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` that is 10% dissociated in an aqueous medium, we can follow these steps:

### Step 1: Determine the dissociation of `AB_2`
The compound `AB_2` dissociates in water as follows:
\[ AB_2 \rightarrow A^+ + 2B^- \]

Given that `AB_2` is 10% dissociated, we can express the degree of dissociation (α) as:
\[ \alpha = \frac{10}{100} = 0.1 \]

### Step 2: Calculate the van't Hoff factor (i)
The van't Hoff factor (i) is calculated using the formula:
\[ i = 1 + \alpha(n - 1) \]
where n is the number of particles produced from one formula unit of the solute.

For `AB_2`, after dissociation, we have:
- 1 ion of `A^+`
- 2 ions of `B^-`

Thus, \( n = 1 + 2 = 3 \).

Now substituting the values:
\[ i = 1 + 0.1(3 - 1) = 1 + 0.1(2) = 1 + 0.2 = 1.2 \]

### Step 3: Calculate the elevation in boiling point (ΔTb)
The elevation in boiling point is given by the formula:
\[ \Delta T_b = K_b \cdot m \cdot i \]

Where:
- \( K_b = 0.52 \) (given)
- \( m = 0.01 \) M (molality, which is approximately equal to molarity for dilute solutions)

Now substituting the values:
\[ \Delta T_b = 0.52 \cdot 0.01 \cdot 1.2 \]

Calculating this gives:
\[ \Delta T_b = 0.52 \cdot 0.012 = 0.00624 \, \text{°C} \]

### Step 4: Calculate the boiling point of the solution (Tb)
The boiling point of the solution can be calculated using the formula:
\[ T_b = T_b^0 + \Delta T_b \]
where \( T_b^0 = 100 \, \text{°C} \) (boiling point of pure water).

Thus:
\[ T_b = 100 + 0.00624 = 100.00624 \, \text{°C} \]

### Step 5: Convert the boiling point to Kelvin
To convert from Celsius to Kelvin, we add 273:
\[ T_b(K) = 100.00624 + 273 = 373.00624 \, \text{K} \]

### Final Answer
The boiling point of the solution is approximately:
\[ T_b \approx 373.006 \, \text{K} \]

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