🤓 Based on our data, we think this question is relevant for Professor Widdifield's class at OAKLAND.

Recall that ** molarity** is the ratio of the moles of solute and the volume of solution (in liters). In other words:

$\overline{){\mathbf{molarity}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}}{\mathbf{L}\mathbf{}\mathbf{solution}}}$

The moles of cocaine hydrochloride is:

molar mass C_{17}H_{22}ClNO_{4} = 339.806 g/mol

$\mathbf{moles}\mathbf{}{\mathbf{C}}_{\mathbf{17}}{\mathbf{H}}_{\mathbf{22}}{\mathbf{ClNO}}_{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\overline{)\mathbf{g}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}}{\mathbf{339}\mathbf{.}\mathbf{806}\mathbf{}\overline{)\mathbf{g}}}$

**moles C _{17}H_{22}ClNO_{4}**

Calculate the volume of the solution: **1 mL = 10 ^{-3} L**

$\mathbf{volume}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{}\overline{)\mathbf{mL}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{L}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}$

**volume = 4.0 x10^{-4} L**

We can now calculate for **molarity**:

The "free-base" form of cocaine (C_{17}H_{21}NO_{4}) and its protonated hydrochloride form (C_{17}H_{22}ClNO_{4}) are shown
below; the free-base form can be converted to the hydrochloride form with one equivalent of HCl. For clarity, not
all the carbon and hydrogen atoms are shown; each vertex
represents a carbon atom with the appropriate number of
hydrogen atoms so that each carbon makes four bonds to
other atoms.

The hydrochloride form of cocaine has a solubility of 1.00 g in 0.400 mL water. Calculate the molarity of a saturated solution of the hydrochloride form of cocaine in water.

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Based on our data, we think this problem is relevant for Professor Widdifield's class at OAKLAND.