Chapter 13 - The Chemistry of Solutes and Solutions - Exercise 13.15 - Calculating teh Boiling Point of a Solution - Page 589: a

$80.35^{\circ}C$

$K_{b}=2.53^{\circ}C\,kg/mol$ $m_{solute}=0.10\,mol/kg$ $\Delta T_{b}= T_{b}(solution)-T_{b}(solvent)$ $=K_{b}\times m_{solute}$ $\implies T_{b}(solution)= T_{b}(solvent)+K_{b}\times m_{solute}$ $=80.10^{\circ} C+ (2.53\times0.10^{\circ})C$ $=80.35^{\circ}C$