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Solutions of hydrogen in palladium may be formed by exposing Pd metal to H gas. The concentration of hydrogen in the palladium depends on the pressure of H 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.

(a) Determine the molarity of this solution (solution density = 10.8 g/cm3).

(b) Determine the molality of this solution (solution density = 10.8 g/cm3).

(c) Determine the percent by mass of hydrogen atoms in this solution (solution density = 10.8 g/cm3).

Answer :

Explanation:

Mass of solution , m= 0.94 g + 215 g = 215.94 g

Volume of the solution = V

Density of solution = d = [tex]10.8 g/cm^3[/tex]

[tex]V=\frac{m}{d}=\frac{215.94 g}{10.8 g/cm^3}=19.99 cm^3=19.99 mL[/tex]

[tex]1 cm^3=1 mL[/tex]

[tex]1 mL=0.001 L[/tex]

[tex]V=0.01999 L[/tex]

Moles of hydrogen = [tex]\frac{0.94 g}{2 g/mol}=0.47 mol[/tex]

Molarity of hydrogen gas:

[tex]Molarity=\frac{Moles}{\text{Volume of solution (L)}}[/tex]

[tex]=\frac{0.47 mol}{0.01999 L}=0.02351 mol/L[/tex]

Molality of hydrogen gas :

Mass of solvent or here it is palladium = 215 g =0.215 kg (1 g = 0.001 kg)

[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]

[tex]=\frac{0.47}{0.215 kg}=2.1860 mol/kg[/tex]

Mass percentage of hydrogen is solution :

[tex]\frac{\text{Mass of hydrogen}}{\text{Mass of solution}}\times 100[/tex]

[tex]=\frac{0.94 g}{215.94 g}\times 100=0.435\%[/tex]

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