If any solution 'A' dimerises in water at 1 atm pressure and the boiling point of this solution is `100.52^@C`. If 2 moles of A is added to 1kg of water and `k_b` of water is `0.52^@C//molal`, calculate the percentage association of A

Options

A. 0.5
B. 0.3
C. 0.25
D. 1

Answer

To solve the problem of calculating the percentage association of solution 'A' that dimerizes in water, we can follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔT)
The boiling point of pure water is 100°C, and the boiling point of the solution is given as 100.52°C.

\[ \Delta T = \text{Boiling point of solution} - \text{Boiling point of pure solvent} = 100.52°C - 100°C = 0.52°C \]

### Step 2: Use the formula for boiling point elevation
The formula for boiling point elevation is given by:

\[ \Delta T = i \cdot K_b \cdot m \]

Where:
- \( \Delta T \) = elevation in boiling point
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_b \) = ebullioscopic constant of the solvent (water in this case)
- \( m \) = molality of the solution

Given:
- \( K_b = 0.52 \, °C \, \text{kg/mol} \)
- \( m = \frac{2 \, \text{moles of A}}{1 \, \text{kg of water}} = 2 \, \text{mol/kg} \)

### Step 3: Substitute known values into the equation
Now we can substitute the known values into the boiling point elevation formula:

\[ 0.52 = i \cdot 0.52 \cdot 2 \]

### Step 4: Solve for the van 't Hoff factor (i)
Rearranging the equation to solve for \( i \):

\[ i = \frac{0.52}{0.52 \cdot 2} = \frac{0.52}{1.04} = 0.5 \]

### Step 5: Set up the association equation
Since solution 'A' dimerizes, we can represent the reaction as:

\[ 2A \rightleftharpoons A_2 \]

Let \( \alpha \) be the degree of association. If 2 moles of 'A' are initially present, at equilibrium:

- Moles of 'A' remaining = \( 2(1 - \alpha) \)
- Moles of dimer \( A_2 \) formed = \( \frac{\alpha}{2} \)

### Step 6: Calculate total moles at equilibrium
The total moles at equilibrium can be expressed as:

\[ \text{Total moles} = \text{Moles of A} + \text{Moles of } A_2 = 2(1 - \alpha) + \frac{\alpha}{2} \]

Simplifying this gives:

\[ \text{Total moles} = 2 - 2\alpha + \frac{\alpha}{2} = 2 - \frac{3\alpha}{2} \]

### Step 7: Relate total moles to van 't Hoff factor
Since \( i \) is the total number of moles at equilibrium, we have:

\[ i = 2 - \frac{3\alpha}{2} \]

We already found \( i = 0.5 \):

\[ 0.5 = 2 - \frac{3\alpha}{2} \]

### Step 8: Solve for α
Rearranging gives:

\[ \frac{3\alpha}{2} = 2 - 0.5 = 1.5 \]

Multiplying through by 2:

\[ 3\alpha = 3 \implies \alpha = 1 \]

### Step 9: Calculate percentage association
The percentage association of 'A' is given by:

\[ \text{Percentage association} = \alpha \times 100 = 1 \times 100 = 100\% \]

### Final Answer
The percentage association of A is **100%**.

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