Boiling order of 0.01 M `AB_2` which is 10% dissociated in aqueous medium `(K_(bH_2O)=0.52)` as `A^(+)` and `B^-`

Options

A. 273.006 K
B. 373.006 K
C. 0.006 K
D. 272.006

Answer

To solve the problem of finding the boiling point of a 0.01 M solution of \( AB_2 \) which is 10% dissociated in an aqueous medium, we will follow these steps:

### Step 1: Determine the degree of dissociation and van 't Hoff factor (i)

Given that \( AB_2 \) dissociates into \( A^{2+} \) and \( 2B^- \):
- The dissociation can be represented as:
\[ AB_2 \rightarrow A^{2+} + 2B^- \]
- For every 1 mole of \( AB_2 \), we get 3 moles of ions (1 mole of \( A^{2+} \) and 2 moles of \( B^- \)).
- The degree of dissociation (α) is given as 10%, or 0.1.

Using the formula for the van 't Hoff factor \( i \):
\[ i = 1 + (n - 1) \cdot \alpha \]
where \( n \) is the number of particles formed upon dissociation. Here, \( n = 3 \):
\[ i = 1 + (3 - 1) \cdot 0.1 = 1 + 0.2 = 1.2 \]

### Step 2: Calculate the molality of the solution

The molarity (M) of the solution is given as 0.01 M. Assuming the density of the solution is approximately that of water (1 g/mL), we can calculate the mass of the solvent (water) in 1 L of solution:
- Mass of the solution = 1000 g
- Moles of \( AB_2 \) in 1 L = 0.01 moles

Since we are assuming the mass of the solute is negligible compared to the solvent, the mass of the solvent (water) is approximately 1000 g.

Now, we can calculate the molality (m):
\[ \text{Molality} (m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.01 \text{ moles}}{1 \text{ kg}} = 0.01 \text{ m} \]

### Step 3: Calculate the elevation in boiling point (\( \Delta T_b \))

The formula for the elevation in boiling point is:
\[ \Delta T_b = i \cdot m \cdot K_b \]
Where:
- \( K_b \) for water is given as 0.52 °C kg/mol.
Substituting the values:
\[ \Delta T_b = 1.2 \cdot 0.01 \cdot 0.52 = 0.00624 \, °C \]

### Step 4: Calculate the boiling point of the solution

The boiling point of pure water is 100 °C (or 373 K). Therefore, the boiling point of the solution is:
\[ T_b = T_{b,\text{solvent}} + \Delta T_b = 100 °C + 0.00624 °C = 100.00624 °C \]

### Final Answer

Thus, the boiling point of the 0.01 M \( AB_2 \) solution, which is 10% dissociated, is approximately:
\[ \boxed{100.00624 \, °C} \]

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