Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605c: 46b


0.174 M

Work Step by Step

Molarity=$ \frac{Number\,of\,moles\,of\,K_{2}CrO_{4}}{Volume\, of\, solution\,in\,L}=\frac{\frac{Mass\,of\,K_{2}CrO_{4}}{Molar\,mass}}{Volume\,of\, solution\,in\,L}$ $=\frac{\frac{5.08g}{194.19g/mol}}{0.150L}=0.174\,M$
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