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shutvik [7]
4 years ago
5

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

epends 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).
Chemistry
1 answer:
Juliette [100K]4 years ago
5 0

Explanation:

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

Volume of the solution = V

Density of solution = d = 10.8 g/cm^3

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

1 cm^3=1 mL

1 mL=0.001 L

V=0.01999 L

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

Molarity of hydrogen gas:

Molarity=\frac{Moles}{\text{Volume of solution (L)}}

=\frac{0.47 mol}{0.01999 L}=0.02351 mol/L

Molality of hydrogen gas :

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

Molality=\frac{Moles}{\text{Mass of solvent(kg)}}

=\frac{0.47}{0.215 kg}=2.1860 mol/kg

Mass percentage of hydrogen is solution :

\frac{\text{Mass of hydrogen}}{\text{Mass of solution}}\times 100

=\frac{0.94 g}{215.94 g}\times 100=0.435\%

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The complete question is as follows: How many moles of a gas at 100 c does it take to fill a 1.00 l flask to a pressure of 152kPa

Answer: There are 0.0489 moles of a gas at 100^{o}C is required to fill a 1.00 l flask to a pressure of 152kPa.

Explanation:

Given: Volume = 1.00 L,    

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Substitute the values into above formula as follows.

PV = nRT\\1.5 atm \times 1.0 L = n \times 0.0821 L atm/mol K \times 373 K\\n = \frac{1.5 atm \times 1.0 L}{0.0821 L atm/mol K \times 373 K}\\n = 0.0489 mol

Thus, we can conclude that there are 0.0489 moles of a gas at 100^{o}C is required to fill a 1.00 l flask to a pressure of 152kPa.

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