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 solve the problem step by step, we will use the ideal gas law and the relationship between density, molar mass, and pressure.

### Step 1: Understand the relationship between density, molar mass, and pressure
The density (ρ) of a gas can be expressed using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Ideal gas constant
- \( T \) = Temperature in Kelvin

For one mole of gas, the equation simplifies to:
\[ V = \frac{RT}{P} \]

### Step 2: Express density in terms of molar mass
Density is defined as mass per unit volume:
\[ \rho = \frac{m}{V} \]

Substituting the expression for volume from the ideal gas equation:
\[ \rho = \frac{m}{\frac{RT}{P}} = \frac{mP}{RT} \]

For one mole of gas, the mass \( m \) is equal to the molar mass \( M \):
\[ \rho = \frac{MP}{RT} \]

### Step 3: Set up the equations for the two gases
Let \( M_g \) be the molar mass of the gaseous molecule and \( M_{N_2} \) be the molar mass of dinitrogen (which is 28 g/mol).

From the problem, we know:
- The density of the gaseous molecule at 2 bar is double that of dinitrogen at 4 bar.

Let \( \rho_{N_2} \) be the density of dinitrogen:
\[ \rho_g = 2\rho_{N_2} \]

Using the density formula for both gases:
1. For the gaseous molecule:
\[ \rho_g = \frac{M_g \cdot 2}{RT} \]

2. For dinitrogen:
\[ \rho_{N_2} = \frac{M_{N_2} \cdot 4}{RT} \]

### Step 4: Substitute and simplify
Substituting the expression for \( \rho_{N_2} \) into the equation for \( \rho_g \):
\[ 2\rho_{N_2} = 2 \cdot \frac{M_{N_2} \cdot 4}{RT} = \frac{M_g \cdot 2}{RT} \]

Now, we can set the two expressions equal to each other:
\[ \frac{M_g \cdot 2}{RT} = 2 \cdot \frac{M_{N_2} \cdot 4}{RT} \]

### Step 5: Cancel out common terms and solve for \( M_g \)
Cancel \( RT \) from both sides:
\[ M_g \cdot 2 = 2 \cdot M_{N_2} \cdot 4 \]

Now simplify:
\[ M_g = 4 \cdot M_{N_2} \]

Substituting \( M_{N_2} = 28 \, \text{g/mol} \):
\[ M_g = 4 \cdot 28 = 112 \, \text{g/mol} \]

### Final Answer
The molar mass of the gaseous molecule is **112 g/mol**.

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