Question

14. When 10.0 grams of Na2S04 is dissolved to 500. mL with water: c. What is the molarity of sodium sulfate? d. What is the molarity of sodium ion in the solution? e. What is the molarity of sulfate ion in the solution? 15. You want to perform a chemical reaction and you need 15.0 grams of silver for the reaction to produce the desired yield of product. How much of a 0.75 M silver nitrate solution will you need to use in this reaction? 16. If you react 25.0ml of 1.5M H2504 with 35.0mL of 1.16M sodium hydroxide, which reactant is the limiting reagent? 17. A solution is prepared by dissolving 24.75 grams of calcium nitrate in 250.00mL of solution. The solution is then diluted using 5.00mL to make 75.00mL of solution. a. What is the calcium nitrate concentration of each solution? b. What is the calcium concentration in each solution? c. What is the nitrate concentration of each solution? d. I if 35.00mL of the dilute solution was evaporated to dryness, what mass of calcium nitrate should be recovered?

Answer 1

14) c) moles of Na₂SO₄ = 10 gms / 142.04 g/mol = 0.07040 mol
Molarity of Na₂SO₄ = 0.07040 mol / (500/1000)L = 0.1408 mol/L = 0.1408 M
d) In Na₂SO₄ there are 2 equivalent Na⁺ ions so
Molarity of sodium ions = 2 * 0.1408 = 0.2816 M
e) In  Na₂SO₄ there is 1 equivalent SO₄^2- ions
Molarity of silphate ions = 0.1408 M

15)In AgNO₃ there is 1 equivalent of Ag⁺ ion and hence molarity of silver is 0.75M
0.75 mol/L = (15gm / 169.87 g/mol) / V in L
0.75 mol/L = 0.0883 mol / V
Volume in L = 0.0883 mol / 0.75 mol/L = 0.1177 L = 117.7 mL of AgNO₃ is required.

16) moles of H₂SO₄ = 1.5 mol/L * (25/1000)L = 0.0375 mol
moles of NaOH = 1.16 mol/L * (35/1000)L = 0.0406 mol
The reactant with less number of moles will be the limiting reactant
Hence H₂SO₄ is the limiting reactant.

17)a) Conc of Ca(NO₃)₂ before dilution = (24.75g/164.088 g/mol) / (250/1000)L = 0.6033 M
moles of Ca(NO₃)₂ = (5/1000)L * 0.6033mol/L = 3.0166*10^-3 mol
Conc of Ca(NO₃)₂ after dilution = (3.0166*10^-3 mol) / (75/1000)L = 0.0402 M
b) There is 1 equivalent of Ca²⁺ ion hence,
Calcium conc before dilution =  0.6033 M
Calcium conc after dilution = 0.0402 M
c) There are 2 equivalents of NO₃^- ions hence,
Nitrate conc before dilution = 0.6033 * 2= 1.2066 M
Nitrate conc after dilution = 0.0402 *2 = 0.0804 M
d) Leftover Ca(NO₃)₂ solution = 75-35 = 40mL
moles leftover = (40/1000)L * 0.0402 mol/L = 1.608*10^-3 mol
mass of Ca(NO₃)₂ to be recovered = 1.608*10^-3 mol *164.088 g/mol = 0.2638 gms