Question

The Henry’s law constant for Ar(g) at 25oC is 1.4 × 10-3 M/atm. What mass (in mg) of Ar is dissolved in a typical 8.0-fl. oz. glass of water at 25oC and at 1.00 atm total pressure? Argon constitutes about 1.0% of the atmosphere.

Answer 1

Let x₂mg of Ar is dissolved in water at 25 ^oC.

x₂ = n₂/(n₁+n₂) = n₂/n₁ (assuming that the number of moles of Ag, n₂ is negligible in comparision with the number of moles , n₁ of the solvent.

The number of moles in one dm³ of water,

n₁ = 1000 g /18 g mol^-1 = 55.5 mol

so, x₂ =n₂/55.5 mol

From Henry's law,

p₂ = kx_2, k is Henry's law constant

k = p₂/x₂

1.4 x 10^-3 = (0.01 ) / (n₂ / 55.5 )

n₂ = 396.4 mol

So 396.4 mol per liter of Ag will dissolve in water at 25 ^oC at 1 atm pressure.