For the equilibrium: BaCl 2 · 2H 2 O(s) ⇌ BaCl 2 (s) + 2H 2 O(g), K p = 1.6 × 10 -3 atm 2 at 27°C. If 1.23 L of air at 27°C in which the partial pressure of water vapour is 0.08 atm is exposed to a la

Question

For the equilibrium: BaCl 2 · 2H 2 O(s) ⇌ BaCl 2 (s) + 2H 2 O(g), K p = 1.6 × 10 -3 atm 2 at 27°C. If 1.23 L of air at 27°C in which the partial pressure of water vapour is 0.08 atm is exposed to a large quantity of BaCl 2 (s), what mass of water vapour will be absorbed?

TestHub · Accepted Answer

K p = P H2O 2 gives equilibrium P H2O = √(1.6 × 10 -3 ) = 0.04 atm Water is absorbed until its partial pressure falls from 0.08 atm to 0.04 atm: Δ P = 0.04 atm at T = 300 K Δ n = Δ PV/RT = (0.04)(1.23)/(0.0821 × 300) = 2.0 × 10 -3 mol Mass = 2.0 × 10 -3 × 18 = 0.036 g = 36 mg This is option (d).

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