What is the percentage dissociaiton of a substance if its vapour densities before and after dissociation are 30 and 15 respectively and I mole of its dissociation to 3 moles of products ?

Options

A. 0.25
B. 0.5
C. 0.65
D. 0.75

Answer

To find the percentage dissociation of the substance, we can follow these steps:

### Step 1: Understand the given data
- Initial vapor density (D) = 30
- Final vapor density (d) = 15
- The reaction is: A ⇌ 3B, meaning 1 mole of A dissociates to form 3 moles of B.

### Step 2: Use the formula for degree of dissociation (α)
The formula for the degree of dissociation (α) is given by:

\[ \alpha = \frac{D - d}{(N - 1) \cdot d} \]

Where:
- D = initial vapor density
- d = final vapor density
- N = number of moles of products formed from 1 mole of reactant (in this case, N = 3)

### Step 3: Substitute the values into the formula
Substituting the values we have:

\[ \alpha = \frac{30 - 15}{(3 - 1) \cdot 15} \]

### Step 4: Simplify the equation
Calculating the numerator and denominator:

- Numerator: \(30 - 15 = 15\)
- Denominator: \((3 - 1) \cdot 15 = 2 \cdot 15 = 30\)

So, we have:

\[ \alpha = \frac{15}{30} = \frac{1}{2} \]

### Step 5: Convert degree of dissociation to percentage
To find the percentage dissociation, we multiply α by 100:

\[ \text{Percentage dissociation} = \alpha \cdot 100 = \frac{1}{2} \cdot 100 = 50\% \]

### Final Answer
The percentage dissociation of the substance is **50%**.

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