A dilute solution contains 'x' moles of solute A in 1 kg of solvent with molal elevation constant `K_(b)`. The solute dimerises in the solution according to the following equation. The degree of association is (a) :
`2A hArrA_(2)`
The degree of assoicition is equal to :

Options

A. `a = ((K_(b)x - /_\T_(b)))/(/_\T_(b)2)`
B. `a = (2(K_(b)x - /_\T_(b)))/(K_(b)x)`
C. `a =2 + (2/_\T_(b))/(K_(b)x)`
D. `a = (/_\T_(b))/(2K_(b)x)`

Answer

To find the degree of association (α) of solute A in a dilute solution, we can follow these steps:

### Step 1: Understand the Dimerization Reaction
The dimerization of solute A is represented by the equation:
\[ 2A \rightleftharpoons A_2 \]
This means that two moles of A combine to form one mole of dimer A₂.

### Step 2: Define Initial and Equilibrium Moles
Let’s assume we start with \( x \) moles of solute A in the solution. At the beginning (initial state), we have:
- Moles of A = \( x \)
- Moles of A₂ = 0

At equilibrium, if α is the degree of association, then:
- Moles of A that dimerize = \( 2 \times \frac{\alpha}{2} = \alpha \) (since 2 moles of A form 1 mole of A₂)
- Moles of A remaining = \( x - \alpha \)
- Moles of A₂ formed = \( \frac{\alpha}{2} \)

### Step 3: Calculate Total Moles at Equilibrium
At equilibrium, the total number of moles will be:
\[ \text{Total moles at equilibrium} = (x - \alpha) + \frac{\alpha}{2} = x - \frac{\alpha}{2} \]

### Step 4: Determine the 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:
\[ I = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} = \frac{x - \frac{\alpha}{2}}{x} \]

### Step 5: Relate the Boiling Point Elevation to the Van't Hoff Factor
The change in boiling point (ΔT_b) is given by the formula:
\[ \Delta T_b = K_b \cdot m \cdot I \]
Where:
- \( K_b \) is the molal elevation constant.
- \( m \) is the molality of the solution.

Since we have \( x \) moles of solute in 1 kg of solvent, the molality \( m \) is equal to \( x \):
\[ \Delta T_b = K_b \cdot x \cdot I \]

### Step 6: Substitute the Value of I
Substituting the expression for I into the boiling point elevation equation:
\[ \Delta T_b = K_b \cdot x \cdot \left(1 - \frac{\alpha}{2x}\right) \]

### Step 7: Rearranging to Solve for α
Rearranging the equation to isolate α:
\[ \Delta T_b = K_b x - \frac{K_b \alpha}{2} \]
\[ \frac{K_b \alpha}{2} = K_b x - \Delta T_b \]
\[ \alpha = \frac{2(K_b x - \Delta T_b)}{K_b} \]

### Final Step: Simplify the Expression for α
Thus, the degree of association (α) can be expressed as:
\[ \alpha = \frac{2(K_b x - \Delta T_b)}{K_b x} \]

### Conclusion
The degree of association α is given by:
\[ \alpha = \frac{2(K_b x - \Delta T_b)}{K_b x} \]