Question

How many sulfate ions are in 184.0 g of $BaSO_4$?
A. 4.750 * 10^(23) \text{ SO}_4^(2-) ions
B. \square * 10^(\square)

Answer 1

To find the number of sulfate ions in 184.0 g of \( BaSO_4 \), we can follow these steps:

1. **Calculate the molar mass of \( BaSO_4 \)**:
- Barium (Ba): approximately 137.33 g/mol
- Sulfur (S): approximately 32.07 g/mol
- Oxygen (O): approximately 16.00 g/mol (and there are 4 oxygen atoms in \( BaSO_4 \))

Molar mass of \( BaSO_4 \):
\[
\text{Molar mass} = 137.33 + 32.07 + (4 \times 16.00)
\]
\[
= 137.33 + 32.07 + 64.00 = 233.40 \text{ g/mol}
\]

2. **Calculate the number of moles of \( BaSO_4 \)** in 184.0 g:
\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{184.0 \text{ g}}{233.40 \text{ g/mol}} \approx 0.788 \text{ moles}
\]

3. **Determine the number of sulfate ions**:
Each formula unit of \( BaSO_4 \) contains one sulfate ion (\( SO_4^{2-} \)). Therefore, the number of sulfate ions will be the same as the number of \( BaSO_4 \) units.

To find the total number of sulfate ions, we multiply the number of moles of \( BaSO_4 \) by Avogadro's number \( (6.022 \times 10^{23} \text{ entities/mol}) \):
\[
\text{Number of sulfate ions} = 0.788 \text{ moles} \times 6.022 \times 10^{23} \text{ ions/mole}
\]
\[
\approx 4.74 \times 10^{23} \text{ SO}_4^{2-} \text{ ions}
\]

Thus, the answer is:
\[
\boxed{4.750 \times 10^{23} \text{ SO}_4^{2-} \text{ ions}}
\]

In the specified format, this corresponds to:
A. \( 4.750 \times 10^{23} \text{ SO}_4^{2-} \text{ ions} \)