Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605d: 63

$85.16 ^{\circ}C$

$Molality\,m=\frac{0.200mol}{0.100kg}= 2.00mol/kg$ $\Delta T_{b}=K_{b}m=$$2.53^{\circ}C\,kg/mol\times2.00\,mol/kg= 5.06^{\circ}C$ $T_{b}(solution)= T_{b}(solvent)+\Delta T_{b}= 80.10^{\circ}C+5.06^{\circ}C= 85.16^{\circ}C$