Assuming we have 38 g HCl (MM = 36 g/mol) and 62 g H_{2}O.

(1) molarity HCl

$\overline{){\mathbf{molarity}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mole}\mathbf{}\mathbf{solute}}{\mathbf{L}\mathbf{}\mathbf{solution}}}\phantom{\rule{0ex}{0ex}}\mathbf{molarity}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{38}\mathbf{}\overline{)\mathbf{g}}\mathbf{}\mathbf{\left(}{\displaystyle \frac{1\mathrm{mol}}{36g}}\mathbf{\right)}}{\mathbf{(}\mathbf{62}\mathbf{}\overline{)\mathbf{g}}\mathbf{)}\mathbf{\left(}{\displaystyle \frac{1\overline{)\mathrm{mL}}}{1.19\overline{)g}}}\mathbf{\right)}\mathbf{\left(}{\displaystyle \frac{1x{10}^{-3}L}{1\overline{)\mathrm{mL}}}}\mathbf{\right)}}$

molarity = 20.26 mol/L or M

(2) molality HCl

$\overline{){\mathbf{molality}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mole}\mathbf{}\mathbf{solute}}{\mathbf{kg}\mathbf{}\mathbf{solvent}}}\phantom{\rule{0ex}{0ex}}\mathbf{molarity}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{38}\mathbf{}\overline{)\mathbf{g}}\mathbf{}\mathbf{\left(}{\displaystyle \frac{1\mathrm{mol}}{36\overline{)g}}}\mathbf{\right)}}{\mathbf{(}\mathbf{62}\mathbf{}\overline{)\mathbf{g}}\mathbf{)}\mathbf{\left(}{\displaystyle \frac{1x{10}^{-3}\mathrm{kg}}{1\overline{)g}}}\mathbf{\right)}}$

Common commercial acids and bases are aqueous solutions with the following properties:

Calculate the molarity, molality, and mole fraction of each of the preceding reagents.

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