# Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605d: 62

$100.26^{\circ}C$

#### Work Step by Step

$K_{b}\, for\,water= 0.512 ^{\circ}C\,kg/mol$ $Molality\,m= \frac{No.\, of\,moles\,of\,solute}{Mass\,of\, solvent\,in\,kg}=\frac{\frac{15.0\,g}{60.\,g/mol}}{0.500\,kg}= 0.50\,mol/kg$ $\Delta T_{b}=K_{b}m=$$0.512 ^{\circ}C\,kg/mol\times0.50\,mol/kg= 0.26^{\circ}C$ $T_{b}(solution)= T_{b}(solvent)+\Delta T_{b}= 100^{\circ}C+0.26^{\circ}C= 100.26^{\circ}C$

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