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 solve the problem step by step, we will use the given information about the dissociation of compound A and the relationship between vapor density and molecular weight.

### Step 1: Understand the dissociation reaction
The dissociation of compound A is given as:
\[ A (g) \rightleftharpoons B(g) + C(g) + D(g) \]

From this reaction, we can see that 1 mole of A dissociates to produce 3 moles of products (B, C, and D).

### Step 2: Identify the degree of dissociation
The degree of dissociation (α) is given as 10%, which can be expressed as:
\[ \alpha = \frac{10}{100} = 0.1 \]

### Step 3: Determine the number of moles of gaseous products
Since 1 mole of A produces 3 moles of products, the total number of moles at equilibrium (n) after dissociation will be:
\[ n = 1 - \alpha + 3\alpha = 1 - 0.1 + 3(0.1) = 1 - 0.1 + 0.3 = 1.2 \]

### Step 4: Use the vapor density formula
The vapor density (d) is related to the molecular weight (M) of a gas by the formula:
\[ \text{Vapor Density} = \frac{M}{2} \]

### Step 5: Apply the vapor density relationship
We are given that the vapor density of the equilibrium mixture is 50. Using the relationship:
\[ d = \frac{M}{2} \]
we can express the molecular weight of the mixture as:
\[ M = 2d = 2 \times 50 = 100 \]

### Step 6: Calculate the initial vapor density
We can use the formula that relates the initial vapor density (D), the equilibrium vapor density (d), and the degree of dissociation (α):
\[ \alpha = \frac{D - d}{n - 1} \times \frac{1}{D} \]

Substituting the known values:
- \( \alpha = 0.1 \)
- \( d = 50 \)
- \( n = 3 \)

We can rearrange the formula:
\[ 0.1 = \frac{D - 50}{3 - 1} \times \frac{1}{D} \]
\[ 0.1 = \frac{D - 50}{2} \times \frac{1}{D} \]
\[ 0.1D = \frac{D - 50}{2} \]
\[ 0.2D = D - 50 \]
\[ 0.8D = 50 \]
\[ D = \frac{50}{0.8} = 62.5 \]

### Step 7: Calculate the molecular weight of compound A
Now, we can find the molecular weight of compound A using the initial vapor density:
\[ M_A = 2D = 2 \times 62.5 = 125 \]

### Final Answer
The molecular weight of compound A is **125 g/mol**.

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