[tex]\tt{\huge{\red{Solution:}}}[/tex]
Step 1: Calculate the number of moles of solute (biphenyl).
[tex]\begin{aligned} \text{moles of solute} & = \frac{\text{mass of solute}}{\text{molar mass of solute}} \\ & = \frac{\text{5.0 g}}{\text{154 g/mol}} \\ & = \text{0.03247 mol} \end{aligned}[/tex]
Step 2: Calculate the molality of the solution.
Note that 100.0 g = 0.1000 kg.
[tex]\begin{aligned} m & = \frac{\text{moles of solute}}{\text{mass of solvent}} \\ & = \frac{\text{0.03247 mol}}{\text{0.1000 kg}} \\ & = \text{0.3247 mol/kg} \end{aligned}[/tex]
Step 3: Calculate the boiling-point elevation.
[tex]\begin{aligned} \Delta T_{\text{b}} & = K_{\text{b}}m \\ & = (2.53^{\circ}\text{C} \cdot \text{kg/mol})(\text{0.3247 mol/kg}) \\ & = 0.82^{\circ}\text{C} \end{aligned}[/tex]
Final Step: 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} + 0.82^{\circ}\text{C} \\ & = \boxed{80.92^{\circ}\text{C}} \end{aligned}[/tex]
Hence, the boiling point of the solution is 80.92°C.
[tex]\\[/tex]
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