A gaseous compound of molecular mass 82.1 dissociates on heating to 400 K as
`X_(2)Y_(4)(g)hArrX_(2)(g)+Y_(2)(g)`
he density of the equilibrium mixture at a pressure of 1 atm and temperature of 400K is `2.0gL^(-1)`. The percentage dissociation of the compound is

Options

A. 0.125
B. 0.485
C. 0.901
D. 0.25

Explanation

To find the percentage dissociation of the gaseous compound \( X_2Y_4 \) that dissociates into \( X_2 \) and \( Y_2 \), we can follow these steps:

### Step 1: Calculate the density of the pure compound
The density of a gas can be calculated using the formula:
\[ D = \frac{PM}{RT} \]
Where:
- \( P \) = pressure (1 atm)
- \( M \) = molecular mass (82.1 g/mol)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (400 K)

Substituting the values:
\[ D = \frac{(1 \, \text{atm}) \times (82.1 \, \text{g/mol})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (400 \, \text{K})} \]
Calculating the denominator:
\[ 0.0821 \times 400 = 32.84 \]
Now calculating the density:
\[ D = \frac{82.1}{32.84} \approx 2.5 \, \text{g/L} \]

### Step 2: Use the equilibrium density to find the percentage dissociation
Given the equilibrium density of the mixture is \( 2.0 \, \text{g/L} \).

Let \( \alpha \) be the degree of dissociation. The reaction is:
\[ X_2Y_4(g) \rightleftharpoons X_2(g) + 2Y_2(g) \]
Initially, we have 1 mole of \( X_2Y_4 \). Upon dissociation:
- Moles of \( X_2 \) = \( 1 - \alpha \)
- Moles of \( Y_2 \) = \( 2\alpha \)

Total moles at equilibrium:
\[ n = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 3: Calculate the density of the equilibrium mixture
Using the formula:
\[ D_{eq} = \frac{P \cdot M_{eq}}{RT} \]
Where \( M_{eq} \) is the molar mass of the equilibrium mixture:
\[ M_{eq} = \frac{(1 - \alpha) \cdot 82.1 + 2\alpha \cdot 82.1}{1 + \alpha} = \frac{82.1 + 82.1\alpha}{1 + \alpha} \]

Substituting into the density formula:
\[ D_{eq} = \frac{(1 \, \text{atm}) \cdot \frac{82.1 + 82.1\alpha}{1 + \alpha}}{(0.0821) \cdot (400)} \]
Setting this equal to the given density:
\[ 2.0 = \frac{82.1 + 82.1\alpha}{32.84(1 + \alpha)} \]

### Step 4: Solve for \( \alpha \)
Cross-multiplying gives:
\[ 2.0 \cdot 32.84(1 + \alpha) = 82.1 + 82.1\alpha \]
\[ 65.68 + 65.68\alpha = 82.1 + 82.1\alpha \]
Rearranging terms:
\[ 65.68 - 82.1 = 82.1\alpha - 65.68\alpha \]
\[ -16.42 = 16.42\alpha \]
\[ \alpha = \frac{-16.42}{16.42} = 1 \]

### Step 5: Calculate the percentage dissociation
Percentage dissociation is given by:
\[ \text{Percentage dissociation} = \alpha \times 100 \]
Substituting \( \alpha \):
\[ \text{Percentage dissociation} = 1 \times 100 = 100\% \]

### Final Answer
The percentage dissociation of the compound is **100%**.