The number of `Cl^(-)` ions in 100 mL of 0.001 M HCl solution is

Options

A. `6.022 xx 10^(23)`
B. `6.022 xx 10^(23)`
C. `6.022 xx 10^(19)`
D. `6.022 xx 10^(24)`

Explanation

To find the number of \( Cl^- \) ions in 100 mL of a 0.001 M HCl solution, we can follow these steps:

### Step 1: Calculate the number of moles of HCl
The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. We can use the formula:

\[ \text{Moles of solute} = \text{Molarity} \times \text{Volume (in liters)} \]

Given:
- Molarity of HCl = 0.001 M
- Volume of solution = 100 mL = 0.1 L (since 1000 mL = 1 L)

Now, substituting the values:

\[ \text{Moles of HCl} = 0.001 \, \text{M} \times 0.1 \, \text{L} = 0.0001 \, \text{moles} \]

### Step 2: Determine the number of moles of \( Cl^- \) ions
Since HCl dissociates completely in solution:

\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]

This means that 1 mole of HCl produces 1 mole of \( Cl^- \) ions. Therefore, the number of moles of \( Cl^- \) ions is also 0.0001 moles.

### Step 3: Calculate the number of \( Cl^- \) ions
To find the number of ions, we use Avogadro's number, which is approximately \( 6.022 \times 10^{23} \) particles per mole.

\[ \text{Number of } Cl^- \text{ ions} = \text{Moles of } Cl^- \times \text{Avogadro's number} \]

Substituting the values:

\[ \text{Number of } Cl^- \text{ ions} = 0.0001 \, \text{moles} \times 6.022 \times 10^{23} \, \text{ions/mole} \]

Calculating this gives:

\[ \text{Number of } Cl^- \text{ ions} = 0.0001 \times 6.022 \times 10^{23} = 6.022 \times 10^{19} \text{ ions} \]

### Final Answer
The number of \( Cl^- \) ions in 100 mL of 0.001 M HCl solution is \( 6.022 \times 10^{19} \) ions.

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