$\overline{){\mathbf{Molarity}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Liters}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

**Calculate the ****molarity of stock solution, M _{1}:**

$\mathbf{Molarity}\mathbf{}\mathbf{=}\frac{\mathbf{}\mathbf{10}\mathbf{}\overline{)\mathbf{mg}\mathbf{}{\mathbf{C}}_{\mathbf{20}}{\mathbf{H}}_{\mathbf{29}}{\mathbf{FO}}_{\mathbf{3}}\mathbf{}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{g}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mg}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{20}}{\mathbf{H}}_{\mathbf{29}}{\mathbf{FO}}_{\mathbf{3}}}{\mathbf{336}\mathbf{.}\mathbf{44}\mathbf{\hspace{0.17em}}\overline{)\mathbf{g}\mathbf{}{\mathbf{C}}_{\mathbf{20}}{\mathbf{H}}_{\mathbf{29}}{\mathbf{FO}}_{\mathbf{3}}}}}{\mathbf{500}\mathbf{}\overline{)\mathbf{mL}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}}$

**= 5.94 x 10 ^{-5} M**

**For dilution:**

$\overline{){{\mathbf{M}}}_{{\mathbf{1}}}{{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{M}}}_{{\mathbf{2}}}{{\mathbf{V}}}_{{\mathbf{2}}}}$

A standard solution is prepared for the analysis of fluoxymesterone (C _{20}H_{29}FO_{3}), an anabolic steroid. A stock solution is first prepared by dissolving 10.0 mg of fluoxymesterone in enough water to give a total volume of 500.0 mL. A 100.0-μL aliquot (portion) of this solution is diluted to a final volume of 100.0 mL. Calculate the concentration of the final solution in terms of molarity.

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