The mass of `1.205 xx 10^(22)` molecules of ammonia is

Options

A. `0.34g`
B. `3.4g`
C. `3.4 kg`
D. `34g`

Explanation

To find the mass of \(1.205 \times 10^{22}\) molecules of ammonia (NH₃), we can follow these steps:

### Step 1: Calculate the molar mass of ammonia (NH₃)
- The atomic mass of nitrogen (N) is 14 g/mol.
- The atomic mass of hydrogen (H) is 1 g/mol.
- Since there are three hydrogen atoms in ammonia, the total mass contribution from hydrogen is \(3 \times 1 = 3\) g/mol.
- Therefore, the molar mass of ammonia is:
\[ \text{Molar mass of NH}_3 = 14 \, \text{g/mol} + 3 \, \text{g/mol} = 17 \, \text{g/mol} \]

### Step 2: Use Avogadro's number to relate molecules to moles
- Avogadro's number is \(6.022 \times 10^{23}\) molecules/mol.
- This means that 1 mole of NH₃ (which is 17 g) contains \(6.022 \times 10^{23}\) molecules.

### Step 3: Set up a proportion to find the mass of the given number of molecules
- We know that:
\[ 6.022 \times 10^{23} \, \text{molecules} \, \text{of NH}_3 \text{ corresponds to } 17 \, \text{g} \]
- We need to find the mass (X) corresponding to \(1.205 \times 10^{22}\) molecules:
\[ \frac{17 \, \text{g}}{6.022 \times 10^{23} \, \text{molecules}} = \frac{X \, \text{g}}{1.205 \times 10^{22} \, \text{molecules}} \]

### Step 4: Cross-multiply to solve for X
- Cross-multiplying gives:
\[ X \times (6.022 \times 10^{23}) = 17 \times (1.205 \times 10^{22}) \]
- Rearranging for X:
\[ X = \frac{17 \times (1.205 \times 10^{22})}{6.022 \times 10^{23}} \]

### Step 5: Calculate the value of X
- Performing the calculation:
\[ X = \frac{17 \times 1.205 \times 10^{22}}{6.022 \times 10^{23}} = \frac{20.485 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.34 \, \text{g} \]

### Conclusion
The mass of \(1.205 \times 10^{22}\) molecules of ammonia is approximately **0.34 g**.

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