How many moles of ions (cations and anions) are contained in 1.2 L of a 1.9 M solution of KCl?

KCl K+ + Cl-)

1232

To determine the number of moles of ions in the given solution, we need to use the concentration (Molarity) of the solution and the volume.

First, let's identify the cation (K+) and anion (Cl-) present in KCl. The chemical formula of KCl indicates that for each formula unit of KCl, one K+ ion and one Cl- ion are produced.

Given:
Molarity of KCl solution (c) = 1.9 M
Volume of the solution (V) = 1.2 L

To find the moles of K+ and Cl- ions in the solution, we can multiply the molarity of the solution by the volume:

moles of K+ = c × V
moles of K+ = 1.9 M × 1.2 L
moles of K+ = 2.28 mol

moles of Cl- = c × V
moles of Cl- = 1.9 M × 1.2 L
moles of Cl- = 2.28 mol

Therefore, in 1.2 L of a 1.9 M solution of KCl, there are 2.28 moles each of K+ and Cl- ions.