What is the concentration of NO3- ions in a solution prepared by dissolving 15.0 g of Ca(NO3)2 in enough water to produce 300. mL of solution?

Answer 1

For this question, we need to (1) calculate the # of moles of Ca(NO₃)₂ used; (2) calculate the molarity (concentration) of the solution in terms of Ca(NO₃)₂ , (3) Realizing that the compound is ionic and for every mole of Ca(NO₃)₂ there are 2 moles of NO₃ ^- ions produced, we can find the concentration in terms of the nitrate ion...

Here goes:

M.Mass of Ca(NO₃)₂ = 164.1; # moles used = 15.0g/164.1 g/mol = 0.0914 mol in 300. ml of solution

Molarity of the solution = 0.0914 mol/0.300 L =0.305 M

Moles of NO₃^- ions in the solution (concentration is 2x.305 = 0.610 M

Hopefully this thought process makes it clear! All the best!