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: 43


$151.25\ g/mol$

Work Step by Step

Boiling point elevation: $61.82-61.70=0.12^{\circ}C$ Molality: $\Delta T_b=K_bm\rightarrow m=0.12^{\circ}C\div 3.63^{\circ}C/m=0.033\ mol/kg$ Number of moles: $0.033\ mol/kg\times (25/1000)kg=8.26\times10^{-4}\ mol$ Molar mass: $0.125\ g\div 8.26\times10^{-4}\ mol=151.25\ g/mol$
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