Question

If 1 mole of NaCl solute is dissolved into 1 kg of water, at what temperature will water boil at 1.013 bar? ($K_b$ of water is $0.52\,\text{K kg mol}^{-1}$)

• A. 373.15 K
• B. 373.67 K
• C. 374.19 K
• D. 373.19 K
• E. 375 K

Hint:

Never forget the Van't Hoff factor (\( i \)) for ionic solutes! If you treat NaCl like sugar (\( i=1 \)), you will get exactly half the correct elevation.

Answer 1

The Correct Option is C

Concept: Boiling point elevation is a colligative property. The increase in boiling point \( \Delta T_b \) is given by: \[ \Delta T_b = i \times K_b \times m \] where:
• \( i \) is the Van't Hoff factor (number of particles the solute dissociates into).
• \( K_b \) is the molal boiling point elevation constant.
• \( m \) is the molality of the solution.

Step 1: Identify the Van't Hoff factor \( i \) for NaCl.
NaCl is a strong electrolyte and dissociates completely: \[ NaCl \rightarrow Na^+ + Cl^- \] Since it produces 2 ions, \( i = 2 \).

Step 2: Calculate the elevation in boiling point \( \Delta T_b \).
Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1 \text{ mol}}{1 \text{ kg}} = 1 \text{ m} \). \[ \Delta T_b = 2 \times 0.52 \times 1 = 1.04 \text{ K} \]

Step 3: Calculate the new boiling point.
Standard boiling point of water \( T_b^\circ = 100^\circ\text{C} = 373.15 \text{ K} \). \[ T_b = T_b^\circ + \Delta T_b = 373.15 + 1.04 = 374.19 \text{ K} \]