The number of atoms in `4.25 g NH_(3)` is approximately:

Options

A. (a) `4xx10^(23)`
B. (b) `2xx10^(23)`
C. (c) `1xx10^(23)`
D. (d) `6xx10^(23)`

Explanation

To find the number of atoms in 4.25 g of ammonia (NH₃), we can follow these steps:

### Step 1: Calculate the molar mass of NH₃
- The molar mass of NH₃ is calculated by adding the atomic masses of nitrogen and hydrogen.
- Nitrogen (N) has an atomic mass of approximately 14 g/mol.
- Hydrogen (H) has an atomic mass of approximately 1 g/mol, and there are three hydrogen atoms in NH₃.

\[ \text{Molar mass of NH}_3 = 1 \times 14 \, \text{g/mol} + 3 \times 1 \, \text{g/mol} = 14 + 3 = 17 \, \text{g/mol} \]

### Step 2: Determine the number of moles in 4.25 g of NH₃
- We can calculate the number of moles using the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- Substituting the values:

\[ \text{Number of moles} = \frac{4.25 \, \text{g}}{17 \, \text{g/mol}} \approx 0.25 \, \text{moles} \]

### Step 3: Calculate the number of molecules in 0.25 moles of NH₃
- Using Avogadro's number, which is approximately \(6.022 \times 10^{23}\) molecules/mol, we can find the number of molecules:

\[ \text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number} \]

- Substituting the values:

\[ \text{Number of molecules} = 0.25 \, \text{moles} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 1.5055 \times 10^{23} \, \text{molecules} \]

### Step 4: Calculate the number of atoms in the molecules of NH₃
- Each molecule of NH₃ contains 4 atoms (1 nitrogen and 3 hydrogen).
- Therefore, the total number of atoms can be calculated as:

\[ \text{Number of atoms} = \text{Number of molecules} \times 4 \]

- Substituting the values:

\[ \text{Number of atoms} = 1.5055 \times 10^{23} \, \text{molecules} \times 4 \approx 6.022 \times 10^{23} \, \text{atoms} \]

### Conclusion
- The number of atoms in 4.25 g of NH₃ is approximately \(6 \times 10^{23}\) atoms.

### Final Answer
The answer is option D: \(6 \times 10^{23}\).

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