Chemistry: The Molecular Science (5th Edition)

by Moore, John W.; Stanitski, Conrad L.

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: 47c

Answer

0.0260 M

Work Step by Step

Molarity of $CaC_{2}O_{4}$=$ \frac{Number\,of\,moles\,of\,CaC_{2}O_{4}}{Volume\, of\, solution\,in\,L}=\frac{\frac{Mass\,in\,grams}{Molar\,mass}}{Volume\,of\, solution\,in\,L}$ $=\frac{\frac{2.50\,g}{128.1\,g/mol}}{0.750\,L}=0.0260\,M$

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