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Step-by-Step Solution
Step 1: Identify the Relevant Concept
The question deals with the elevation of boiling point, a colligative property. When an electrolyte dissociates in water, the total number of particles affects the boiling point elevation through the van’t Hoff factor.
Step 2: Write the Dissociation Equation
Given the compound AB2 dissociates in water according to:
AB2 → A+ + 2 B−
The total number of ions (n) produced per formula unit of AB2 is 3 (1 A+ plus 2 B–).
Step 3: Calculate the van’t Hoff Factor ( i )
Let \alpha be the degree of dissociation. We are given \alpha = 0.1 (i.e., 10%).
The van’t Hoff factor i for an electrolyte that can give n particles when fully dissociated is calculated using:
i = 1 + (n - 1) \alpha
Here, n = 3 and \alpha = 0.1 , so
i = 1 + (3 - 1) \times 0.1 = 1 + 2 \times 0.1 = 1.2.
Step 4: Use the Elevation of Boiling Point Formula
The elevation in boiling point \Delta T_b is given by:
\Delta T_b = i \, K_b \, m
where
i is the van’t Hoff factor,
K_b is the molal elevation constant (0.5 K·kg·mol−1), and
m is the molality of the solution (10.0 molal).
Substitute the known values:
\Delta T_b = 1.2 \times 0.5 \times 10.0 = 6.0 \; (\text{K}).
Step 5: Calculate the Boiling Point of the Solution
The normal boiling point of pure water is given as 100°C. Therefore, the boiling point of the solution, T_b , is:
T_b = 100^{\circ}\mathrm{C} + 6^{\circ}\mathrm{C} = 106^{\circ}\mathrm{C}.
Final Answer
The boiling point of the solution is approximately 106°C.
