Number of aluminium ions in 0.055 g of aluminium oxide is

Options

A. `3.24 xx 10^(23)`
B. `6.48 xx 10^(24)`
C. `3.24 xx 10^(20)`
D. `6.48 xx 10^(20)`

Explanation

To find the number of aluminum ions in 0.055 g of aluminum oxide (Al₂O₃), we can follow these steps:

### Step 1: Calculate the Molar Mass of Aluminum Oxide (Al₂O₃)
The molar mass of Al₂O₃ can be calculated using the atomic masses of aluminum (Al) and oxygen (O):
- Atomic mass of Al = 27 g/mol
- Atomic mass of O = 16 g/mol

The molar mass of Al₂O₃ is calculated as follows:
\[ \text{Molar mass of Al₂O₃} = (2 \times \text{Atomic mass of Al}) + (3 \times \text{Atomic mass of O}) \]
\[ = (2 \times 27) + (3 \times 16) \]
\[ = 54 + 48 = 102 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of Al₂O₃
To find the number of moles of Al₂O₃ in 0.055 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ = \frac{0.055 \text{ g}}{102 \text{ g/mol}} \approx 0.000539 \text{ moles} \]

### Step 3: Calculate the Number of Formula Units of Al₂O₃
Using Avogadro's number (6.022 × 10²³), we can find the number of formula units of Al₂O₃:
\[ \text{Number of formula units} = \text{Number of moles} \times \text{Avogadro's number} \]
\[ = 0.000539 \text{ moles} \times 6.022 \times 10^{23} \text{ units/mol} \approx 3.24 \times 10^{20} \text{ units} \]

### Step 4: Calculate the Number of Aluminum Ions
Each formula unit of Al₂O₃ contains 2 aluminum ions (Al³⁺). Therefore, the total number of aluminum ions is:
\[ \text{Number of Al ions} = 2 \times \text{Number of formula units} \]
\[ = 2 \times 3.24 \times 10^{20} \approx 6.48 \times 10^{20} \text{ ions} \]

### Final Answer
The number of aluminum ions in 0.055 g of aluminum oxide is approximately \( 6.48 \times 10^{20} \) ions.

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