At `t^@`C temperature, the observed vapour density of A is 17.5 for the reaction `2A_((g)) hArr ƒ B_((g)) + 2C_((g))`. If molecular weight of A is 48, then the percentage dissociation of A will be :-

Options

A. `70.27%`
B. `74.28%`
C. `37.14%`
D. `85.71%`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the Reaction
The reaction given is:
\[ 2A_{(g)} \rightleftharpoons B_{(g)} + 2C_{(g)} \]

### Step 2: Define Variables
Let:
- \( \alpha \) = degree of dissociation of A
- Initial moles of A = 1 mole
- Initial moles of B = 0
- Initial moles of C = 0

### Step 3: Determine Moles at Equilibrium
At equilibrium, the moles will be:
- Moles of A = \( 1 - 2\alpha \)
- Moles of B = \( \alpha \)
- Moles of C = \( 2\alpha \)

### Step 4: Calculate Total Moles at Equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = (1 - 2\alpha) + \alpha + 2\alpha = 1 + \alpha \]

### Step 5: Relate Vapor Density to Moles
The vapor density (D) is related to the molecular weight (M) of the substance:
\[ D = \frac{M}{2} \]
Given that the molecular weight of A is 48, the initial vapor density \( D \) is:
\[ D = \frac{48}{2} = 24 \]

### Step 6: Use Observed Vapor Density
The observed vapor density is given as 17.5. We can set up the relationship:
\[ \frac{D_{initial}}{D_{observed}} = \frac{\text{Initial moles}}{\text{Total moles at equilibrium}} \]
Substituting the values we have:
\[ \frac{24}{17.5} = \frac{1}{1 + \alpha} \]

### Step 7: Solve for \( \alpha \)
Cross-multiplying gives:
\[ 24(1 + \alpha) = 17.5 \]
Expanding and rearranging:
\[ 24 + 24\alpha = 17.5 \]
\[ 24\alpha = 17.5 - 24 \]
\[ 24\alpha = -6.5 \]
\[ \alpha = \frac{-6.5}{24} \]
This should be corrected as:
\[ 24 + 24\alpha = 17.5 \]
\[ 24\alpha = 17.5 - 24 \]
\[ 24\alpha = -6.5 \]
This indicates a mistake in signs; we should have:
\[ 24\alpha = 24 - 17.5 \]
\[ 24\alpha = 6.5 \]
\[ \alpha = \frac{6.5}{24} \]
\[ \alpha = 0.2708 \]

### Step 8: Calculate Percentage Dissociation
Percentage dissociation of A:
\[ \text{Percentage dissociation} = \alpha \times 100 \]
\[ \text{Percentage dissociation} = 0.2708 \times 100 = 27.08\% \]

### Final Answer
The percentage dissociation of A is approximately **27.08%**.