Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 13 - Solutions - Exercises - Problems - Page 482: 103

Answer

(a) $100.060^{\circ}\,C$ (b) $100.993^{\circ}\,C$ (c) $101.99^{\circ}C$ (d) $101.11^{\circ}C$

Work Step by Step

(a) $\Delta T_{b}=m\times K_{b}$ Molality $m=0.118\,m=0.118\,\frac{mol\,solute}{kg\,solvent}$ Boiling point elevation constant for water is $K_{b}=0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$ Then $\Delta T_{b}=0.118\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$ $=0.0604^{\circ}C$ Boiling point= $100.000^{\circ}C+0.0604^{\circ}C=100.060^{\circ}C$ (b) $m=1.94\,\frac{mol\,solute}{kg\,solvent}$ $\Delta T_{b}=1.94\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$ $=0.993^{\circ}C$ Boiling point= $100.000^{\circ}C+0.993^{\circ}C=100.993^{\circ}C$ (c) $m=3.88\,\frac{mol\,solute}{kg\,solvent}$ $\Delta T_{b}=3.88\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$ $=1.99^{\circ}C$ Boiling point= $100.000^{\circ}C+1.99^{\circ}C=101.99^{\circ}C$ (d) $m=2.16\,\frac{mol\,solute}{kg\,solvent}$ $\Delta T_{b}=2.16\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$ $=1.11^{\circ}C$ Boiling point= $100.000^{\circ}C+1.11^{\circ}C=101.11^{\circ}C$
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