Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 13 - Problems: 13.56

Answer

a) .944 M C$_{12}$H$_{22}$O$_{11}$ b) .167 M LiNO$_{3}$

Work Step by Step

a) Step 1) 32.3 g C$_{12}$H$_{22}$O$_{11}$ ($\frac{1 mol}{342.3g}$)=.0944 mol C$_{12}$H$_{22}$O$_{11}$) Step 2) 100 mL ($\frac{1 L}{1000mL}$)= .1 L H$_{2}$O Step 3) $\frac{.0944 mol C_{12}H_{22}O_{11}}{.100 L}$= .944 M C$_{12}$H$_{22}$O$_{11}$ b) Step 1) 5.8 g LiNO$_{3}$ ($\frac{1 mol}{68.95 g}$)= .0841 g LiNO$_{3}$ Step 2) (505 mL)($\frac{1 L}{1000 mL}$) = .505 L Step 3) .0841g LiNO$_{3}$/.505 L = .167 M LiNO$_{3}$
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.