Question

Calculate the number of sulphate ions present in 22. 5 cubic centimeters of 2 M aluminium sulphate solution

Answer 1

To find the number of sulfate ions in a given volume of a 2 M aluminium sulfate solution, we can follow these steps:

### 1. Determine the amount of aluminium sulfate in moles

First, we know that:
- Molarity (M) = moles of solute / liters of solution

Given that the molarity of the solution is 2 M, and we have 22.5 cubic centimeters (cc) of the solution, we need to convert cc to liters:

\[
22.5 \, \text{cc} = 22.5 \, \text{mL} = \frac{22.5}{1000} \, \text{L} = 0.0225 \, \text{L}
\]

Now, we calculate the moles of aluminium sulfate \((\text{Al}_2(\text{SO}_4)_3)\) in this volume:

\[
\text{Moles of } \text{Al}_2(\text{SO}_4)_3 = \text{Molarity} \times \text{Volume in L} = 2 \, \text{mol/L} \times 0.0225 \, \text{L} = 0.045 \, \text{mol}
\]

### 2. Determine the number of sulfate ions produced

From the formula of aluminium sulfate, we can see that:

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

For every 1 mole of aluminium sulfate, 3 moles of sulfate ions are produced. Therefore, the number of moles of sulfate ions from 0.045 moles of aluminium sulfate is:

\[
\text{Moles of } \text{SO}_4^{2-} = 0.045 \, \text{mol} \times 3 = 0.135 \, \text{mol}
\]

### 3. Convert moles of sulfate ions to number of ions

To find the number of sulfate ions, we use Avogadro's number \((6.022 \times 10^{23} \, \text{ions/mol})\):

\[
\text{Number of SO}_4^{2-} \text{ ions} = 0.135 \, \text{mol} \times 6.022 \times 10^{23} \, \text{ions/mol}
\]

Calculating this gives:

\[
\text{Number of SO}_4^{2-} \text{ ions} = 0.135 \times 6.022 \times 10^{23} \approx 8.12 \times 10^{22} \text{ ions}
\]

### Final Answer:

The number of sulfate ions present in 22.5 cubic centimeters of 2 M aluminium sulfate solution is approximately \( \boxed{8.12 \times 10^{22}} \) sulfate ions.