To solve this problem, we need to determine the boiling point elevation of the solution.$\\$Given:$\\$• Mass of solute $= 6.5$ g$\\$• Mass of solvent (water) $= 100$ g$\\$• Vapour pressure of solution at $100^\circ C = 732$ mm Hg$\\$• Vapour pressure of pure water at $100^\circ C = 760$ mm Hg$\\$• Boiling point elevation constant $K_b = 0.52$ $^\circ C \cdot$kg/mol$\\$First, calculate the mole fraction of the solute using Raoult's Law:$\\ P_0 - P = P_0 \cdot X_{solute} \\$Where:$\\ P_0 = 760$ mm Hg (vapour pressure of pure water)$\\ P = 732$ mm Hg (vapour pressure of solution)$\\$Substitute the values:$\\ 760 - 732 = 760 \cdot X_{solute} \\ 28 = 760 \cdot X_{solute} \\ X_{solute} = \frac{28}{760} \\ X_{solute} \approx 0.0368 \\$Next, calculate the molality of the solution:$\\$Mole fraction of solute $X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \\$Assuming the molar mass of the solute is $M,$ then:$\\ n_{solute} = \frac{6.5}{M} \\ n_{solvent} = \frac{100}{18}$ (moles of water)$\\$Substitute into the mole fraction equation:$\\ 0.0368 = \frac{\frac{6.5}{M}}{\frac{6.5}{M} + \frac{100}{18}} \\$Solve for $M: \\ 0.0368 \left(\frac{6.5}{M} + \frac{100}{18}\right) = \frac{6.5}{M} \\ 0.0368 \cdot \frac{6.5}{M} + 0.0368 \cdot \frac{100}{18} = \frac{6.5}{M} \\ 0.0368 \cdot \frac{6.5}{M} + 0.2044 = \frac{6.5}{M} \\ 0.2044 = \frac{6.5}{M} - 0.0368 \cdot \frac{6.5}{M} \\ 0.2044 = \frac{6.5(1 - 0.0368)}{M} \\ 0.2044 = \frac{6.5 \cdot 0.9632}{M} \\ M = \frac{6.5 \cdot 0.9632}{0.2044} \\ M \approx 30.6$ g/mol$\newline$Now, calculate the molality m$\newline m = \frac{n_{solute}}{mass \ of \ solvent \ in \ kg} \\ m = \frac{6.5/30.6}{0.1} \\ m \approx 2.12$ mol/kg$\\$Calculate the boiling point elevation $\Delta T_b: \\ \Delta T_b = K_b \cdot m \\ \Delta T_b = 0.52 \cdot 2.12 \\ \Delta T_b \approx 1.10^\circ C \\$The boiling point of the solution is:$\\ 100^\circ C + 1.10^\circ C = 101.10^\circ C \\$Since the options are given in whole numbers, the closest option is $102^\circ C. \\$Therefore, the boiling point of the solution is $102^\circ C,$ which corresponds to Option 1.