At equilibrium of the reaction ,
`N_(2)O_(4)(g) hArr 2NO_(2)(g)`
the observed molecular weight of `N_92)O_(4)` is 80g `mol^(-1)` at 350K. The percentage dissociation of `N_(2)O_(4)(g)` at 350K is

Options

A. `10%`
B. `15%`
C. `20%`
D. `18%`

Explanation

To solve the problem, we need to determine the percentage dissociation of \( N_2O_4 \) at equilibrium. Let's break down the solution step by step.

### Step 1: Understanding the Reaction
The reaction is given as:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]
At equilibrium, some of the \( N_2O_4 \) dissociates into \( NO_2 \).

### Step 2: Define Initial and Final Conditions
Let:
- The initial amount of \( N_2O_4 \) be \( n \) moles.
- The amount of \( N_2O_4 \) that dissociates be \( x \) moles.

At equilibrium:
- The amount of \( N_2O_4 \) remaining = \( n - x \)
- The amount of \( NO_2 \) formed = \( 2x \)

### Step 3: Calculate Molar Masses
The molar mass of \( N_2O_4 \) is calculated as follows:
- Molar mass of \( N_2O_4 \) = \( 2 \times 14 + 4 \times 16 = 28 + 64 = 92 \, g/mol \)

The molar mass of \( NO_2 \) is:
- Molar mass of \( NO_2 \) = \( 14 + 2 \times 16 = 14 + 32 = 46 \, g/mol \)

### Step 4: Calculate the Average Molar Mass at Equilibrium
The average molar mass at equilibrium can be expressed as:
\[ \text{Average Molar Mass} = \frac{\text{Total Mass}}{\text{Total Moles}} \]
At equilibrium, the total mass is:
\[ \text{Total Mass} = (n - x) \times 92 + 2x \times 46 \]
The total moles at equilibrium are:
\[ \text{Total Moles} = (n - x) + 2x = n + x \]
Thus, the average molar mass at equilibrium is:
\[ \text{Average Molar Mass} = \frac{(n - x) \times 92 + 2x \times 46}{n + x} \]

### Step 5: Set Up the Equation
Given that the observed average molar mass at equilibrium is 80 g/mol, we can set up the equation:
\[ \frac{(n - x) \times 92 + 2x \times 46}{n + x} = 80 \]

### Step 6: Solve the Equation
Cross-multiplying gives:
\[ (n - x) \times 92 + 2x \times 46 = 80(n + x) \]
Expanding both sides:
\[ 92n - 92x + 92x = 80n + 80x \]
This simplifies to:
\[ 92n = 80n + 80x \]
Rearranging gives:
\[ 12n = 80x \]
Thus:
\[ x = \frac{12n}{80} = \frac{3n}{20} \]

### Step 7: Calculate Percentage Dissociation
The percentage dissociation is given by:
\[ \text{Percentage Dissociation} = \frac{x}{n} \times 100 = \frac{\frac{3n}{20}}{n} \times 100 = \frac{3}{20} \times 100 = 15\% \]

### Final Answer
The percentage dissociation of \( N_2O_4 \) at 350 K is **15%**.

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