A dilute solution contains `m` mol of solute `A` in 1 kg of a solvent with molal elevation constant `K_(b)`. The solute dimerises in solution as
`2AhArrA_(2)`
Select the correct expression of equilibrium constant `'K'` for the dimer formation. (assume molarity `=` molality)

Options

A. A) `K=(K_(b)(K_(b)m-DeltaT_(b)))/((K_(b)m+2DeltaT_(b))^(2))`
B. B) `K=(alpha)/(2m(1-alpha)^(2))`
C. C) `K=(K_(b)(K_(b)m-DeltaT_(b)))/((2DeltaT_(b)-K_(b)m)^(2))`
D. D) `K=(2(K_(b)m-DeltaT_(b)))/(K_(b)m)`

Explanation

To solve the problem, we need to derive the expression for the equilibrium constant \( K \) for the dimerization of solute \( A \) into \( A_2 \) based on the information provided.

### Step-by-Step Solution:

1. **Understanding the Dimerization Reaction**:
The dimerization reaction is given as:
\[ 2A \rightleftharpoons A_2 \]
This means that two moles of solute \( A \) combine to form one mole of dimer \( A_2 \).

2. **Initial Moles**:
Given that there are \( m \) moles of solute \( A \) in 1 kg of solvent, at the start of the reaction (time \( t = 0 \)):
- Moles of \( A \) = \( m \)
- Moles of \( A_2 \) = 0

3. **Degree of Dimerization**:
Let \( \alpha \) be the degree of dimerization. At equilibrium:
- Moles of \( A_2 \) formed = \( \frac{m \alpha}{2} \) (since 2 moles of \( A \) give 1 mole of \( A_2 \))
- Moles of \( A \) remaining = \( m - m \alpha = m(1 - \alpha) \)

4. **Equilibrium Concentrations**:
At equilibrium, the concentrations can be expressed as:
- Concentration of \( A_2 \) = \( \frac{m \alpha}{2} \)
- Concentration of \( A \) = \( m(1 - \alpha) \)

5. **Expression for Equilibrium Constant \( K \)**:
The equilibrium constant \( K \) for the dimerization can be expressed as:
\[ K = \frac{[A_2]}{[A]^2} \]
Substituting the equilibrium concentrations:
\[ K = \frac{\frac{m \alpha}{2}}{(m(1 - \alpha))^2} \]
Simplifying this expression:
\[ K = \frac{m \alpha}{2m^2(1 - \alpha)^2} = \frac{\alpha}{2m(1 - \alpha)^2} \]

6. **Using the Van 't Hoff Factor**:
The van 't Hoff factor \( i \) for the dimerization can be given by:
\[ i = 1 + \frac{1}{n - 1} \alpha \]
Here, \( n = 2 \) (since 2 moles of \( A \) form 1 mole of \( A_2 \)):
\[ i = 1 + \frac{1}{2 - 1} \alpha = 1 + \alpha \]

7. **Change in Boiling Point**:
The change in boiling point \( \Delta T_B \) is given by:
\[ \Delta T_B = K_B \cdot m \cdot i \]
Substituting \( i \):
\[ \Delta T_B = K_B \cdot m \cdot (1 + \alpha) \]

8. **Rearranging for \( \alpha \)**:
From the above equation, we can express \( \alpha \):
\[ \alpha = \frac{\Delta T_B}{K_B m} - 1 \]

9. **Substituting \( \alpha \) back into \( K \)**:
Substitute this expression for \( \alpha \) back into the equilibrium constant expression derived in step 5:
\[ K = \frac{\left(\frac{\Delta T_B}{K_B m} - 1\right)}{2m\left(1 - \left(\frac{\Delta T_B}{K_B m} - 1\right)\right)^2} \]

10. **Final Expression**:
After simplification, we will arrive at the final expressions for \( K \).