A solution containing $10 \mathrm{~g}$ of an electrolyte $\mathrm{AB}_2$ in $100 \mathrm{~g}$ of water boils at $100.52^{\circ} \mathrm{C}$. The degree of ionization of the electrolyte $(\alpha)$ is _________ $\times 10^{-1}$. (nearest integer)

[Given : Molar mass of $\mathrm{AB}_2=200 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~K}_{\mathrm{b}}$ (molal boiling point elevation const. of water) $=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, boiling point of water $=100^{\circ} \mathrm{C} ; \mathrm{AB}_2$ ionises as $\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}]$

<p>To find the degree of ionization $(\alpha)$, we need to use the boiling point elevation formula and the van't Hoff factor. The formulas we will use are:</p> <p> <p>$\Delta T_b = i \cdot K_b \cdot m$</p> </p> <p>where: <p>$\Delta T_b$ is the boiling point elevation,</p> <p>$i$ is the van't Hoff factor,</p> <p>$K_b$ is the molal boiling point elevation constant, and</p> <p>$m$ is the molality of the solution.</p> </p> <p>Given data: <ul> <li>$$\Delta T_b = 100.52^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C} = 0.52^{\circ} \mathrm{C}$$</li> <li>Mass of $\mathrm{AB}_2$ = 10 g</li> <li>Mass of water = 100 g = 0.1 kg</li> <li>Molar mass of $\mathrm{AB}_2$ = 200 g/mol</li> <li>$K_b$ = 0.52 K kg mol^-1</li> </ul> </p> <p>First, we calculate the molality (m): <p> <p>$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$</p> <p>$= \frac{\frac{10}{200}}{0.1} \:\mathrm{mol\ kg}^{-1}$</p> <p>$= 0.5 \:\mathrm{mol\ kg}^{-1}$</p> </p> <p>Next, using the formula for boiling point elevation: <p> <p>$0.52 = i \cdot 0.52 \cdot 0.5$</p> <p>$i = \frac{0.52}{0.52 \cdot 0.5}$</p> <p>$i = \frac{1}{0.5}$</p> <p>$i = 2$</p> </p> <p>Now, the van't Hoff factor $i$ is related to the degree of ionization $(\alpha)$. For the electrolyte $\mathrm{AB}_2$, which ionizes as $\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}$, the van't Hoff factor $i$ can be expressed as: </p> <p> <p>$i = 1 + (n-1)\alpha$</p> </p> <p>where <ul> <li>$n$ is the number of ions produced (which is 3 for $\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}$)</li> <li>$\alpha$ is the degree of ionization </ul> </p> <p>Substituting $i = 2$ into the equation: <p> <p>$2 = 1 + 2 \alpha$</p> <p>$2 - 1 = 2 \alpha$</p> <p>$\alpha = \frac{1}{2}$</p> <p>$\alpha = 0.5$</p> </p> <p>The degree of ionization $(\alpha)$ is therefore <p>$0.5 \times 10^{-1} = 5 \times 10^{-2}$.</p> </p> <p>Hence, the degree of ionization of the electrolyte $(\alpha)$ is approximately 5 $\times 10^{-1}$ (the nearest integer is 5).</p>