The Henry's law constant for helium in water at 20.0˚C and 0.25 atm is 3.9 × 10^-4M/atm. The solubility of helium in water at 20.0˚C at an atmospheric pressure of 0.25 atm is _____ M. A) 0.97 × 10^-3 B) 9.8 × 10^-5 C) 9.6 × 10^-4 D) 0.98 × 10^-5 E) 0.39 × 10^-4

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Quizplus · Accepted Answer

The solubility is calculated using Henry's law: Solubility = k * P, where k is the Henry's law constant (3.9 × 10^-4 M/atm) and P is the pressure (0.25 atm). Solubility = 3.9 × 10^-4 M/atm * 0.25 atm = 9.8 × 10^-5 M.

The Henry's Law Constant for Helium in Water at 20