Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 13 Solutions and Their Behavior - Study Questions - Page 505b: 32


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Work Step by Step

Boiling point elevation: $104.4-100.00=4.4^{\circ} C$ Molality: $\Delta T_b=K_bm\rightarrow 4.4^{\circ}C=0.5121^{\circ}C/m\times m\rightarrow m=8.59\ m$ Number of moles of glycerol: $8.59\ mol/kg\times0.735\ kg=6.32\ mol$ Mass of glycerol added: $6.32\ mol\times92.09\ g/mol=581.56\ g$ Number of moles of water: $735\ g\div18.015\ g/mol=40.80\ mol$ Mole fraction of solute: $6.32/(40.80+6.32)=0.134$
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