If `D_(T)and D_(o)` are the theoretical and observed vapour densities at a definite temperature and `alpha` be the degree of dissociation of a substance ,then ,`alpha` in the terms of `D_(o),D_(T)` and n (number of moles of products formed from 1 mole reactant ) is calculated by the formula :

Options

A. (a) `alpha=(D_(o)-D_(T))/((1-n)D_(T))`
B. (b) `alpha=(D_(T)-D_(o))/((n-1)D_(T))`
C. (c) `alpha=(D_(T)-D_(o))/((n-1)D_(o))`
D. (d) `alpha=(D-D_(T))/((n-1)D_(T))`

Answer

To find the degree of dissociation (α) in terms of the theoretical vapor density (D_T), observed vapor density (D_O), and the number of moles of products formed from one mole of reactant (n), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Reaction**:
Assume the dissociation reaction is represented as:
\[ A_n \rightleftharpoons nA \]
where one mole of \( A_n \) dissociates to form \( n \) moles of \( A \).

2. **Initial and Equilibrium Moles**:
- Initially, we have \( a \) moles of \( A_n \).
- At equilibrium, the number of moles of \( A_n \) remaining will be \( a(1 - \alpha) \) and the number of moles of \( A \) formed will be \( n\alpha a \).

3. **Total Moles at Equilibrium**:
The total number of moles at equilibrium can be expressed as:
\[ \text{Total moles} = a(1 - \alpha) + n\alpha a = a(1 - \alpha + n\alpha) = a(1 + (n - 1)\alpha) \]

4. **Van't Hoff Factor (i)**:
The Van't Hoff factor \( i \) is defined as the ratio of the total number of moles at equilibrium to the initial number of moles:
\[ i = \frac{a(1 + (n - 1)\alpha)}{a} = 1 + (n - 1)\alpha \]

5. **Relating Molecular Mass and Vapor Density**:
The vapor density \( D \) is related to the molar mass \( M \) by the formula:
\[ D = \frac{M}{2} \]
Therefore, we can express the molar masses in terms of vapor densities:
\[ M_T = 2D_T \quad \text{and} \quad M_O = 2D_O \]

6. **Substituting into the Van't Hoff Factor**:
The abnormal molecular mass can also be expressed in terms of vapor densities:
\[ i = \frac{M_T}{M_O} = \frac{2D_T}{2D_O} = \frac{D_T}{D_O} \]
Thus, we have:
\[ 1 + (n - 1)\alpha = \frac{D_T}{D_O} \]

7. **Solving for α**:
Rearranging the equation to isolate \( \alpha \):
\[ (n - 1)\alpha = \frac{D_T}{D_O} - 1 \]
\[ \alpha = \frac{\frac{D_T}{D_O} - 1}{n - 1} \]

8. **Final Expression**:
To express \( \alpha \) in terms of \( D_T \), \( D_O \), and \( n \):
\[ \alpha = \frac{D_T - D_O}{D_O(n - 1)} \]

### Final Answer:
Thus, the degree of dissociation \( \alpha \) can be calculated using the formula:
\[ \alpha = \frac{D_T - D_O}{D_O(n - 1)} \]

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