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 - Page 569: 13.49a



Work Step by Step

First, we need to find moles of methanol and moles of acetonitrile. Also, from the periodic table, molar mass of methanol = 32.04 g/mol. Molar mass of acetonitrile = 41.05 g/mol $\frac{22.5\;mL\;methanol}{} (\frac{0.791\;g}{1 \;mL})(\frac{1\;mol}{32.04\;g})=0.5555\;mol\;methanol$ $\frac{98.7\;mL\;acetonitrile}{}(\frac{0.786\;g}{1\;mL})(\frac{1\;mol}{41.05\;g})=1.890\;mol\;acetonitrile$ Now, we will find the mole fraction of methanol. That is, the number of moles of methanol divided by total moles. $X_{methanol}=\frac{0.5555\;mol\;methanol}{0.5555\;mol\;methanol+1.890\;mol\;acetonitrile}=0.227$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.