1. Use elevation in boiling point

For a solution, $\Delta T_b = K_b \sum i m$ where $i$ is the van't Hoff factor and $m$ is molality.

Since ionization is $100\%$:

• For $\mathrm{NaCl}$: $i=2$
• For $\mathrm{MgCl_2}$: $i=3$

2. Assume 100 g of sea water

Given composition by mass of solution:

• $29.25\%\ \mathrm{NaCl} \Rightarrow 29.25\,\text{g}$
• $19\%\ \mathrm{MgCl_2} \Rightarrow 19\,\text{g}$

So mass of water: $100 - 29.25 - 19 = 51.75\,\text{g} = 0.05175\,\text{kg}$

3. Moles of solutes

For $\mathrm{NaCl}$: $n_{\mathrm{NaCl}}=\frac{29.25}{58.5}=0.5\,\text{mol}$

For $\mathrm{MgCl_2}$: $n_{\mathrm{MgCl_2}}=\frac{19}{95}=0.2\,\text{mol}$

4. Calculate effective molality contribution

Using $\sum i n$ first: $i n \text{ for NaCl} = 2 \times 0.5 = 1.0$ $i n \text{ for MgCl}_2 = 3 \times 0.2 = 0.6$

Total effective moles of particles: $1.0 + 0.6 = 1.6$

Now effective molality: $m_{\text{eff}} = \frac{1.6}{0.05175} \approx 30.92\,\text{mol kg}^{-1}$

5. Boiling point elevation

$\Delta T_b = 0.52 \times 30.92 \approx 16.08\,\text{K}$

So normal boiling point of sea water: $100 + 16.08 = 116.08^\circ\text{C}$

Nearest integer: $\boxed{116}$

6. Comparison with stored answer

Derived answer = $116$

Stored correct answer = $116$

Hence, they agree.