In a 1 lit. Container following equilibrium is estabilished with equal moles of `NO_(2)(g)` & `N_(2)O_(4)(g)`.
`N_(2)O_(4)(g) hArr 2NO_(2)(g) hArr 2NO_(2)(g)` at equilibrium `M_(avg.)=(184)/(3)`, then ratio of `K_(c)` & total initial mole is .

Options

A. 3
B. `3//2`
C. `27//4`
D. 6

Explanation

To solve the problem step by step, we will analyze the equilibrium established in the container and calculate the required ratio of Kc to the total initial moles.

### Step 1: Write the Equilibrium Reaction
The equilibrium reaction given is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

### Step 2: Define Initial Moles
Let the initial moles of \( N_2O_4 \) and \( NO_2 \) be \( B \) moles each. Therefore, the total initial moles in the container is:
\[ \text{Total Initial Moles} = B + B = 2B \]

### Step 3: Define Change in Moles at Equilibrium
At equilibrium, let \( X \) moles of \( N_2O_4 \) dissociate to form \( 2X \) moles of \( NO_2 \). Thus, the moles at equilibrium will be:
- Moles of \( N_2O_4 \) at equilibrium: \( B - X \)
- Moles of \( NO_2 \) at equilibrium: \( B + 2X \)

### Step 4: Average Molar Mass
The average molar mass \( M_{avg} \) is given as:
\[ M_{avg} = \frac{184}{3} \]

Using the average molar mass formula:
\[ M_{avg} = \frac{\text{Total Mass}}{\text{Total Moles}} \]

### Step 5: Calculate Total Mass
The total mass can be expressed as:
\[ \text{Total Mass} = (B - X) \cdot 92 + (B + 2X) \cdot 46 \]
Where:
- \( 92 \) g/mol is the molar mass of \( N_2O_4 \)
- \( 46 \) g/mol is the molar mass of \( NO_2 \)

### Step 6: Set Up the Equation
Substituting the total mass into the average molar mass equation:
\[ \frac{(B - X) \cdot 92 + (B + 2X) \cdot 46}{(B - X) + (B + 2X)} = \frac{184}{3} \]

### Step 7: Simplify the Equation
The denominator simplifies to:
\[ 2B + X \]
So, the equation becomes:
\[ \frac{(B - X) \cdot 92 + (B + 2X) \cdot 46}{2B + X} = \frac{184}{3} \]

### Step 8: Cross-Multiply and Solve for X
Cross-multiplying gives:
\[ 3[(B - X) \cdot 92 + (B + 2X) \cdot 46] = 184(2B + X) \]

Expanding and simplifying will yield a value for \( X \).

### Step 9: Substitute X Back
Once \( X \) is found, substitute it back to find the equilibrium concentrations:
- \( [N_2O_4] = B - X \)
- \( [NO_2] = B + 2X \)

### Step 10: Calculate Kc
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[NO_2]^2}{[N_2O_4]} \]

Substituting the equilibrium concentrations into the expression will yield \( K_c \).

### Step 11: Find the Ratio of Kc to Total Initial Moles
Finally, calculate the ratio:
\[ \text{Ratio} = \frac{K_c}{\text{Total Initial Moles}} = \frac{K_c}{2B} \]

### Conclusion
After performing the calculations, we find that the ratio of \( K_c \) to the total initial moles is \( \frac{3}{2} \).