Question

What is the calcium concentration of a solution formed by adding \(1 \mathrm{~mol}\) of \(\mathrm{CaCl}_2\) to \(1 \mathrm{~L}\) of distilled water at \(298 \mathrm{~K}\) ? 1\. \(1 M\) 2\. \(1 \mathrm{~m}\) 3\. \(2 M\) 4\. \(2 m\)

Short answer

1. \(1 M\)

Step-by-step solution

Writing the dissolution equation for CaCl2

When CaCl2 dissolves in water, it dissociates into its constituent ions. The balanced equation for this process is: \[CaCl_2 (s) \rightarrow Ca^{2+} (aq) + 2Cl^- (aq)\] This means that for every 1 mole of CaCl2, 1 mole of calcium ions (Ca^2+) and 2 moles of chloride ions (Cl^-) are produced in the solution.

Calculating the initial moles and volume of the solution

We are given that 1 mol of CaCl2 is added to 1 L of distilled water. Thus, we have: Initial moles of CaCl2 = 1 mol We also have the volume of the solution: V = 1 L

Calculating moles of calcium ions

As we know from step 1, 1 mole of CaCl2 produces 1 mole of calcium ions (Ca^2+). Therefore, the moles of calcium ions in the solution will be equal to the initial moles of CaCl2. Moles of Ca^2+ = Initial moles of CaCl2 = 1 mol

Calculating the molarity of calcium ions

Molarity (M) is defined as the number of moles of solute per liter of solution. We have the moles of calcium ions and the volume of the solution, so we can calculate the molarity as: M = (Moles of Ca^2+) / V M = (1 mol) / (1 L) M = 1 mol/L So, the calcium concentration in the solution is 1 M. Looking at the given options, our answer is: 1. \(1 M\)