Acetone, \(\mathrm{CH}_{3} \mathrm{COCH}_{3},\) is a nonelectrolyte; hypochlorous acid, \(\mathrm{HClO},\) is a weak electrolyte; and ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), is a strong electrolyte. (a) What are the solutes present in aqueous solutions of each compound? (b) If $0.1 \mathrm{~mol}$ of each compound is dissolved in solution, which one contains \(0.2 \mathrm{~mol}\) of solute particles, which contains \(0.1 \mathrm{~mol}\) of solute particles, and which contains somewhere between 0.1 and 0,2 mol of solute particles?

To determine the solutes present in the aqueous solutions of the three compounds, we need to know their behavior as nonelectrolytes, weak electrolytes, or strong electrolytes. Acetone is a nonelectrolyte, meaning it does not dissociate into ions in an aqueous solution. Therefore, the solute present in this solution will be unchanged acetone molecules, \(\mathrm{CH}_{3}\mathrm{COCH}_{3}\). Hypochlorous acid is a weak electrolyte, meaning it partially dissociates into ions in an aqueous solution. The solute present in this solution will be a mixture of undissociated hypochlorous acid, \(\mathrm{HClO}\), and its dissociated ions, \(\mathrm{H}^{+}\) and \(\mathrm{ClO}^{-}\). Ammonium chloride is a strong electrolyte, meaning it fully dissociates into ions in an aqueous solution. The solute present in this solution will be its dissociated ions, \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{Cl}^{-}\).

Now that we know the solutes present in each solution, we can find the number of solute particles in each. (a) For acetone: Since it does not dissociate, 0.1 mol of acetone in solution will contain 0.1 mol of solute particles. (b) For hypochlorous acid: Since it is a weak electrolyte and partially dissociates, 0.1 mol of the acid in solution will produce an amount of solute particles between 0.1 mol and 0.2 mol, depending on the degree of dissociation of the acid. (c) For ammonium chloride: Since it is a strong electrolyte and fully dissociates, 0.1 mol of the salt will produce 0.1 mol \(\mathrm{NH}_{4}^{+}\) ions and 0.1 mol \(\mathrm{Cl}^{-}\) ions. Together, these make up a total of 0.1 + 0.1 = 0.2 mol of solute particles in the solution.