A solution is prepared by dissolving 0.50 mol of NaCl in 1.00 kg of water. If the boiling point elevation constant (K_b) for water is 0.512 °C·kg/mol, what is the boiling point of the solution? (Assume complete dissociation of NaCl and the normal boiling point of water is 100.00 °C.)

• A. 100.00 °C
• B. 100.26 °C
• C. 100.51 °C
• D. 101.02 °C

Answer: (D)

Verified step by step guidance

1. Identify the key concept involved: boiling point elevation, which is a colligative property. The boiling point of a solvent increases when a solute is dissolved in it, depending on the number of dissolved particles.
2. Calculate the molality (m) of the solution using the formula: \(m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\). Here, moles of NaCl = 0.50 mol and solvent mass = 1.00 kg.
3. Determine the van't Hoff factor (\(i\)) for NaCl. Since NaCl dissociates completely into Na\(^+\) and Cl\(^-\) ions, \(i = 2\).
4. Use the boiling point elevation formula: \(\Delta T_b = i \times K_b \times m\), where \(K_b\) is the boiling point elevation constant (0.512 °C·kg/mol for water).
5. Calculate the new boiling point by adding the boiling point elevation to the normal boiling point of water: \(\text{Boiling point} = 100.00\,^{\circ}\mathrm{C} + \Delta T_b\).