Q34 / 180 · COMEDK 2025 (10 May Shift 3) · Chemistry
An aqueous solution of an electrolyte $\text{A}_3\text{B}$ is prepared by dissolving 0.5625 moles in 750 ml of water and is found to be 80% ionised. If $\text{K}_\text{b}$ for water is $0.52\text{ K kg mol}^{-1}$, calculate the boiling point of the solution at 1.0 atm pressure.
(Assume the boiling point of pure water is 373 K).
(Assume the boiling point of pure water is 373 K).
- A373.18 K
- B371.68 K
- Correct answerC374.33 K
- D377.2 K
Solution
Molality: $m = \dfrac{0.5625}{0.75} = 0.75$ m
$A_3B$ dissociates into 4 ions, $n = 4$, $\alpha = 0.8$
Van't Hoff factor: $i = 1 + (n-1)\alpha = 1 + 3 \times 0.8 = 3.4$
$\Delta T_b = i \cdot K_b \cdot m = 3.4 \times 0.52 \times 0.75 = 1.326$ K
$T_b = 373 + 1.326 \approx 374.33$ K
$A_3B$ dissociates into 4 ions, $n = 4$, $\alpha = 0.8$
Van't Hoff factor: $i = 1 + (n-1)\alpha = 1 + 3 \times 0.8 = 3.4$
$\Delta T_b = i \cdot K_b \cdot m = 3.4 \times 0.52 \times 0.75 = 1.326$ K
$T_b = 373 + 1.326 \approx 374.33$ K
Hence, the correct answer is option: 'C'.