Calculate the molarity of a solution of `CaCl_(2)` if on chemical analysis it is found that 500 mL of `CaCl_(2)` solution contain `1.505 xx 10^(23) Cl^(-)` ions.

Explanation

To calculate the molarity of a solution of \( \text{CaCl}_2 \) given that 500 mL of the solution contains \( 1.505 \times 10^{23} \) chloride ions, we can follow these steps:

### Step 1: Understand the dissociation of \( \text{CaCl}_2 \)
When \( \text{CaCl}_2 \) dissolves in water, it dissociates into one calcium ion (\( \text{Ca}^{2+} \)) and two chloride ions (\( \text{Cl}^- \)):
\[ \text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^- \]
This means that for every mole of \( \text{CaCl}_2 \), there are 2 moles of \( \text{Cl}^- \) ions produced.

### Step 2: Calculate the number of moles of \( \text{Cl}^- \) ions
We know that 1 mole of any substance contains Avogadro's number of entities, which is approximately \( 6.022 \times 10^{23} \). From the problem, we have:
\[ \text{Number of } \text{Cl}^- \text{ ions} = 1.505 \times 10^{23} \]

### Step 3: Calculate the number of moles of \( \text{CaCl}_2 \)
Since 2 moles of \( \text{Cl}^- \) ions come from 1 mole of \( \text{CaCl}_2 \), we can set up the following relationship:
\[ \text{Moles of } \text{CaCl}_2 = \frac{\text{Number of } \text{Cl}^- \text{ ions}}{2} \]
Substituting the value we have:
\[ \text{Moles of } \text{CaCl}_2 = \frac{1.505 \times 10^{23}}{2 \times 6.022 \times 10^{23}} \]
Calculating this gives:
\[ \text{Moles of } \text{CaCl}_2 = \frac{1.505 \times 10^{23}}{1.2044 \times 10^{24}} \approx 0.125 \text{ moles} \]

### Step 4: Convert volume from mL to L
The volume of the solution is given as 500 mL. To convert this to liters:
\[ \text{Volume in L} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L} \]

### Step 5: Calculate the molarity
Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:
\[ \text{Molarity} = \frac{\text{Number of moles of } \text{CaCl}_2}{\text{Volume in L}} \]
Substituting the values we have:
\[ \text{Molarity} = \frac{0.125 \text{ moles}}{0.5 \text{ L}} = 0.25 \text{ M} \]

### Final Answer:
The molarity of the \( \text{CaCl}_2 \) solution is \( 0.25 \text{ M} \).

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