## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

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

#### Answer

$85.16 ^{\circ}C$

#### Work Step by Step

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

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