## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179f: 73

#### Answer

$104.03\ g/mol$

#### Work Step by Step

Number of moles of the base: $0.509\ M\times 36.04\ mL=18.34\ mmol$ From stoichiometry: $9.17\ mmol$ of acid. Molar mass: $954\ mg\div 9.17\ mmol=104.03\ g/mol$

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