# Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605c: 47c

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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