Help! How would I answer the following chemistry questions? Thank you so much!!

Question

a) Determine the molarity and mass percent of a solution in mol/L that is prepared by completely dissolving 3.56g of SrBr₂ (molar mass is: 247.428 g/mol). The density of water is 1.00 g/mL.

b) A chemist adds 141 mL of a solution of K₂SO₄ to 378mL of 0.67 M CaCl_2. Assume the reaction goes to 100% completion and K₂SO₄ is the limiting reagent. If 17.74g of CaSO₄ (molar mass is: 136.14 g/mol) is produced, what was the concentration, in mol/L, of the K₂SO₄ solution that was added.

c) You are asked to make two nitrate containing stock solutions, Solution A and Solution B. To make Solution A, you mix 308 mL of 0.748 mol/L iron(III) nitrate with 153 mL of 0.888 mol/L potassium nitrate. What is the concentration of nitrate ions in Solution A in mol/L?

d) Making Solution B requires you to mix 731 mL of 0.627 mol/L magnesium nitrate with 759 mL of water. What is the concentration of nitrate ions in stock solution B in mol/L?

Answer 1

Too many questions asked at once. I'll start and see how far I get.

a). moles SrBr2 = 3.56 g x 1 mol / 247 g = 0.0144 moles

Molarity = moles / liter and there is some unknown volume of water. Since they don't give the volume, we CANNOT find the molarity. Knowing the density of water doesn't help in this situation.

b). K₂SO₄ + CaCl₂ ==> 2KCl + CaSO₄(s)

17.74 g CaSO4 x 1 mol CaSO4 / 136.14 g = 0.1303 moles CaSO4 formed

moles K2SO4 used = 0.1303 mol CaSO4 x 1 mol K2SO4 / 1 mol CaSO4 = 0.1303 mols K2SO4 used

Concentration of K2SO4 = 0.1303 moles / 141 ml x 1000 ml / L = 0.9241 mol / L

c). Soln A = Fe(NO₃)₃ iron(III) nitrate + KNO3 potassium nitrate

For Fe(NO₃)₃: 0.308 L x 0.748 mol/L x 3 mol NO₃^- / mol Fe(NO₃)₃ = 0.691 mols NO₃^-

For KNO₃: 0.153 L x 0.888 x 1 mol NO₃^- / mol KNO₃ = 0.136 mols NO₃^-

Total mols NO3- = 0.691 + 0.136 = 0.827 mols NO3-

Total volume = 308 ml + 153 ml = 461 ml = 0461 L

[NO3-] = 0.827 mol / 0.461 L = 1.79 mol / L