The student is now told that the four solids, in no particular order, are barium chloride (BaCl2), sugar (C6H12O6), butanoic acid (C3H7COOH), 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.

Out of those choices, the chemical that ionizes the most is NaBr. It is like table salt (NaCl) in its properties. Solutbility 121 grams in 100 mL

You might think "Well if NaBr is such a good conductor, maybe a Metal combined with a non metal is the key and BaCl2 should be next." And in this case you would be right. 36 grams will dissolve in 100 mL

C3H7COOH is an acid and it is soluble in water. It is an acid and the last H splits off. It is not quite as good as the top two, but good enough all the same.

Sugar is probably the least soluble but it does form a suspension. It would be at the bottom of the last.

The order of conductivities is: Sugar < Butanoic acid < Sodium bromide < Barium chloride

Explanation:

The conductivity of a solution is related to the concentration of ions. The higher the concentration of ions, the higher the conductivity.

Barium chloride is a strong electrolyte, according to the following equation.

BaCl₂ (aq) ⇒ Ba²⁺ (aq) + 2 Cl⁻ (aq)

Each mole of BaCl₂ produces 3 moles of ions. Thus, if the solution is 0.20 M BaCl₂, it will be 0.60 M in ions.

Sugar (C₆H₁₂O₆) is a non-electrolyte. Thus, the concentration of ions will be zero.

Butanoic acid is a weak electrolyte, according to the following equation.

C₃H₇COOH (aq) ⇄ C₃H₇COO⁻ (aq) + H⁺ (aq)

Due to its weak nature, the concentration of ions will be lower than 0.20 M.

Sodium bromide is a strong electrolyte, according to the following equation.

NaBr (aq) ⇒ Na⁺ (aq) + Br⁻ (aq)

Each mole of NaBr produces 2 moles of ions. Thus, if the solution is 0.20 M BaCl₂, it will be 0.40 M in ions.

The order of conductivities is: Sugar < Butanoic acid < Sodium bromide < Barium chloride