To find the degree of ionization

, we need to use the boiling point elevation formula and the van't Hoff factor. The formulas we will use are:

where:

is the boiling point elevation,

is the van't Hoff factor,

is the molal boiling point elevation constant, and

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):

Next, using the formula for boiling point elevation:

Now, the van't Hoff factor $i$ is related to the degree of ionization

. For the electrolyte , which ionizes as , the van't Hoff factor $i$ can be expressed as:

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:

The degree of ionization

is therefore

.

Hence, the degree of ionization of the electrolyte

is approximately 5 (the nearest integer is 5).