Answer to Question #222588 in Chemistry for MJD

Question

A solution contains 0.71 mol of aluminum chloride in a 910 mL solution. Calculate the molarity of all the ions (Al and Cl) in the solution.

Answer 1

Solution:

Convert mL to L:

1 L = 1000 mL

Thus, (910 mL solution) × (1 L / 1000 mL) = 0.91 L solution

One formula unit of aluminum chloride (AlCl₃) contains one aluminum cation (Al³⁺) and three chloride anions (Cl⁻).

This means that when 1 mol of aluminum chloride is dissolved in water, 1 mol of aqueous aluminum ions and 3 mol of aqueous chloride ions are formed.

AlCl₃(aq) → Al³⁺(aq) + 3Cl⁻(aq)

From this it follows that the number of moles of Al³⁺ and Cl^- present in this solution will be:

Moles of Al³⁺ = Moles of AlCl₃ = 0.71 mol Al³⁺

Moles of Cl⁻ = 3 × Moles of AlCl₃ = 3 × (0.71 mol) = 2.13 mol Cl⁻

Molarity = Moles / Solution volume

Therefore,

Molarity of Al³⁺ = (0.71 mol Al³⁺) / (0.91 L) = 0.78 mol/L Al³⁺ = 0.78 M Al³⁺

Molarity of Cl⁻ = (2.13 mol Cl⁻) / (0.91 L) = 2.34 mol/L Cl⁻ = 2.34 M Cl⁻

Molarity of AlCl₃ = (0.71 mol AlCl₃) / (0.91 L) = 0.78 mol/L AlCl₃ = 0.78 M AlCl₃

Answer:

Molarity of Al³⁺ is 0.78 M

Molarity of Cl⁻ is 2.34 M