A sample of mixture of A(g),B(g) and C(g) under equilibrium has a mean molecular weight (observed ) is 80.
The equilibrium is `underset((mol.wt.=100))(A(g))hArrunderset((mol.wt.=60))(B(g))+underset((mol.wt.=40))(C(g))`
Find the degree of dissociation `alpha` for A(g)

Options

A. `0.25`
B. `0.5`
C. `0.75`
D. `0.8`

Explanation

To solve the problem of finding the degree of dissociation (α) for A(g) in the equilibrium reaction:

\[ A(g) \rightleftharpoons B(g) + C(g) \]

given the mean molecular weight of the mixture is 80, we can follow these steps:

### Step 1: Define the Initial Conditions
Let:
- The initial number of moles of A = \( c \)
- The initial number of moles of B = 0
- The initial number of moles of C = 0

### Step 2: Define the Change in Moles at Equilibrium
At equilibrium, if α is the degree of dissociation of A, then:
- Moles of A at equilibrium = \( c(1 - \alpha) \)
- Moles of B at equilibrium = \( c\alpha \)
- Moles of C at equilibrium = \( c\alpha \)

### Step 3: Calculate Total Moles at Equilibrium
Total moles at equilibrium = Moles of A + Moles of B + Moles of C
\[ = c(1 - \alpha) + c\alpha + c\alpha = c(1 + \alpha) \]

### Step 4: Calculate the Mean Molecular Weight
The mean molecular weight (MW) of the mixture can be calculated using the formula:
\[ \text{Mean Molecular Weight} = \frac{\text{Total Mass}}{\text{Total Moles}} \]

The total mass of the mixture at equilibrium is:
\[ \text{Total Mass} = \text{Moles of A} \times \text{Molecular Weight of A} + \text{Moles of B} \times \text{Molecular Weight of B} + \text{Moles of C} \times \text{Molecular Weight of C} \]
\[ = c(1 - \alpha) \times 100 + c\alpha \times 60 + c\alpha \times 40 \]
\[ = c(100(1 - \alpha) + 60\alpha + 40\alpha) \]
\[ = c(100 - 100\alpha + 100\alpha) = c(100) \]

Thus, the total mass = \( 100c \).

Now, substituting into the mean molecular weight formula:
\[ 80 = \frac{100c}{c(1 + \alpha)} \]

### Step 5: Simplify the Equation
Cancelling \( c \) from numerator and denominator:
\[ 80 = \frac{100}{1 + \alpha} \]

### Step 6: Solve for α
Rearranging the equation:
\[ 1 + \alpha = \frac{100}{80} \]
\[ 1 + \alpha = 1.25 \]
\[ \alpha = 1.25 - 1 = 0.25 \]

### Conclusion
The degree of dissociation \( \alpha \) for A(g) is:
\[ \alpha = 0.25 \quad \text{or} \quad \frac{1}{4} \]