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

Correct Answer: 5

To find the degree of ionization (\(\alpha\)) for \(AB_2\), we follow these steps:
1. Calculate the molality (\(m\)) of the solution:
\[ \text{Mass of }AB_2 = 10 \, \text{g} \]
\[ \text{Molar mass of }AB_2 = 200 \, \text{g mol}^{-1} \]
\[ \text{Moles of }AB_2 = \frac{10}{200} = 0.05 \, \text{mol} \]
Mass of water = 100 g = 0.1 kg
\[ m = \frac{0.05 \, \text{mol}}{0.1 \, \text{kg}} = 0.5 \, \text{mol kg}^{-1} \]
2. Use boiling point elevation formula:
\(\Delta T_b = i \cdot K_b \cdot m\) where \(\Delta T_b = 100.52 - 100 = 0.52^\circ \text{C}\)
\[ i = 1 + \alpha \cdot (n-1) \] for dissociation, where \(n\) is the number of ions formed.
For \(AB_2 \rightarrow A^{2+} + 2B^-\), \(n = 3\).
3. Calculate the van 't Hoff factor (\(i\)):
\[ i = \frac{\Delta T_b}{K_b \cdot m} = \frac{0.52}{0.52 \cdot 0.5} = 2 \]
4. Solve for \(\alpha\):
\[ 2 = 1 + 2\alpha \]
\[ 2\alpha = 2 - 1 \]
\[ \alpha = \frac{1}{2} = 0.5 \]
5. Express \(\alpha\) as the nearest integer when multiplied by \(10^{-1}\):
\[ \alpha \times 10^{-1} = 0.5 \times 10^{-1} = 5 \]
The computed value \(5\) is within the expected range of \(5,5\).

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