SOLUTION:

Step 1: Calculate the mass of solute (NaCl).

Assume that the mass of solution is 100 g.

[tex]\begin{aligned} \%(w/w) & = \frac{mass_{\text{solute}}}{mass_{\text{solution}}} \times 100\% \\ \frac{\%(w/w)}{100\%} & = \frac{mass_{\text{solute}}}{mass_{\text{solution}}} \times \frac{100\%}{100\%} \\ \frac{\%(w/w)}{100\%} & = \frac{mass_{\text{solute}}}{mass_{\text{solution}}} \\ \frac{mass_{\text{solute}}}{mass_{\text{solution}}} & = \frac{\%(w/w)}{100\%} \\ mass_{\text{solute}} & = \frac{\%(w/w)}{100\%} \times mass_{\text{solution}} \\ & = \frac{3.5\%}{100\%} \times \text{100 g} \\ & = \text{3.5 g} \end{aligned}[/tex]

Step 2: Calculate the number of moles of solute.

The molar mass of NaCl is 58.44 g/mol.

[tex]\begin{aligned} n_{\text{solute}} & = \frac{mass_{\text{solute}}}{MM_{\text{solute}}} \\ & = \frac{\text{3.5 g}}{\text{58.44 g/mol}} \\ & = \text{0.05989 mol} \end{aligned}[/tex]

Step 3: Calculate the mass of solvent.

[tex]\begin{aligned} mass_{\text{solvent}} & = mass_{\text{solution}} - mass_{\text{solute}} \\ & = \text{100 g} - \text{3.5 g} \\ & = \text{96.5 g} \\ & = \text{0.0965 kg} \end{aligned}[/tex]

Step 4: Calculate the molality of the solution.

[tex]\begin{aligned} m & = \frac{n_{\text{solute}}}{mass_{\text{solvent}}} \\ & = \frac{\text{0.05989 mol}}{\text{0.0965 kg}} \\ & = \text{0.6206 mol/kg} \\ & = 0.6206 \: m \end{aligned}[/tex]

Step 5: Determine the van't Hoff factor.

Since NaCl is a strong electrolyte, it will dissociate completely into one Na⁺ and one Cl⁻ ions. Then, the van't Hoff factor is

i = 2

Step 6: Calculate the boiling-point elevation.

[tex]\begin{aligned} \Delta T_{\text{b}} & = iK_{\text{b}}m \\ & = (2)(0.52^{\circ}\text{C/}m)(0.6206 \: m) \\ & = 0.65^{\circ}\text{C} \end{aligned}[/tex]

Step 7: Calculate the boiling point of the solution.

[tex]\begin{aligned} T_{\text{b}} & = T^{\circ}_{\text{b}} + \Delta T_{\text{b}} \\ & = 100^{\circ}\text{C} + 0.65^{\circ}\text{C} \\ & = \boxed{100.65^{\circ}\text{C}} \end{aligned}[/tex]

Hence, the boiling point of seawater is 100.65°C.

[tex]\\[/tex]

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