Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 16 Principles of Chemical Reactivity: The Chemistry of Acids and Bases - 16-7 Calculations with Equilibrium Constants - Case Study - Questions - Page 614: 1


$3.46\times10^{-7}\ mol$

Work Step by Step

The molar mass of atropine is: $289.37\ g/mol$ Therefore the number of moles of 100 mg is: $(100/1000)g \div 289.37\ g/mol=3.46\times10^{-7}\ mol$
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