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 505d: 72


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Work Step by Step

In 1 kg of the solution: Mass of acetic acid: $5/100\times 1000\ g=50\ g$ Mass of water: $950\ g$ The number of moles of acetic acid: $50\ g/60.05\ g/mol=0.833\ mol$ The number of moles of water: $950\ g\div 18.015\ g/mol=52.74\ mol$ Mole fraction: $0.833/(0.833+52.74)=0.016$ Molality: $0.833\ mol/0.95\ kg=0.877\ mol/kg$ ppm concentration: $5/100\times 10^6=50\ 000\ ppm$ The molarity can't be calculated because, from the information provided, it's not possible to calculate the volume of the 1 kg calculation basis, for that, the density of the solution would be required.
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