Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 13 - Properties of Solutions - Exercises: 13.49b

Answer

$7.16\;m\;CH_3OH$

Work Step by Step

molality = mol solute / kg solvent Since methanol is dissolved in acetonitrile, methanol is the solute and acetonitrile is the solvent. First, calculate moles of methanol (molar mass = 32.04 g/mol). $\frac{22.5\;mL\;CH_3OH}{}(\frac{0.791\;g}{1\;mL})(\frac{1\;mol}{32.04\;g})=0.5555\;mol\;CH_3OH$ Next, calculate kg acetonitrile. $\frac{98.7\;mL\;CH_3CN}{}(\frac{0.786\;g}{1\;mL})(\frac{1\;kg}{1000\;g})=0.7758\;kg\;CH_3CN$ Next, calculate molality. $\frac{0.5555\;mol\;CH_3OH}{0.7758\;kg\;CH_3CN}=7.16\;m\;CH_3OH$
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