The number of aluminium ions present in 54g of aluminium (atomic weight 27) is

Options

A. 2
B. 18
C. `1.1 xx 10^(24)`
D. `1.2 xx 10^(24)`

Explanation

To find the number of aluminum ions present in 54g of aluminum, we can follow these steps:

### Step 1: Determine the number of moles of aluminum
To calculate the number of moles, we use the formula:

\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

Here, the given mass of aluminum is 54g and the atomic weight (molar mass) of aluminum is 27 g/mol.

\[ \text{Number of moles} = \frac{54 \, \text{g}}{27 \, \text{g/mol}} = 2 \, \text{moles} \]

### Step 2: Calculate the number of aluminum ions
We know that 1 mole of any substance contains Avogadro's number of entities (atoms, molecules, ions, etc.), which is approximately \(6.022 \times 10^{23}\).

Since we have calculated that there are 2 moles of aluminum, we can find the total number of aluminum ions by multiplying the number of moles by Avogadro's number:

\[ \text{Number of aluminum ions} = \text{Number of moles} \times \text{Avogadro's number} \]

\[ \text{Number of aluminum ions} = 2 \, \text{moles} \times 6.022 \times 10^{23} \, \text{ions/mole} \]

\[ \text{Number of aluminum ions} = 1.2044 \times 10^{24} \, \text{ions} \]

### Step 3: Round the answer
Rounding \(1.2044 \times 10^{24}\) to two significant figures gives us:

\[ \text{Number of aluminum ions} \approx 1.2 \times 10^{24} \, \text{ions} \]

### Final Answer
Thus, the number of aluminum ions present in 54g of aluminum is approximately \(1.2 \times 10^{24}\) ions.

---