Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 115: 3.39c


There are $ 1.85 \times 10^{19}$ allicin molecules in 5.00 mg of that substance.

Work Step by Step

1. Identify the conversion factor: - Mole to molecules: $\frac{6.022 \times 10^{23} \space molecules}{1 \space mole}$ 2. As we have calculated in 3.39b, there is a total of $ 3.08 \times 10^{-5} \space mole$ of allicin in 5.00 mg of that substance. $3.08 \times 10^{-5} \space mole \times \frac{6.022 \times 10^{23} \space molecules}{1 \space mole} = 1.85 \times 10^{19} \space molecules$
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