JEE Main 2024 — Solutions Question with Solution

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):

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

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

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

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 $${\mathrm{AB}}_2\to {\mathrm{A}}^{2+}+2{\mathrm{B}}^{-}$$, 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 $${\mathrm{AB}}_2\to {\mathrm{A}}^{2+}+2{\mathrm{B}}^{-}$$)
• $$\alpha$$ 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).