Determine the molarity of these solutions:

(Assume that all are aqueous)

a) 3.67 moles of Na₂SO₃ dissolved to make 3.04 liters of solution.

b) 0.629 moles of Al₂O₃ to make 1.500 liters of solution.

c) 4.783 grams of Na₂CO₃ to make 10.00 liters of solution.

d) 0.897 grams of (NH₄) ₂CO₃ to make 250 mL of solution.

e) 0.0348 grams of PbCl₂ to form 45.0 mL of solution.

a. 1.20M

b. 0.42M

c. 4.51x10⁻³ M

d. 0.037M

e. 2.78x10⁻³M

Explanation:

Molarity shows the moles of solute that occupy 1L of solution:

a. 3.04 L of solution have 3.67 moles of solute

1 L of solution has (3.67.1) / 3.04 = 1.20M

b. 1.5 L of solution have 0.629 moles of solute

1L of solution has (0.629.1) / 1.5 = 0.42M

c. Molar mass Na₂CO₃ = 105.98

Mass / Molar mass = 4.783 g / 105.98 g/m = 0.045 moles

10 L of solution have 0.045 moles

1L of solution has (1. 0.045) / 10 = 4.51x10⁻³ M

d. Molar Mass (NH₄) ₂CO₃ = 96.09 g/m

Mass / Molar mass = 0.897 g / 96.09 g/m = 9.33x10⁻³ moles

250mL have 9.33x10⁻³ moles of solute

1000 mL have (1000. 9.33x10⁻³) / 250 = 0.037M

e. Molar Mass PbCl₂ = 278.1 g/m

Mass / Molar mass = 0.0348 g / 278.1 g/m = 1.25x10⁻⁴ moles

45 mL of solution have 1.25x10⁻⁴ moles of salt

1000 mL of solution have (1000. 1.25x10⁻⁴) / 45 = 2.78x10⁻³M