How many `Na^+` ions are present in 100 mL of 0.25 M of NaCl solution ?

Options

A. `0.025 xx 10^23`
B. `1.505xx10^22`
C. `15xx10^22`
D. `2.5xx10^23`

Explanation

To find the number of `Na^+` ions present in 100 mL of a 0.25 M NaCl solution, we can follow these steps:

### Step 1: Understand the given data
We have:
- Molarity (M) of NaCl = 0.25 M
- Volume of solution = 100 mL

### Step 2: Convert volume from mL to L
Since molarity is expressed in moles per liter, we need to convert the volume from milliliters to liters.
\[ \text{Volume in liters} = \frac{100 \text{ mL}}{1000} = 0.1 \text{ L} \]

### Step 3: Calculate the number of moles of NaCl
Using the molarity formula:
\[ \text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in L}} \]
Rearranging the formula to find the number of moles:
\[ \text{Number of moles of NaCl} = \text{Molarity} \times \text{Volume in L} \]
Substituting the values:
\[ \text{Number of moles of NaCl} = 0.25 \, \text{mol/L} \times 0.1 \, \text{L} = 0.025 \, \text{mol} \]

### Step 4: Determine the number of `Na^+` ions
From the dissociation of NaCl in solution:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
This means that 1 mole of NaCl produces 1 mole of `Na^+` ions. Therefore, the number of moles of `Na^+` ions is the same as the number of moles of NaCl:
\[ \text{Number of moles of } Na^+ = 0.025 \, \text{mol} \]

### Step 5: Convert moles of `Na^+` ions to number of ions
Using Avogadro's number:
\[ \text{Number of particles} = \text{Number of moles} \times 6.022 \times 10^{23} \text{ ions/mol} \]
Substituting the values:
\[ \text{Number of } Na^+ \text{ ions} = 0.025 \, \text{mol} \times 6.022 \times 10^{23} \, \text{ions/mol} \]
Calculating this gives:
\[ \text{Number of } Na^+ \text{ ions} \approx 1.5055 \times 10^{22} \text{ ions} \]

### Final Answer
The number of `Na^+` ions present in 100 mL of 0.25 M NaCl solution is approximately \( 1.505 \times 10^{22} \) ions.

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