`0.2` mole of `HCI and 0.2` mole of barium chloride were dissolved in water to produce a `500mL` solution. The molarity of the `CI^-` ions is :

Options

A. `0.06 M`
B. `0.09 M`
C. `1.2 M`
D. `0.80 M`

Answer

To find the molarity of the \( Cl^- \) ions in the solution, we can follow these steps:

### Step 1: Identify the sources of \( Cl^- \) ions
We have two compounds contributing \( Cl^- \) ions:
1. Hydrochloric acid (HCl)
2. Barium chloride (BaCl\(_2\))

### Step 2: Calculate the moles of \( Cl^- \) from HCl
From the dissociation of HCl:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
For every mole of HCl, we get 1 mole of \( Cl^- \). Therefore, from 0.2 moles of HCl:
\[ \text{Moles of } Cl^- \text{ from HCl} = 0.2 \text{ moles} \]

### Step 3: Calculate the moles of \( Cl^- \) from BaCl\(_2\)
From the dissociation of barium chloride:
\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2 \text{Cl}^- \]
For every mole of BaCl\(_2\), we get 2 moles of \( Cl^- \). Therefore, from 0.2 moles of BaCl\(_2\):
\[ \text{Moles of } Cl^- \text{ from BaCl}_2 = 0.2 \times 2 = 0.4 \text{ moles} \]

### Step 4: Calculate the total moles of \( Cl^- \)
Now, we can find the total moles of \( Cl^- \) ions in the solution:
\[ \text{Total moles of } Cl^- = \text{Moles from HCl} + \text{Moles from BaCl}_2 = 0.2 + 0.4 = 0.6 \text{ moles} \]

### Step 5: Calculate the volume of the solution in liters
The total volume of the solution is given as 500 mL. We need to convert this to liters:
\[ \text{Volume in liters} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L} \]

### Step 6: Calculate the molarity of \( Cl^- \)
Molarity (M) is calculated using the formula:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Substituting the values we have:
\[ \text{Molarity of } Cl^- = \frac{0.6 \text{ moles}}{0.5 \text{ L}} = 1.2 \text{ M} \]

### Final Answer
The molarity of the \( Cl^- \) ions is \( 1.2 \, \text{M} \).

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