To determine the number of atoms contained in 85 grams of ammonia (NH₃), we need to follow several steps involving the molar mass, moles, and Avogadro's number. Here’s the step-by-step solution:
### Step 1: Find the Molar Mass of Ammonia (NH₃)
Ammonia consists of one nitrogen atom (N) and three hydrogen atoms (H). We need to calculate its molar mass by adding the atomic masses of nitrogen and hydrogen.
- Molar mass of Nitrogen (N) = 14.01 g/mol
- Molar mass of Hydrogen (H) = 1.01 g/mol
[tex]\[
\text{Molar mass of NH}_3 = \text{molar mass of N} + 3 \times \text{molar mass of H}
\][/tex]
[tex]\[
\text{Molar mass of NH}_3 = 14.01 + 3 \times 1.01 = 14.01 + 3.03 = 17.04 \, \text{g/mol}
\][/tex]
### Step 2: Calculate the Number of Moles of Ammonia in 85 Grams
Using the molar mass, we calculate the number of moles of ammonia:
[tex]\[
\text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}}
\][/tex]
[tex]\[
\text{Number of moles of NH}_3 = \frac{85 \, \text{g}}{17.04 \, \text{g/mol}} \approx 4.988 \, \text{mol}
\][/tex]
### Step 3: Determine the Number of Molecules of Ammonia
We use Avogadro's number ([tex]\(6.022 \times 10^{23}\)[/tex]) to convert moles into the number of molecules.
[tex]\[
\text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number}
\][/tex]
[tex]\[
\text{Number of molecules of NH}_3 = 4.988 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 3.004 \times 10^{24} \, \text{molecules}
\][/tex]
### Step 4: Calculate the Total Number of Atoms in Ammonia
Each molecule of ammonia (NH₃) contains 4 atoms (1 nitrogen and 3 hydrogen atoms).
[tex]\[
\text{Total number of atoms} = \text{Number of molecules} \times 4
\][/tex]
[tex]\[
\text{Total number of atoms} = 3.004 \times 10^{24} \times 4 \approx 1.202 \times 10^{25} \, \text{atoms}
\][/tex]
### Conclusion
The number of atoms in 85 grams of ammonia is approximately [tex]\(1.2 \times 10^{25}\)[/tex].
Hence, the correct answer is:
[tex]\[ \boxed{1.2 \times 10^{25} \text{ atoms}} \][/tex]
So, the correct option is:
[tex]\[ \text{B } 1.2 \times 10^{25} \text{ atoms} \][/tex]
