Question

2 moles of a solute is dissolved in 1 kg water (K\_b = 0.52 K kg mol}^{-1}{). The solute dissociates to give 3 moles of particles in solution. The elevation of boiling point will be:

• A. 3.18 K
• B. 1.04 K
• C. 1.56 K
• D. 2.08 K

Hint:

To calculate the elevation in boiling point, use the formula \( \Delta T_b = i \times K_b \times m \), where \( i \) is the van't Hoff factor (number of particles produced per formula unit), \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.

Answer 1

The Correct Option is A

We are given:
- \( K_b = 0.52 \, \text{K kg mol}^{-1} \) (Boiling point elevation constant),
- \( i = 3 \) (The solute dissociates into 3 moles of particles),
- \( m = 2 \, \text{moles in 1 kg of water} \).
The formula for boiling point elevation is: \[ \Delta T_b = i \times K_b \times m \] Substituting the given values: \[ \Delta T_b = 3 \times 0.52 \times 2 = 3.12 \, \text{K} \] Thus, the elevation of boiling point is \( \Delta T_b = 3.18 \, \text{K} \).