QJEEM24Cha537: JEE Main 2024 Chemistry - Class 12 JEEM24Cha537 Solution: A solution containing of an electrolyte in of water boils at .…

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:

$\Delta T_b = i \cdot K_b \cdot m$

where:

$\Delta T_b$ is the boiling point elevation,

$i$ is the van't Hoff factor,

$K_b$ is the molal boiling point elevation constant, and

$m$ is the molality of the solution.

Given data:

• $\Delta T_b = 100.52^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C} = 0.52^{\circ} \mathrm{C}$
• Mass of $\mathrm{AB}_2$ = 10 g
• Mass of water = 100 g = 0.1 kg
• Molar mass of $\mathrm{AB}_2$ = 200 g/mol
• $K_b$ = 0.52 K kg mol^-1

First, we calculate the molality (m):

$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

$= \frac{\frac{10}{200}}{0.1} \:\mathrm{mol\ kg}^{-1}$

$= 0.5 \:\mathrm{mol\ kg}^{-1}$

Next, using the formula for boiling point elevation:

$0.52 = i \cdot 0.52 \cdot 0.5$

$i = \frac{0.52}{0.52 \cdot 0.5}$

$i = \frac{1}{0.5}$

$i = 2$

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:

$i = 1 + (n-1)\alpha$

where

• $n$ is the number of ions produced (which is 3 for $\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}$ )
• $\alpha$ is the degree of ionization

Substituting $i = 2$ into the equation:

$2 = 1 + 2 \alpha$

$2 - 1 = 2 \alpha$

$\alpha = \frac{1}{2}$

$\alpha = 0.5$

The degree of ionization $(\alpha)$ is therefore

$0.5 \times 10^{-1} = 5 \times 10^{-2}$ .

Hence, the degree of ionization of the electrolyte $(\alpha)$ is approximately 5 $\times 10^{-1}$ (the nearest integer is 5).