SOLUTION:
Step 1: List the given values.
[tex]\begin{aligned} & mass_{\text{solute}} = \text{100 g} \\ & MM_{\text{solute}} = \text{128.174 g/mol} \\ & mass_{\text{solvent}} = \text{480 g = 0.480 kg} \\ & K_{\text{b}} = 2.53^{\circ}\text{C/}m \\ & T^{\circ}_{\text{b}} = 80.1^{\circ}\text{C} \end{aligned}[/tex]
Step 2: Calculate the number of moles of solute.
[tex]\begin{aligned} n_{\text{solute}} & = \frac{mass_{\text{solute}}}{MM_{\text{solute}}} \\ & = \frac{\text{100 g}}{\text{128.174 g/mol}} \\ & = \text{0.780189 mol} \end{aligned}[/tex]
Step 3: Calculate the molality of solution.
[tex]\begin{aligned} m & = \frac{n_{\text{solute}}}{mass_{\text{solvent}}} \\ & = \frac{\text{0.780189 mol}}{\text{0.480 kg}} \\ & = \text{1.62539 mol/kg} \\ & = 1.62539 \: m \end{aligned}[/tex]
Step 4: Calculate the boiling-point elevation.
[tex]\begin{aligned} \Delta T_{\text{b}} & = K_{\text{b}}m \\ & = (2.53^{\circ}\text{C/}m)(1.62539 \: m) \\ & = 4.11^{\circ}\text{C} \end{aligned}[/tex]
Step 5: Calculate the boiling point of the solution.
[tex]\begin{aligned} T_{\text{b}} & = T^{\circ}_{\text{b}} + \Delta T_{\text{b}} \\ & = 80.1^{\circ}\text{C} + 4.11^{\circ}\text{C} \\ & = \boxed{84.21^{\circ}\text{C}} \end{aligned}[/tex]
Hence, the boiling point of the solution is 84.21°C.
[tex]\\[/tex]
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