23 g of `NO_(2)` contains same number of molecules as

Options

A. 8 g of oxygen
B. 28 g of carbon monoxide
C. 16 g of `SO_(2)`
D. 22 g of `CO_(2)`

Answer

To solve the question, we need to find out which of the given options contains the same number of molecules as 23 g of \( NO_2 \).

### Step-by-Step Solution:

1. **Calculate the Moles of \( NO_2 \)**:
The formula to calculate the number of moles is:
\[ n = \frac{w}{M} \]
where \( w \) is the weight of the substance and \( M \) is the molar mass.

For \( NO_2 \):
- Weight \( w = 23 \, g \)
- Molar mass \( M = 14 \, (N) + 16 \times 2 \, (O) = 14 + 32 = 46 \, g/mol \)

Now, substituting the values:
\[ n_{NO_2} = \frac{23 \, g}{46 \, g/mol} = 0.5 \, mol \]

2. **Calculate the Number of Molecules in \( NO_2 \)**:
The number of molecules can be calculated using Avogadro's number (\( N_a \)):
\[ \text{Number of molecules} = n \times N_a \]
where \( N_a = 6.022 \times 10^{23} \, molecules/mol \).

Therefore:
\[ \text{Number of molecules in } NO_2 = 0.5 \, mol \times 6.022 \times 10^{23} \, molecules/mol = 3.011 \times 10^{23} \, molecules \]

3. **Check Each Option**:
We need to find which of the options has the same number of molecules.

- **Option A: 8 g of \( O_2 \)**:
\[ n_{O_2} = \frac{8 \, g}{32 \, g/mol} = 0.25 \, mol \]
\[ \text{Number of molecules} = 0.25 \, mol \times 6.022 \times 10^{23} \, molecules/mol = 1.505 \times 10^{23} \, molecules \]

- **Option B: 28 g of \( CO \)**:
\[ n_{CO} = \frac{28 \, g}{28 \, g/mol} = 1 \, mol \]
\[ \text{Number of molecules} = 1 \, mol \times 6.022 \times 10^{23} \, molecules/mol = 6.022 \times 10^{23} \, molecules \]

- **Option C: 16 g of \( AsO_2 \)**:
\[ n_{AsO_2} = \frac{16 \, g}{64 \, g/mol} = 0.25 \, mol \]
\[ \text{Number of molecules} = 0.25 \, mol \times 6.022 \times 10^{23} \, molecules/mol = 1.505 \times 10^{23} \, molecules \]

- **Option D: 22 g of \( CO_2 \)**:
\[ n_{CO_2} = \frac{22 \, g}{44 \, g/mol} = 0.5 \, mol \]
\[ \text{Number of molecules} = 0.5 \, mol \times 6.022 \times 10^{23} \, molecules/mol = 3.011 \times 10^{23} \, molecules \]

4. **Conclusion**:
The only option that has the same number of molecules as 23 g of \( NO_2 \) (which is \( 3.011 \times 10^{23} \, molecules \)) is **Option D: 22 g of \( CO_2 \)**.

### Final Answer:
**22 g of \( CO_2 \)** contains the same number of molecules as 23 g of \( NO_2 \).