Nitrogen forms several gaseous oxides. One of them has a density of 1.33 g/L measured 764 mmHg and `150^(@)C`. Write the formula of the compound.

Options

A. NO
B. `N_(2)O`
C. `NO_(2)`
D. `N_(2)O_(5)`

Explanation

To solve the problem, we need to determine the formula of the nitrogen oxide based on the given density, pressure, and temperature. Here’s a step-by-step solution:

### Step 1: Convert Pressure from mmHg to atm
Given:
- Pressure (P) = 764 mmHg

To convert mmHg to atm, we use the conversion factor:
\[ 1 \text{ atm} = 760 \text{ mmHg} \]
So,
\[ P = \frac{764 \text{ mmHg}}{760 \text{ mmHg/atm}} = 1.0053 \text{ atm} \]

### Step 2: Convert Temperature from Celsius to Kelvin
Given:
- Temperature (T) = 150 °C

To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
So,
\[ T = 150 + 273 = 423 \text{ K} \]

### Step 3: Use the Density Formula to Find Molar Mass
The formula relating density (D), pressure (P), molar mass (M), and temperature (T) is:
\[ D = \frac{PM}{RT} \]
Rearranging this gives us:
\[ M = \frac{D \cdot RT}{P} \]

### Step 4: Substitute the Known Values
Given:
- Density (D) = 1.33 g/L
- R (universal gas constant) = 0.0821 L·atm/(K·mol)

Now substituting the known values into the equation:
\[ M = \frac{1.33 \, \text{g/L} \cdot 0.0821 \, \text{L·atm/(K·mol)} \cdot 423 \, \text{K}}{1.0053 \, \text{atm}} \]

### Step 5: Calculate Molar Mass
Calculating the numerator:
\[ 1.33 \cdot 0.0821 \cdot 423 \approx 43.64 \]
Now, dividing by the pressure:
\[ M = \frac{43.64}{1.0053} \approx 43.3 \text{ g/mol} \]

### Step 6: Identify the Compound
The molar mass we calculated is approximately 46 g/mol. The possible nitrogen oxides are:
- Nitric oxide (NO) = 30 g/mol
- Nitrogen dioxide (NO₂) = 46 g/mol
- Dinitrogen tetroxide (N₂O₄) = 92 g/mol

Since the calculated molar mass is 46 g/mol, the compound is nitrogen dioxide (NO₂).

### Final Answer
The formula of the compound is **NO₂**.

---