If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling of the solution is_____ molal. (Rounded-off to the nearest integer)

Answer

To solve the problem, we need to find the molality of the solution that results in a 2.5 K rise in boiling point when the compound AB dissociates to the extent of 75%. We will use the formula for elevation in boiling point.

### Step-by-Step Solution:

1. **Understand the Formula**:
The elevation in boiling point is given by the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \(\Delta T_b\) = change in boiling point (2.5 K)
- \(i\) = van 't Hoff factor
- \(K_b\) = ebullioscopic constant of the solvent (for water, \(K_b = 0.52 \, \text{K kg/mol}\))
- \(m\) = molality of the solution

2. **Calculate the van 't Hoff Factor (i)**:
Since the compound AB dissociates into two ions (A and B), the dissociation can be represented as:
\[ AB \rightarrow A^+ + B^- \]
Given that it dissociates to 75%, we can calculate \(i\) using the formula:
\[ i = 1 + (n - 1) \cdot \alpha \]
where \(n\) is the number of particles the solute dissociates into (which is 2 for AB), and \(\alpha\) is the degree of dissociation (75% or 0.75).

Plugging in the values:
\[ i = 1 + (2 - 1) \cdot 0.75 = 1 + 0.75 = 1.75 \]

3. **Substitute Values into the Elevation Formula**:
Now substitute \(\Delta T_b\), \(i\), and \(K_b\) into the elevation formula:
\[ 2.5 = 1.75 \cdot 0.52 \cdot m \]

4. **Solve for m (molality)**:
Rearranging the equation to find \(m\):
\[ m = \frac{2.5}{1.75 \cdot 0.52} \]

Calculate the denominator:
\[ 1.75 \cdot 0.52 = 0.91 \]

Now, calculate \(m\):
\[ m = \frac{2.5}{0.91} \approx 2.747 \]

5. **Round to the Nearest Integer**:
The nearest integer to 2.747 is 3.

### Final Answer:
The molality of the solution is approximately **3 molal**.

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