At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen `(N_(2))` at 4 bar. The molar mass of gaseous molecule is:

Options

A. `56 g mol^(-1)`
B. `112 g mol^(-1)`
C. `224 g mol^(-1)`
D. `28 g mol^(-1)`

Explanation

To find the molar mass of the gaseous molecule, we can follow these steps:

### Step 1: Understand the relationship between density, pressure, and molar mass.
Using the ideal gas equation, we can express the density of a gas in terms of its pressure and molar mass:
\[ \text{Density} (D) = \frac{PM}{RT} \]
Where:
- \( P \) = pressure of the gas
- \( M \) = molar mass of the gas
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Set up the equations for the two gases.
Let:
- \( D_1 \) = density of the gaseous molecule at 2 bar
- \( D_2 \) = density of dinitrogen (\( N_2 \)) at 4 bar

From the problem statement, we know:
\[ D_1 = 2D_2 \]
We also have:
\[ D_1 = \frac{P_1 M_1}{RT} \quad \text{(for the gaseous molecule)} \]
\[ D_2 = \frac{P_2 M_2}{RT} \quad \text{(for dinitrogen)} \]
Where:
- \( P_1 = 2 \) bar (pressure of the gaseous molecule)
- \( P_2 = 4 \) bar (pressure of dinitrogen)
- \( M_2 = 28 \, \text{g/mol} \) (molar mass of dinitrogen)

### Step 3: Substitute the known values into the equations.
Substituting the values into the equations:
\[ D_1 = \frac{2 \cdot M_1}{RT} \]
\[ D_2 = \frac{4 \cdot 28}{RT} \]

### Step 4: Relate the densities.
Using the relationship \( D_1 = 2D_2 \):
\[ \frac{2M_1}{RT} = 2 \left( \frac{4 \cdot 28}{RT} \right) \]

### Step 5: Simplify the equation.
Cancel \( RT \) from both sides:
\[ 2M_1 = 2 \cdot 4 \cdot 28 \]
\[ M_1 = 4 \cdot 28 \]
\[ M_1 = 112 \, \text{g/mol} \]

### Conclusion:
The molar mass of the gaseous molecule is \( 112 \, \text{g/mol} \).

---