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)
$[ \text{Given: Molar mass of } AB_2 = 200 \, \text{g mol}^{-1}, \, K_b \, (\text{molal boiling point elevation const. of water}) = 0.52 \, \text{K kg mol}^{-1}, \text{ boiling point of water} = 100^\circ \text{C}; \, AB_2 \text{ ionises as } AB_2 \rightarrow A^{2+} + 2B^- ]$

Question

$W$ g of a non-volatile electrolyte solid solute of molar mass $M\,\text{g mol}^{-1}$ when dissolved in $100\,\text{mL}$ water decreases vapour pressure of water from $640\,\text{mm Hg}$ to $600\,\text{mm Hg}$. If aqueous solution of the electrolyte boils at $375\,\text{K}$ and $K_b$ for water is $0.52\,\text{K kg mol}^{-1}$, then the mole fraction of the electrolyte $(x_2)$ in the solution can be expressed as ___.
(Given: density of water $=1\,\text{g mL}^{-1}$, boiling point of water $=373\,\text{K}$)

Answer 1

The degree of ionization ($\alpha$) for $AB_2$ is determined through the following calculations:
1. Molality ($m$) of the solution: Mass of $AB_2 = 10 \, \text{g}$, Molar mass of $AB_2 = 200 \, \text{g mol}^{-1}$. Therefore, Moles of $AB_2 = \frac{10}{200} = 0.05 \, \text{mol}$. With a water mass of 100 g (0.1 kg), the molality is $m = \frac{0.05 \, \text{mol}}{0.1 \, \text{kg}} = 0.5 \, \text{mol kg}^{-1}$.
2. Boiling point elevation formula: $\Delta T_b = i \cdot K_b \cdot m$, where $\Delta T_b = 100.52 - 100 = 0.52^\circ \text{C}$. The van 't Hoff factor ($i$) for dissociation is given by $i = 1 + \alpha \cdot (n-1)$, with $n$ representing the number of ions formed. For $AB_2 \rightarrow A^{2+} + 2B^-$, $n = 3$.
3. Calculation of 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. Solving for $\alpha$: From $2 = 1 + 2\alpha$, we get $2\alpha = 2 - 1$, resulting in $\alpha = \frac{1}{2} = 0.5$.
5. Expressing $\alpha$ multiplied by $10^{-1}$ as the nearest integer: $\alpha \times 10^{-1} = 0.5 \times 10^{-1} = 5$. The calculated value of $5$ falls within the expected range of $5,5$.

Correct Answer: 5

Solution and Explanation

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