Solubility of N2 in water at 25°C is 6.8 × 10⁻⁴ mol/L at 1 atm partial pressure. By Henry's law, at 5 atm partial pressure of N2:
A6.8 × 10⁻⁴
B6.8 × 10⁻⁵
C3.4 × 10⁻³
D6.8 × 10⁻²
Answer & Solution
Correct answer: C. 3.4 × 10⁻³
By Henry's law: C ∝ P. At 5 atm: C = 5 × 6.8 × 10⁻⁴ = 3.4 × 10⁻³ mol/L.
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