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.