0.1 molal aqueous solution of an electrolyte `AB_(3)` is 90% ionised. The boiling point of the solution at 1 atm is (`K_(b(H_(2)O)) = 0.52 kg " mol"^(-1)`)

Options

A. 273.19 K
B. 374.92 K
C. 376. 4 K
D. 373. 19 K

Answer

To solve the problem step by step, we will calculate the boiling point of a 0.1 molal aqueous solution of an electrolyte \( AB_3 \) that is 90% ionized.

### Step 1: Determine the degree of ionization (\( \alpha \))
The problem states that the electrolyte \( AB_3 \) is 90% ionized. Therefore, we can express the degree of ionization as:
\[ \alpha = 0.9 \]

### Step 2: Write the dissociation equation
The dissociation of the electrolyte \( AB_3 \) can be represented as:
\[ AB_3 \rightarrow A^{3+} + 3B^{-} \]
From this dissociation, we can see that for every 1 mole of \( AB_3 \) that dissociates, it produces 1 mole of \( A^{3+} \) and 3 moles of \( B^{-} \).

### Step 3: Calculate the total number of moles at equilibrium
Initially, we have 1 mole of \( AB_3 \). After dissociation, the number of moles at equilibrium can be calculated as:
\[ \text{Total moles at equilibrium} = 1 - \alpha + \alpha + 3\alpha = 1 + 3\alpha \]
Substituting \( \alpha = 0.9 \):
\[ \text{Total moles at equilibrium} = 1 + 3(0.9) = 1 + 2.7 = 3.7 \]

### Step 4: Calculate the Van't Hoff factor (\( i \))
The Van't Hoff factor \( i \) is defined as the ratio of the total number of moles at equilibrium to the number of moles initially present:
\[ i = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} = \frac{1 + 3\alpha}{1} = 1 + 3\alpha \]
Substituting \( \alpha = 0.9 \):
\[ i = 1 + 3(0.9) = 3.7 \]

### Step 5: Calculate the elevation in boiling point (\( \Delta T_b \))
The formula for the elevation in boiling point is given by:
\[ \Delta T_b = K_b \cdot m \cdot i \]
Where:
- \( K_b = 0.52 \, \text{K kg mol}^{-1} \) (given)
- \( m = 0.1 \, \text{mol/kg} \) (given)
- \( i = 3.7 \) (calculated)

Substituting the values:
\[ \Delta T_b = 0.52 \cdot 0.1 \cdot 3.7 = 0.1924 \, \text{K} \]

### Step 6: Calculate the boiling point of the solution
The boiling point of pure water is \( 100^\circ C \) or \( 373 \, \text{K} \). Therefore, the boiling point of the solution can be calculated as:
\[ T_b = T_{b,\text{pure}} + \Delta T_b \]
Substituting the values:
\[ T_b = 373 \, \text{K} + 0.1924 \, \text{K} = 373.1924 \, \text{K} \]

### Final Answer
The boiling point of the solution is approximately:
\[ \boxed{373.19 \, \text{K}} \]

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