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

0.167 M

Molarity=$ \frac{Number\,of\,moles\,of\,C_{6}H_{12}O_{6}}{Volume\, of\, solution\,in\,L}=\frac{\frac{Mass\,of\,C_{6}H_{12}O_{6}}{Molar\,mass}}{Volume\,of\, solution\,in\,L}$ $=\frac{\frac{15.0\,g}{180.g/mol}}{500.\times10^{-3}L}=0.167\,M$