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 \( \text{Al}_2(\text{SO}_4)_3 \)
We know that molarity (M) is defined as the number of moles of solute per liter of solution. The formula to calculate the number of moles from molarity is:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume in liters} \]

Given:
- Molarity = 0.15 M
- Volume = 100 mL = 0.1 L (since 1000 mL = 1 L)

Now, substituting the values:

\[ \text{Number of moles of } \text{Al}_2(\text{SO}_4)_3 = 0.15 \, \text{mol/L} \times 0.1 \, \text{L} = 0.015 \, \text{moles} \]

### Step 2: Determine the number of moles of \( \text{Al}^{3+} \) ions produced
From the dissociation of \( \text{Al}_2(\text{SO}_4)_3 \):

\[ \text{Al}_2(\text{SO}_4)_3 \rightarrow 2 \text{Al}^{3+} + 3 \text{SO}_4^{2-} \]

This means that 1 mole of \( \text{Al}_2(\text{SO}_4)_3 \) produces 2 moles of \( \text{Al}^{3+} \).

Thus, the number of moles of \( \text{Al}^{3+} \) ions produced from 0.015 moles of \( \text{Al}_2(\text{SO}_4)_3 \) is:

\[ \text{Number of moles of } \text{Al}^{3+} = 0.015 \, \text{moles} \times 2 = 0.030 \, \text{moles} \]

### Step 3: Convert moles of \( \text{Al}^{3+} \) to number of ions
Using Avogadro's number \( (6.022 \times 10^{23} \, \text{ions/mole}) \), we can find the total number of \( \text{Al}^{3+} \) ions:

\[ \text{Number of } \text{Al}^{3+} \text{ ions} = 0.030 \, \text{moles} \times 6.022 \times 10^{23} \, \text{ions/mole} \]

Calculating this gives:

\[ \text{Number of } \text{Al}^{3+} \text{ ions} = 0.030 \times 6.022 \times 10^{23} \approx 1.8066 \times 10^{22} \text{ ions} \]

### 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:

\[ \text{1.8} \times 10^{22} \text{ ions} \]

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