Question

An aqueous solution contains 20g of a non-volatile strong electrolyte \( {A}_2{B} \) (Molar mass = 60 g mol\(^{-1}\)) in 1 kg of water. If the electrolyte is 100% dissociated at this concentration, what is the boiling point of the solution? (Kb of water is 0.52 K kg mol\(^{-1}\))

• A. 372.482 K
• B. 374.56 K
• C. 373.52 K
• D. 371.44 K
• E. 374.02 K

Hint:

To calculate the boiling point elevation, remember to use the van't Hoff factor for dissociation, and ensure the mass of solvent is in kilograms.

Answer 1

The Correct Option is C

The formula for the boiling point elevation is:

\[ \Delta T_b = i \cdot K_b \cdot m \]

Where:

• \( i \) is the van't Hoff factor (number of particles the electrolyte dissociates into).
• \( K_b \) is the ebullioscopic constant of the solvent (water in this case).
• \( m \) is the molality of the solution.

Step 1: Calculate the molality of the solution.

Molality \( m \) is defined as:

\[ m = \frac{\text{mol of solute}}{\text{mass of solvent in kg}} \]

The number of moles of solute is:

\[ \text{mol of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{20 \text{ g}}{60 \text{ g/mol}} = 0.3333 \text{ mol} \]

Since the solvent mass is 1 kg, the molality is:

\[ m = \frac{0.3333 \text{ mol}}{1 \text{ kg}} = 0.3333 \text{ mol/kg} \]

Step 2: Since the electrolyte \( \text{A}_2\text{B} \) dissociates into 3 ions (\( 2\text{A}^+ \) and \( \text{B}^- \)), the van't Hoff factor \( i = 3 \).

Step 3: Now, calculate the change in boiling point:

\[ \Delta T_b = i \cdot K_b \cdot m = 3 \cdot 0.52 \text{ K kg mol}^{-1} \cdot 0.3333 \text{ mol/kg} = 0.51996 \text{ K} \]

Step 4: The normal boiling point of water is 373.15 K, so the new boiling point is:

\[ T_b = 373.15 \text{ K} + 0.51996 \text{ K} = 373.52 \text{ K} \]

Thus, the boiling point of the solution is \( 373.52 \text{ K} \).