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 point of the solution is ________ molal. (Rounded-off to the nearest integer) [K$_b$=0.52 K kg mol⁻¹]
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For colligative properties, always calculate the van 't Hoff factor 'i' first if the solute is an electrolyte. The formula $i = 1 + (n-1)\alpha$ is crucial, where 'n' is the number of ions produced per formula unit and '$\alpha$' is the degree of dissociation.
Step 1: Use the boiling point elevation formula
The elevation in boiling point is given by:
\[
\Delta T_b = i \, K_b \, m
\]
where
$\Delta T_b$ = elevation in boiling point,
$i$ = van’t Hoff factor,
$K_b$ = ebullioscopic constant,
$m$ = molality.
Step 2: Determine the van’t Hoff factor
The compound AB dissociates as:
\[
\text{AB} \rightleftharpoons \text{A}^+ + \text{B}^-
\]
Total number of particles formed on complete dissociation:
\[
n = 2
\]
Degree of dissociation:
\[
\alpha = 75% = 0.75
\]
The van’t Hoff factor is given by:
\[
i = 1 + (n - 1)\alpha
\]
\[
i = 1 + (2 - 1)(0.75) = 1 + 0.75 = 1.75
\]
Step 3: Substitute values into the formula
Given:
\[
\Delta T_b = 2.5\ \text{K}, \quad K_b = 0.52\ \text{K kg mol}^{-1}
\]
\[
2.5 = 1.75 \times 0.52 \times m
\]
\[
2.5 = 0.91\, m
\]
Step 4: Calculate molality
\[
m = \frac{2.5}{0.91} \approx 2.75\ \text{molal}
\]
Step 5: Round off
Rounded to the nearest integer:
\[
\boxed{m = 3\ \text{molal}}
\]
