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Solutions of hydrogen in palladium may be formed by exposing pd metal to h2 gas. the concentration of hydrogen in the palladium depends on the pressure of h2 gas applied, but in a more complex fashion than can be described by henry's law. under certain conditions, 0.95 g of hydrogen gas is dissolved in 275 g of palladium metal. (the density of the resulting solution is 11.7 g/cm3.) (a) determine the molarity of this solution.

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Dejanras
Answer is: molarity of this solution is 19,92M.
m(H₂) = 0,95 g.
m(Pd) = 275 g.
d(solution) = 11,7 g/cm³.
m(solution) = 0,95 g + 275 g.
m(solution) = 275,95 g.
V(solution) = m(solution) ÷ d(solution).
V(solution) = 275,95 g ÷ 11,7 g/cm³.
V(solution) = 23,585 cm³ = 0,023585 L.
n(H₂) = m(H₂) ÷ M(H₂).
n(H₂) = 0,95 g ÷ 2,016 g/mol.
n(H₂) = 0,47 mol.
c(solution) = 0,47 mol ÷ 0,023585 L.
c(solution) = 19,92 mol/L = 19,92 M.
dsdrajlin

Answer:

20 M

Explanation:

Given data

mass of H₂ (solute): 0.95 g

mass of Pd (solvent): 275 g

density of the solution (ρ): 11.7 g/cm³

The mass of the solution is:

m(H₂) + m(Pd) = 0.95 g + 275 g = 276 g

The volume of the solution is:

276 g × (1 cm³/11.7 g) = 23.6 cm³ = 23.6 mL = 23.6 × 10⁻³ L

The molar mass of H₂ is 2.02 g/mol. The moles of H₂ are:

0.95 g × (1 mol/ 2.02 g) = 0.47 mol

The molarity of the solution is:

M = n(H₂)/V(solution) = 0.47 mol / 23.6 × 10⁻³ L = 20 M

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