What should be the boiling point of `1.0 molal` aqueous `KCl` solution (assuming complete dissociation of KCl) if `K_(b)^(H_(2)O)` is `0.52 K m^(-1)`?

Options

A. `100.52^(@)C`
B. `101.04^(@)C`
C. `99.48^(@)C`
D. `98.96^(@)C`

Explanation

To find the boiling point of a 1.0 molal aqueous KCl solution, we will follow these steps:

### Step 1: Understand the dissociation of KCl
KCl dissociates completely in water into potassium ions (K⁺) and chloride ions (Cl⁻). Therefore, for every mole of KCl that dissolves, it produces 2 moles of ions:
\[ \text{KCl} \rightarrow \text{K}^+ + \text{Cl}^- \]

### Step 2: Determine the van 't Hoff factor (i)
Since KCl dissociates into 2 ions, the van 't Hoff factor (i) is:
\[ i = 2 \]

### Step 3: Use the boiling point elevation formula
The boiling point elevation (\( \Delta T_b \)) can be calculated using the formula:
\[ \Delta T_b = K_b \times m \times i \]
where:
- \( K_b \) is the ebullioscopic constant of water, given as \( 0.52 \, \text{K kg}^{-1} \).
- \( m \) is the molality of the solution, which is \( 1.0 \, \text{molal} \).
- \( i \) is the van 't Hoff factor, which we found to be \( 2 \).

### Step 4: Plug in the values
Now, substituting the values into the formula:
\[ \Delta T_b = 0.52 \, \text{K kg}^{-1} \times 1.0 \, \text{molal} \times 2 \]
\[ \Delta T_b = 0.52 \times 1.0 \times 2 = 1.04 \, \text{K} \]

### Step 5: Calculate the boiling point of the solution
The boiling point of the solution can be found using the formula:
\[ T_b = T_{b0} + \Delta T_b \]
where \( T_{b0} \) is the boiling point of pure water, which is \( 100 \, \text{°C} \):
\[ T_b = 100 \, \text{°C} + 1.04 \, \text{K} \]
\[ T_b = 101.04 \, \text{°C} \]

### Final Answer
The boiling point of the 1.0 molal aqueous KCl solution is \( 101.04 \, \text{°C} \).

---