The molal elevation constant of water =`0.52 K m^(-1)`. The boiling point of `1.0 molal` aqueous `KCl` solution (assuming complete dissociation of `KCl`) should be

Options

A. `100.52^(@)C`
B. `101.04^(@)C`
C. `99.46^(@)C`
D. `98.96^(@)C`

Answer

To solve the problem of finding the boiling point of a 1.0 molal aqueous KCl solution, we will use the formula for boiling point elevation. Here are the steps to arrive at the solution:

### Step 1: Understand the Formula
The boiling point elevation can be calculated using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \(\Delta T_b\) = elevation in boiling point
- \(i\) = Van't Hoff factor (number of particles the solute dissociates into)
- \(K_b\) = molal elevation constant of the solvent (water in this case)
- \(m\) = molality of the solution

### Step 2: Identify the Values
From the question, we have:
- \(K_b = 0.52 \, K \cdot m^{-1}\)
- \(m = 1.0 \, molal\)
- For KCl, which dissociates into K\(^+\) and Cl\(^-\), the Van't Hoff factor \(i = 2\).

### Step 3: Calculate the Elevation in Boiling Point
Using the values identified:
\[ \Delta T_b = i \cdot K_b \cdot m = 2 \cdot 0.52 \cdot 1.0 \]
Calculating this gives:
\[ \Delta T_b = 1.04 \, K \]

### Step 4: Determine the New Boiling Point
The boiling point of pure water is 100°C. Therefore, the boiling point of the KCl solution will be:
\[ T_b = T_{b, \text{pure}} + \Delta T_b = 100°C + 1.04°C = 101.04°C \]

### Final Answer
The boiling point of the 1.0 molal aqueous KCl solution is:
\[ \boxed{101.04 \, °C} \]

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