cm³, we use the conversion factor 1 cm³ = 0.1 dL:

87.6 dL × 1 cm³/0.1 dL = 876 cm³

Now, we can calculate the mass of the solution:

mass = density × volume
mass = 1.09 g/cm³ × 876 cm³ = 954.84 g

Next, we need to determine the mass of barium nitrate within the solution. We are told that the solution is 18.7% barium nitrate by mass.

mass of barium nitrate = 18.7% × mass of solution
mass of barium nitrate = 0.187 × 954.84 g = 178.62 g

Now, let's convert the mass of barium nitrate to moles of nitrate ions. The molar mass of barium nitrate can be calculated by summing the molar masses of barium and nitrate.

Molar mass of Ba(NO3)2 = molar mass of Ba + 2 × molar mass of NO3
Molar mass of Ba = 137.33 g/mol
Molar mass of NO3 = 62.00 g/mol

Molar mass of Ba(NO3)2 = 137.33 g/mol + 2 × 62.00 g/mol
Molar mass of Ba(NO3)2 = 261.33 g/mol

Now we can calculate the number of moles of barium nitrate:

moles of Ba(NO3)2 = mass of barium nitrate / molar mass of Ba(NO3)2
moles of Ba(NO3)2 = 178.62 g / 261.33 g/mol = 0.683 mol

Finally, since each formula unit of barium nitrate contains 2 nitrate ions, we can calculate the number of nitrate ions:

number of nitrate ions = 2 × moles of Ba(NO3)2
number of nitrate ions = 2 × 0.683 mol = 1.366 mol

Therefore, there are 1.366 moles of nitrate ions in 87.6 dL of the given barium nitrate solution.