Chemistry: A Molecular Approach (3rd Edition)

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

Chapter 4 - Sections 4.1-4.9 - Exercises - Problems by Topic - Page 188: 58c


$6.0\times10^{-2}$$L$ $C_{2}H_{6}O$

Work Step by Step

Use molarity as a conversion factor in order to convert moles into liters. $1.2\times10^{-2}$ $mols$ $C_{2}H_{6}O$ $\times\frac{1 L C_{2}H_{6}O}{0.200 mol C_{2}H_{6}O}$$=6.0\times10^{-2}$$L$ $C_{2}H_{6}O$
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