For the equilibrium: SrCl₂·6H₂O(s) ⇌ SrCl₂·2H₂O(s) + 4H₂O(g), K_p = 8.1 × 10⁻⁷ atm⁴ at 27°C. If 1.642L of air saturated with water vapour at 27°C is exposed to a large quantity of SrCl₂·2H₂O(s), what

Question

For the equilibrium: SrCl₂·6H₂O(s) ⇌ SrCl₂·2H₂O(s) + 4H₂O(g), K_p = 8.1 × 10⁻⁷ atm⁴ at 27°C. If 1.642L of air saturated with water vapour at 27°C is exposed to a large quantity of SrCl₂·2H₂O(s), what mass of water vapour will be absorbed? Saturated vapour pressure of water at 27°C = 30.4 torr.

Prashant Kotian · Accepted Answer

Correct answer: A. K_p = (P_H₂O)⁴ = 8.1 × 10⁻⁷ atm⁴ => P_H₂O = 0.03 atm = 22.8 torr. Initial water vapour pressure is 30.4 torr. The pressure must drop to 22.8 torr, so ΔP = 30.4 - 22.8 = 7.6 torr = 0.01 atm. Moles of H₂O absorbed = ΔPV/RT = (0.01 × 1.642) / (0.0821 × 300) = 0.01642 / 24.63 = 6.66 × 10⁻⁴ mol. Mass = 6.66 × 10⁻⁴ × 18 g = 0.012 g = 12 mg. Therefore, correct answer is A.

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