Chemistry and Chemical Reactivity (9th Edition)

by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.

Published by Cengage Learning

ISBN 10: 1133949649

ISBN 13: 978-1-13394-964-0

Chapter 13 Solutions and Their Behavior - Study Questions - Page 505c: 56

Answer

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Work Step by Step

a) Weight percent: $1130/(1130+7250)\times 100\%=13.5\%$ Number of moles of NaCl: $1130\ g\div 58.44\ g/mol=19.35\ mol$ Number of moles of water: $7250\ g\div 18.015\ g/mol=402.44\ mol$ b) Mole fraction: $19.35/(19.35+402.44)=0.046$ c) Molality: $19.35\ mol/(7250/1000)kg=2.67\ m$

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