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: 13.51c

Answer

$4.44\times10^{-2}\;mol\;C_6H_{12}O_6$

Work Step by Step

Since we know that the solution is 6.45% glucose by mass, we can easily find the mass of glucose (molar mass = 180.16 g/mol) in solution. $124.0\;g\times6.45\%=7.998\;g\;C_6H_{12}O_6$ Next, convert to moles. $\frac{7.998\;g\;C_6H_{12}O_6}{}(\frac{1\;mol}{180.16\;g})=4.44\times10^{-2}\;mol\;C_6H_{12}O_6$
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