1 2 pts Calculate the molarity of all the ions present in ea

1. (2 pts) Calculate the molarity of all the ions present in each of the following solutions of strong electrolytes a 0.100 g of MgCia in 100.0 ml of solution b. 55.1 mg of NH.Br in 500.0 mL of solution. c 6.43 g of Na2S in 1.00 L of solution. a: 0.610 g of AICIs in 250.0 mL of solution:

Solution

(a)

Mass of MgCl₂ = 0.100 g.

Molar mass of MgCl₂ = 24 + 2 (35.5) = 95 g/mol

Moles of MgCl₂ = mass / molar mass = 0.100 / 95.0 = 0.00105 mol

Volume of solution = 100.0 mL = 0.100 L

Molarity of MgCl₂ = moles / volume = 0.00105 / 0.100 = 0.0105 M

MgCl₂ (aq.) -----------> Mg²⁺ (aq.) + 2 Cl- (aq.)

[Mg²⁺] = [MgCl₂] = 0.0105 M

[Cl^-] = 2 * [MgCl₂] = 2 * 0.0105 = 0.0210 M

(b)

Mass of NH₄Br = 55.1 mg = 0.0551 g.

Molar mass of NH₄Br = 7 + 4 + 80 = 91 g/mol

Moles of NH₄Br = mass / molar mass = 0.0551 / 91 = 0.000605 mol

Volume of solution = 500.0 mL = 0.500 L

Molarity of NH₄Br = moles / volume = 0.000605 / 0.500 = 0.00121 M

NH₄Br (aq.) ------------> NH₄⁺ (aq.) + Br^- (aq.)
[NH₄⁺] = [Br^-] = [NH₄Br] = 0.00121 M

(c)

Mass of Na₂S = 6.43 g.

Molar mass of Na₂S = 2 (23) + 1 (32) = 78 g/mol

Moles of Na₂S = 6.43 / 78 = 0.0824 mol

Volume of solution = 1.00 L

Molarity of Na₂S = 0.0824 / 1.00 = 0.0824 M

Na₂S (aq.) -------------> 2 Na⁺ (aq.) + S^2- (aq.)

[Na⁺] = 2 * [Na₂S] = 2 * 0.0824 = 0.165 M

[S^2-] = [Na₂S] = 0.0824 M

(d)

Mass of AlCl₃ = 0.610 g.

Molar mass of AlCl₃ = 27 + 3 (35.5) = 133.5 g/mol

Moles of AlCl₃ = 0.610 / 133.5 = 0.00457 mol

Volume of solution = 250.0 mL = 0.250 L

Molarity of AlCl₃ = 0.610 /0.250 = 2.44 M

AlCl₃ (aq.) --------------> Al³⁺ (aq.) + 3 Cl^- (aq.)

[Al³⁺] = [AlCl₃] = 2.44 M

[Cl^-] = 3 * [AlCl₃] = 3 * 2.44 = 7.32 M