The largest number of N-atoms are found in

Options

A. 50 g `NO_(2)`
B. 150 mL of liquid pyridine `(C_(5)H_(5)N` density `=0.983` g/mL)
C. 25 g of `N_(2)O`
D. `0.5` mol of `N_(2)`

Explanation

To solve the problem of finding which option contains the largest number of nitrogen (N) atoms, we will analyze each option step by step.

### Step 1: Calculate the number of N atoms in 50 g of NO2

1. **Determine the molar mass of NO2**:
- Nitrogen (N) = 14 g/mol
- Oxygen (O) = 16 g/mol
- Since there are 2 oxygen atoms in NO2, the molar mass is calculated as:
\[ \text{Molar mass of NO2} = 14 + 2 \times 16 = 14 + 32 = 46 \text{ g/mol} \]

2. **Calculate the number of moles in 50 g of NO2**:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{50 \text{ g}}{46 \text{ g/mol}} \approx 1.087 \text{ moles} \]

3. **Calculate the number of N atoms**:
- Each molecule of NO2 contains 1 nitrogen atom.
- Therefore, the number of nitrogen atoms in 50 g of NO2 is:
\[ \text{Number of N atoms} = 1.087 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \approx 6.54 \times 10^{23} \text{ N atoms} \]

### Step 2: Calculate the number of N atoms in 150 ml of liquid pyridine

1. **Calculate the mass of pyridine using density**:
- Density = 0.983 g/ml
\[ \text{Mass} = \text{Density} \times \text{Volume} = 0.983 \text{ g/ml} \times 150 \text{ ml} = 147.45 \text{ g} \]

2. **Determine the molar mass of pyridine (C5H5N)**:
- Carbon (C) = 12 g/mol, Hydrogen (H) = 1 g/mol, Nitrogen (N) = 14 g/mol
- Molar mass = \(5 \times 12 + 5 \times 1 + 14 = 60 + 5 + 14 = 79 \text{ g/mol}\)

3. **Calculate the number of moles in 147.45 g of pyridine**:
\[ \text{Number of moles} = \frac{147.45 \text{ g}}{79 \text{ g/mol}} \approx 1.86 \text{ moles} \]

4. **Calculate the number of N atoms**:
- Each molecule of pyridine contains 1 nitrogen atom.
\[ \text{Number of N atoms} = 1.86 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \approx 11.24 \times 10^{23} \text{ N atoms} \]

### Step 3: Calculate the number of N atoms in 25 g of N2

1. **Determine the molar mass of N2**:
- Molar mass = \(2 \times 14 = 28 \text{ g/mol}\)

2. **Calculate the number of moles in 25 g of N2**:
\[ \text{Number of moles} = \frac{25 \text{ g}}{28 \text{ g/mol}} \approx 0.893 \text{ moles} \]

3. **Calculate the number of N atoms**:
- Each molecule of N2 contains 2 nitrogen atoms.
\[ \text{Number of N atoms} = 0.893 \text{ moles} \times 2 \times 6.022 \times 10^{23} \text{ atoms/mole} \approx 6.84 \times 10^{23} \text{ N atoms} \]

### Step 4: Calculate the number of N atoms in 0.5 moles of N2

1. **Calculate the number of N atoms**:
- In 0.5 moles of N2:
\[ \text{Number of N atoms} = 0.5 \text{ moles} \times 2 \times 6.022 \times 10^{23} \text{ atoms/mole} = 6.022 \times 10^{23} \text{ N atoms} \]

### Conclusion

Now, we can summarize the results:
- **A**: 50 g of NO2 → \(6.54 \times 10^{23}\) N atoms
- **B**: 150 ml of pyridine → \(11.24 \times 10^{23}\) N atoms
- **C**: 25 g of N2 → \(6.84 \times 10^{23}\) N atoms
- **D**: 0.5 moles of N2 → \(6.022 \times 10^{23}\) N atoms

The option with the largest number of N atoms is **B: 150 ml of liquid pyridine**.