The number of F ions in 4.2 gm `AlF_(3)` is (Al = 27, F = 19)

Options

A. `0.05`
B. `9.03 xx 10^(22)`
C. `3.01 xx 10^(22)`
D. `3.01 xx 10^(22)`

Answer

To find the number of fluoride ions (F⁻) in 4.2 grams of aluminum fluoride (AlF₃), we can follow these steps:

### Step 1: Calculate the molar mass of AlF₃
The molar mass of AlF₃ can be calculated using the atomic masses of aluminum (Al) and fluorine (F):
- Atomic mass of Al = 27 g/mol
- Atomic mass of F = 19 g/mol

The formula for the molar mass of AlF₃ is:
\[ \text{Molar mass of AlF}_3 = \text{mass of Al} + 3 \times \text{mass of F} = 27 + 3 \times 19 \]
Calculating this gives:
\[ \text{Molar mass of AlF}_3 = 27 + 57 = 84 \text{ g/mol} \]

### Step 2: Calculate the number of moles of AlF₃ in 4.2 grams
Using the molar mass calculated in Step 1, we can find the number of moles of AlF₃ in 4.2 grams:
\[ \text{Number of moles of AlF}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{4.2 \text{ g}}{84 \text{ g/mol}} = 0.05 \text{ moles} \]

### Step 3: Determine the number of fluoride ions in AlF₃
Each formula unit of AlF₃ produces 3 fluoride ions (F⁻). Therefore, the number of fluoride ions can be calculated as:
\[ \text{Number of F}^- \text{ ions} = \text{Number of moles of AlF}_3 \times 3 \times \text{Avogadro's number} \]
Where Avogadro's number is \(6.02 \times 10^{23}\) ions/mol.
Substituting the values:
\[ \text{Number of F}^- \text{ ions} = 0.05 \text{ moles} \times 3 \times 6.02 \times 10^{23} \text{ ions/mol} \]
Calculating this gives:
\[ \text{Number of F}^- \text{ ions} = 0.15 \times 6.02 \times 10^{23} \approx 9.03 \times 10^{22} \text{ F}^- \text{ ions} \]

### Final Answer
The number of F⁻ ions in 4.2 grams of AlF₃ is approximately \(9.03 \times 10^{22}\).

---