$\overline{){\mathbf{molar}}{\mathbf{}}{\mathbf{mass}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{grams}}{\mathbf{moles}}}\phantom{\rule{0ex}{0ex}}\mathbf{moles}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{grams}}{\mathbf{molar}\mathbf{}\mathbf{mass}}\phantom{\rule{0ex}{0ex}}\mathbf{moles}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{0}\mathbf{.}\mathbf{94}\mathbf{}\overline{)\mathbf{g}}}{\mathbf{2}\mathbf{.}\mathbf{016}\mathbf{}{\displaystyle \frac{\overline{)\mathbf{g}}}{\mathbf{mol}}}}$

**moles = 0.4663 mol**

$\mathbf{volume}\mathbf{}\mathbf{=}\mathbf{}\mathbf{215}\mathbf{}\overline{)\mathbf{g}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{){\mathbf{cm}}^{\mathbf{3}}}}{\mathbf{0}\mathbf{.}\mathbf{94}\mathbf{}\overline{)\mathbf{g}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}{\mathbf{1}\mathbf{}\overline{){\mathbf{cm}}^{\mathbf{3}}}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{L}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}$

**volume = 0.2287 L**

Solutions of hydrogen in palladium may be formed by exposing Pd metal to H_{2 }gas. The concentration of hydrogen in the palladium depends on the pressure of H_{2} gas applied, but in a more complex fashion than can be described by Henry’s law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal (solution density = 10.8 g cm^{3}).

(a) Determine the molarity of this solution.

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