The number of `AI^(3+)` ions in 100 mL of 0.15 M solution of `AI_2(SO_4)_3` are:

Options

A. `18 xx 10^(22)` ions
B. `1.8 xx 10^(22)` ions
C. `18 xx 10^(24)` ions
D. `1.8 xx 10^(27)` ions

Answer

To find the number of \( \text{Al}^{3+} \) ions in 100 mL of a 0.15 M solution of \( \text{Al}_2(\text{SO}_4)_3 \), we can follow these steps:

### Step 1: Calculate the number of moles of \( \text{Al}_2(\text{SO}_4)_3 \)

Using the formula for molarity:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

Given:
- Molarity \( = 0.15 \, \text{M} \)
- Volume \( = 100 \, \text{mL} = 0.1 \, \text{L} \)

We can rearrange the formula to find the moles of solute:
\[ \text{moles of solute} = \text{Molarity} \times \text{Volume} \]
\[ \text{moles of } \text{Al}_2(\text{SO}_4)_3 = 0.15 \, \text{M} \times 0.1 \, \text{L} = 0.015 \, \text{moles} \]

### Step 2: Determine the number of moles of \( \text{Al}^{3+} \) ions

From the chemical formula \( \text{Al}_2(\text{SO}_4)_3 \), we see that each formula unit produces 2 \( \text{Al}^{3+} \) ions. Therefore, the number of moles of \( \text{Al}^{3+} \) ions can be calculated as:
\[ \text{moles of } \text{Al}^{3+} = 2 \times \text{moles of } \text{Al}_2(\text{SO}_4)_3 \]
\[ \text{moles of } \text{Al}^{3+} = 2 \times 0.015 \, \text{moles} = 0.030 \, \text{moles} \]

### Step 3: Calculate the number of \( \text{Al}^{3+} \) ions

To find the total number of ions, we use Avogadro's number, which is approximately \( 6.022 \times 10^{23} \) ions/mole:
\[ \text{Number of } \text{Al}^{3+} \text{ ions} = \text{moles of } \text{Al}^{3+} \times \text{Avogadro's number} \]
\[ \text{Number of } \text{Al}^{3+} \text{ ions} = 0.030 \, \text{moles} \times 6.022 \times 10^{23} \, \text{ions/mole} \]
\[ \text{Number of } \text{Al}^{3+} \text{ ions} = 1.8066 \times 10^{22} \, \text{ions} \approx 1.81 \times 10^{22} \, \text{ions} \]

### Final Answer
The number of \( \text{Al}^{3+} \) ions in 100 mL of 0.15 M solution of \( \text{Al}_2(\text{SO}_4)_3 \) is approximately \( 1.81 \times 10^{22} \) ions.

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