When 0.44 g of a colourless oxide of nitrogen occupies 224 mL at 1520 mm Hg and `274^(@)C`, then the oxide is

Options

A. `N_(2)O_(5)`
B. `N_(2)O_(3)`
C. `NO_(2)`
D. `N_(2)O`

Explanation

To determine the colorless oxide of nitrogen based on the given data, we can follow these steps:

### Step 1: Gather the Given Data
We have the following information:
- Mass of the oxide (W) = 0.44 g
- Volume (V) = 224 mL
- Pressure (P) = 1520 mm Hg
- Temperature (T) = 274 °C

### Step 2: Convert Units
1. **Convert Volume from mL to L**:
\[ V = 224 \, \text{mL} = \frac{224}{1000} \, \text{L} = 0.224 \, \text{L} \]

2. **Convert Temperature from °C to K**:
\[ T = 274 \, \text{°C} + 273 = 547 \, \text{K} \]

3. **Convert Pressure from mm Hg to atm**:
\[ P = \frac{1520 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 2 \, \text{atm} \]

### Step 3: Use the Ideal Gas Equation
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( n \) = number of moles
- \( R \) = gas constant = 0.0821 L·atm/(K·mol)

### Step 4: Rearranging the Ideal Gas Equation
We can rearrange the equation to find the number of moles \( n \):
\[ n = \frac{PV}{RT} \]

### Step 5: Substitute the Values
Substituting the values we have:
\[ n = \frac{(2 \, \text{atm}) \times (0.224 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (547 \, \text{K})} \]

### Step 6: Calculate the Number of Moles
Calculating the denominator:
\[ 0.0821 \times 547 = 44.94 \, \text{L·atm/(K·mol)} \]
Now substituting back:
\[ n = \frac{(2) \times (0.224)}{44.94} \approx \frac{0.448}{44.94} \approx 0.00996 \, \text{mol} \]

### Step 7: Calculate Molar Mass
The molar mass \( M \) can be calculated using the formula:
\[ M = \frac{W}{n} \]
Substituting the values:
\[ M = \frac{0.44 \, \text{g}}{0.00996 \, \text{mol}} \approx 44.1 \, \text{g/mol} \]

### Step 8: Identify the Oxide
Now we need to identify which nitrogen oxide corresponds to a molar mass of approximately 44 g/mol.

- **N2O**:
\[ M = (2 \times 14) + (1 \times 16) = 28 + 16 = 44 \, \text{g/mol} \]
- **N2O3**:
\[ M = (2 \times 14) + (3 \times 16) = 28 + 48 = 76 \, \text{g/mol} \]
- **NO2**:
\[ M = (1 \times 14) + (2 \times 16) = 14 + 32 = 46 \, \text{g/mol} \]
- **N2O5**:
\[ M = (2 \times 14) + (5 \times 16) = 28 + 80 = 108 \, \text{g/mol} \]

### Conclusion
The oxide of nitrogen that corresponds to a molar mass of approximately 44 g/mol is **N2O**.