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

0.174 M

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}$ $=\frac{\frac{5.08g}{194.19g/mol}}{0.150L}=0.174\,M$