Find boiling point elevation when 3g of NaCl dissolves in 100g water (Kb=0.52)

Question

Find the boiling point elevation when 3 g of NaCl is dissolved in 100 g of water. Given: $K_b = 0.52$ K·kg/mol, and NaCl completely dissociates into Na⁺ and Cl⁻ ions.

Solution — Step by Step

\Delta T_b = i \cdot K_b \cdot m

where:

$\Delta T_b$ = boiling point elevation (in K or °C)
$i$ = van’t Hoff factor (number of particles per formula unit after dissociation)
$K_b$ = molal boiling point elevation constant of solvent (0.52 K·kg/mol for water)
$m$ = molality of the solution (mol of solute per kg of solvent)

NaCl is a strong electrolyte. In dilute aqueous solution, it completely dissociates:

\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-

One formula unit gives 2 ions. Therefore, $i = \mathbf{2}$ .

Molar mass of NaCl: $M = 23 + 35.5 = 58.5$ g/mol.

Moles of NaCl: $n = \frac{3}{58.5} = 0.0513$ mol (approx.)

Mass of solvent = 100 g = 0.1 kg.

Molality: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0513}{0.1} = 0.513$ mol/kg

\Delta T_b = i \cdot K_b \cdot m = 2 \times 0.52 \times 0.513

\Delta T_b = 2 \times 0.267 = \mathbf{0.533 \text{ K}}

The boiling point of the solution = $100 + 0.533 = 100.533°C$ .

Why This Works

Boiling point elevation is a colligative property — it depends only on the number of solute particles, not their chemical identity. More particles mean a lower vapour pressure (Raoult’s law), which requires a higher temperature to reach atmospheric pressure (the definition of the boiling point).

NaCl dissociates into 2 ions, effectively doubling the number of solute particles compared to a non-electrolyte. Hence $i = 2$ doubles the boiling point elevation.

Quick check: without the van’t Hoff factor ( $i = 1$ ), $\Delta T_b = 0.267$ K. With $i = 2$ , it doubles to $0.533$ K. This doubling is the most important step students miss in electrolyte problems.

Alternative Method

You can also use the simplified formula directly for complete dissociation: effective molality = $i \times m$ = actual moles of particles per kg of solvent.

Effective molality = $2 \times 0.513 = 1.026$ mol/kg.

$\Delta T_b = K_b \times m_{\text{eff}} = 0.52 \times 1.026 = 0.533$ K. Same answer.

Common Mistake

The most frequent error: forgetting the van’t Hoff factor $i$ for electrolytes. Students directly use $\Delta T_b = K_b \times m$ without multiplying by 2 for NaCl. This gives 0.267 K instead of 0.533 K — exactly half the correct answer.

Also, many students use the mass of solution (100 g + 3 g = 103 g) as the solvent mass in the molality calculation. Molality uses only the solvent mass (100 g), not total solution mass. Total mass is used for mass percentage calculations, not molality.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next