An unknown compound A dissociates at `500^(@)C` to give products as follows -
`A ( g) hArr B(g) + C(g) + D(g)`
Vapour density of the equilibrium mixture is 50 when it dissociates to the extent to 10% . What will be the molecular weight of compound A-

Options

A. 120
B. 130
C. 134
D. 140

Answer

To find the molecular weight of compound A, we can follow these steps:

### Step 1: Understand the dissociation reaction
The compound A dissociates into three products:
\[ A(g) \rightleftharpoons B(g) + C(g) + D(g) \]
From this reaction, we can see that 1 mole of A produces 1 mole of B, 1 mole of C, and 1 mole of D, resulting in a total of 3 moles of products from 1 mole of A.

### Step 2: Define the extent of dissociation
Given that the extent of dissociation is 10%, we can denote this as:
\[ \alpha = 0.1 \]
This means that if we start with 1 mole of A, 0.1 moles of A will dissociate.

### Step 3: Calculate the moles at equilibrium
Initially, we have:
- Moles of A = 1
- Moles of B = 0
- Moles of C = 0
- Moles of D = 0

At equilibrium, the moles will be:
- Moles of A = \( 1 - 0.1 = 0.9 \)
- Moles of B = 0.1
- Moles of C = 0.1
- Moles of D = 0.1

Total moles at equilibrium:
\[ \text{Total moles} = 0.9 + 0.1 + 0.1 + 0.1 = 1.2 \]

### Step 4: Use the vapor density to find the molecular weight
The vapor density \( d \) of the equilibrium mixture is given as 50. The formula for vapor density is:
\[ d = \frac{\text{Molecular weight of mixture}}{2} \]

### Step 5: Calculate the molecular weight of the mixture
The molecular weight of the mixture can be calculated as:
\[ \text{Molecular weight of mixture} = d \times 2 = 50 \times 2 = 100 \text{ g/mol} \]

### Step 6: Relate the molecular weight of A to the molecular weight of the mixture
Using the formula for vapor density in terms of the molecular weight of A:
\[ D = \frac{M_A}{2} \]
Where \( D \) is the theoretical vapor density of A, and we can relate it to the vapor density of the mixture:
\[ D = \frac{M_A}{2} \]
And the relationship:
\[ D = d \left( 1 + (N - 1) \alpha \right) \]
Where \( N = 3 \) (the number of moles produced from A).

Substituting the values:
\[ 100 = \frac{M_A}{2} \left( 1 + (3 - 1) \times 0.1 \right) \]
\[ 100 = \frac{M_A}{2} \left( 1 + 0.2 \right) \]
\[ 100 = \frac{M_A}{2} \times 1.2 \]

### Step 7: Solve for \( M_A \)
Rearranging gives:
\[ M_A = \frac{100 \times 2}{1.2} = \frac{200}{1.2} = 166.67 \text{ g/mol} \]

### Step 8: Final answer
The molecular weight of compound A is approximately:
\[ M_A \approx 166.67 \text{ g/mol} \]