Question

What is the boiling point of solution of 0.1 m KCl? 
K$_b$ of water is 0.52 K kg mol$^{-1}$, $\alpha = 100%$ (water boils at 373 K)

• A. $100.104~\text{K}$
• B. $373.104~\text{K}$
• C. $273.104~\text{K}$
• D. $373.052~\text{K}$

Hint:

Boiling Point Elevation.

• Colligative property that depends on number of particles.
• $\Delta T_b = iK_bm$
• For ionic solutes, dissociation increases the van’t Hoff factor $i$.

Answer 1

The Correct Option is B

The elevation in boiling point is given by the formula: \[ \Delta T_b = i \cdot K_b \cdot m \] Where:

• $i$ is the van't Hoff factor. For KCl, $i = 2$ (dissociates into K$^+$ and Cl$^-$).
• $K_b = 0.52~\text{K kg mol}^{-1}$
• $m = 0.1~\text{mol/kg}$

Substitute the values: \[ \Delta T_b = 2 \times 0.52 \times 0.1 = 0.104~\text{K} \] Boiling point of the solution: \[ T_b = 373~\text{K} + 0.104~\text{K} = 373.104~\text{K} \]

Answer 2

Step 1: Understanding the problem
We need to find the boiling point of a 0.1 molal (m) solution of KCl in water.
Given data:
- Molal concentration, m = 0.1 mol/kg
- Boiling point elevation constant for water, K_b = 0.52 K·kg/mol
- Degree of ionization, α = 100% (complete dissociation)
- Normal boiling point of water = 373 K

Step 2: Calculate van't Hoff factor (i)
KCl dissociates completely into K⁺ and Cl⁻ ions, so the number of particles formed = 2
Therefore, i = 1 + (n - 1) × α = 1 + (2 - 1) × 1 = 2

Step 3: Calculate boiling point elevation (ΔT_b)
\[ \Delta T_b = i \times K_b \times m = 2 \times 0.52 \times 0.1 = 0.104 \text{ K} \]

Step 4: Calculate new boiling point
\[ T_{\text{boiling}} = 373 \text{ K} + 0.104 \text{ K} = 373.104 \text{ K} \]

Step 5: Conclusion
The boiling point of the 0.1 m KCl solution is 373.104 K.