At the equilibrium of the reaction ,
`N_(2)O_(4)(g)rArr2NO_(2)(g)`, the observed molar mass of `N_(2)O_(4)` is 77.70 g . The percentage dissociation of `N_(2)O_(4)` is :-

Options

A. (a)28 .4
B. (b)46 .7
C. (c)22.4
D. (d)18 .4

Answer

To solve the problem of finding the percentage dissociation of \( N_2O_4 \) at equilibrium, we can follow these steps:

### Step 1: Write the Reaction and Initial Conditions
The reaction is given as:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]
At the start (time = 0), we assume we have 1 mole of \( N_2O_4 \) and 0 moles of \( NO_2 \).

### Step 2: Define the Change in Concentration
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). At equilibrium:
- The moles of \( N_2O_4 \) remaining will be \( 1 - \alpha \).
- The moles of \( NO_2 \) formed will be \( 2\alpha \).

### Step 3: Calculate the Normal Molar Mass of \( N_2O_4 \)
The normal molar mass of \( N_2O_4 \) can be calculated as:
\[ \text{Molar mass of } N_2O_4 = (2 \times 14) + (4 \times 16) = 28 + 64 = 92 \, \text{g/mol} \]

### Step 4: Use the Given Observed Molar Mass
The observed molar mass of \( N_2O_4 \) at equilibrium is given as 77.70 g/mol.

### Step 5: Calculate the Van't Hoff Factor (i)
The Van't Hoff factor \( i \) can be calculated using the formula:
\[ i = \frac{\text{Normal Molar Mass}}{\text{Observed Molar Mass}} = \frac{92}{77.70} \approx 1.184 \]

### Step 6: Set Up the Equation for the Van't Hoff Factor
The Van't Hoff factor can also be expressed as:
\[ i = \frac{1 - \alpha + 2\alpha}{1} = 1 - \alpha + 2\alpha = 1 + \alpha \]
Setting this equal to the calculated \( i \):
\[ 1 + \alpha = 1.184 \]

### Step 7: Solve for \( \alpha \)
Rearranging the equation gives:
\[ \alpha = 1.184 - 1 = 0.184 \]

### Step 8: Calculate the Percentage Dissociation
To find the percentage dissociation, we multiply \( \alpha \) by 100:
\[ \text{Percentage Dissociation} = \alpha \times 100 = 0.184 \times 100 = 18.4\% \]

### Final Answer
The percentage dissociation of \( N_2O_4 \) is **18.4%**.

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