Question

A solution containing $10 \, \text{g}$ of an electrolyte $AB_2$ in $100 \, \text{g}$ of water boils at $100.52^\circ \text{C}$. The degree of ionization of the electrolyte ($\alpha$) is _____ $\times 10^{-1}$. (nearest integer)
$[ \text{Given: Molar mass of } AB_2 = 200 \, \text{g mol}^{-1}, \, K_b \, (\text{molal boiling point elevation const. of water}) = 0.52 \, \text{K kg mol}^{-1}, \text{ boiling point of water} = 100^\circ \text{C}; \, AB_2 \text{ ionises as } AB_2 \rightarrow A^{2+} + 2B^- ]$

Answer 1

Ionization of AB$_2$:
$\text{AB}_2 \rightarrow \text{A}^{2+} + 2\text{B}^-.$
The van't Hoff factor ($i$) is: $i = 1 + (3 - 1)\alpha = 1 + 2\alpha.$
Boiling point elevation:$\Delta T_b = K_b \cdot m \cdot i,$
where $m = \frac{\text{Mass of solute}}{\text{Molar mass of solute} \cdot \text{Mass of solvent (kg)}}.$
Substitute values:
$m = \frac{10}{200 \cdot 0.1} = 0.5 \, \text{mol/kg}.$
$\Delta T_b = 0.52 = 0.52 \cdot 0.5 \cdot (1 + 2\alpha).$
Simplify:
$1 = 1 + 2\alpha \implies 2\alpha = 1 \implies \alpha = 0.5.$
Convert to nearest integer:
$\alpha \times 10 = 5.$
Final Answer: 5

Answer 2

Ionization of AB$_2$:
$\text{AB}_2 \rightarrow \text{A}^{2+} + 2\text{B}^-.$
The van't Hoff factor ($i$) is: $i = 1 + (3 - 1)\alpha = 1 + 2\alpha.$
Boiling point elevation:$\Delta T_b = K_b \cdot m \cdot i,$
where $m = \frac{\text{Mass of solute}}{\text{Molar mass of solute} \cdot \text{Mass of solvent (kg)}}.$
Substitute values:
$m = \frac{10}{200 \cdot 0.1} = 0.5 \, \text{mol/kg}.$
$\Delta T_b = 0.52 = 0.52 \cdot 0.5 \cdot (1 + 2\alpha).$
Simplify:
$1 = 1 + 2\alpha \implies 2\alpha = 1 \implies \alpha = 0.5.$
Convert to nearest integer:
$\alpha \times 10 = 5.$
Final Answer: 5