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: 53b

Answer

0.123 M $ C _{6}H_{12}0_{6} $

Work Step by Step

$\frac{ 28.33 grams C _{6}H_{12}0_{6} }{1.28 L}\times\frac{1 molC _{6}H_{12}0_{6} }{180.16 gramsC _{6}H_{12}0_{6} } $
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