thank you VERY much for your replies, DrBob.

Another question, however:

How many moles of ions -both cations and anions- are present in solution when 0.75 mol of potassium chloride is dissolved in 0.25L of distilled water?

How do I work this answer to be: 1.5mol?

KCl(s) + H2O ==> K+(aq) + Cl-(aq)

Since there is 1 K ion per KCl molecule and 1 Cl ion per KCl molecule, there will be 0.75 mols K ion and 0.75 mols Cl ion in solution. I gather from the question that this is what you want. Actually, there are a few mols of ions (H and OH) from the water but water only ionizes to a very small extent and I don't think that is the intent of the question. For example, the (H+) in H2 is 10-7 molar

I must have missed the O in H2O on the last line of the above response. To expand a little more, there would be only 10-7mol of H ion and the same of OH ion in pure water. That's not much H ion and OH ion.

To find the number of moles of ions in a solution, you need to consider the dissociation of the compound in water. In this case, you have 0.75 mol of potassium chloride (KCl) dissolved in 0.25 L of water.

When KCl dissolves in water, it dissociates into its constituent ions, K+ and Cl-. Since there is 1 K ion and 1 Cl ion per KCl molecule, when 0.75 mol of KCl is dissolved, you will have 0.75 mol of K+ ions and 0.75 mol of Cl- ions in the solution.

So, the total number of moles of ions (both cations and anions) in solution would be the sum of the moles of K+ and Cl- ions:
Total moles of ions = moles of K+ ions + moles of Cl- ions = 0.75 mol + 0.75 mol = 1.5 mol

Therefore, there are 1.5 moles of ions (both cations and anions) present in the solution when 0.75 mol of potassium chloride is dissolved in 0.25 L of distilled water.