Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 12 - Sections 12.1-12.8 - Exercises - Problems by Topic - Page 590: 63b

Answer

0.444 m

Work Step by Step

Molality=$ \frac{Number\,of\,moles\,of\,solute}{mass\,of\,solvent\,in\,kg}=\frac{\frac{28.4\,g}{180.\,g/mol}}{0.355\,kg}$ $=0.444\,mol/kg= 0.444\,m$
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