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Step-by-Step Solution
Step 1: Identify the colligative property and given data
The question involves the elevation of boiling point ( \Delta T_b ), which is
a colligative property. We are given:
Elevation of boiling point, \Delta T_b = 2.5\ \text{K}
Boiling point elevation constant, K_b = 0.52\ \text{K kg mol}^{-1}
Dissociation extent (degree of dissociation), \alpha = 75\% = 0.75
The compound is a binary salt AB that dissociates into two ions:
AB \rightarrow A^+ + B^-
Step 2: Determine the van't Hoff factor
Since AB dissociates into 2 ions, the total number of particles n=2 . The van't Hoff factor
i for an electrolyte dissociating to an extent \alpha is given by:
i = 1 + \alpha(n - 1)
Here, n = 2 and \alpha = 0.75 , so
i = 1 + 0.75 \times (2 - 1) = 1 + 0.75 = 1.75.
Step 3: Apply the formula for elevation of boiling point
The colligative property relation for the elevation of boiling point is:
\Delta T_b = i \, K_b \, m
where
m is the molality of the solution. Rearrange to solve for m :
m = \frac{\Delta T_b}{i \, K_b}.
Step 4: Substitute the given values
Substitute \Delta T_b = 2.5\ \text{K} , K_b = 0.52\ \text{K kg mol}^{-1} , and i = 1.75 into the expression:
m = \frac{2.5}{1.75 \times 0.52}.
Compute:
1.75 \times 0.52 = 0.91
approximately, so
m = \frac{2.5}{0.91} \approx 2.74.
Step 5: Final answer (rounded to the nearest integer)
Rounding 2.74 to the nearest integer gives 3 . Hence, the required molality of the solution is
3\ \text{molal} .
