The student is now told that the four solids, in no particular order, are aluminum chloride (AlCl3), sugar (C6H12O6), benzoic acid (C6H5COOH), and sodium bromide (NaBr). Assuming that conductivity is correlated to the number of ions in solution, rank the four substances based on how well a 0.20 M solution in water will conduct electricity. Rank from most conductive to least conductive. To rank items as equivalent, overlap them. View Available Hint (s)

AlCl₃, NaBr, C₆H₅COOH, C₆H₁₂O₆.

Explanation:

The more ions in solution, the greater the conductivity of a solution because these charged particles can carry electrons in the solution.

In the first place, when AlCl₃ dissolves in water it produces 4 ions:

AlCl₃ (aq) ⇄ Al³⁺ (aq) + 3 Cl⁻ (aq)

If the solution is 0.20M in AlCl₃, it will be 4 x 0.20M = 0.80M in ions.

Secondly, NaBr is a strong electrolyte (complete ionization) so it produces 2 moles of ions per each mole of NaBr dissolved:

NaBr (aq) ⇄ Na⁺ (aq) + Br⁻ (aq)

If the solution is 0.20M in NaBr, it will be 2 x 0.20M = 0.40M in ions.

Then, benzoic acid is a weak electrolyte (partial ionization) so it will produce less than 2 moles of ions per each mole of benzoic acid dissolved:

C₆H₅COOH (aq) ⇄ C₆H₅COO⁻ (aq) + H⁺ (aq)

If the solution is 0.20M in benzoic acid, the solution will be far below 0.40M in ions.

Finally, sugar has only nonpolar covalent bonds so it will produce no ions in solution, thus being a poor electricity conductor.