The solubility of N₂ gas in water at 25 °C and 1 bar is $6.85 \times 10^{-4}$ mol L⁻¹. If the partial pressure of N₂ in the atmosphere is 0.75 bar, the molarity of dissolved N₂ is:
A$6.85 \times 10^{-4}$ mol L⁻¹
B$1.37 \times 10^{-3}$ mol L⁻¹
C$9.13 \times 10^{-4}$ mol L⁻¹
D$5.14 \times 10^{-4}$ mol L⁻¹
Answer & Solution
Correct answer: D. $5.14 \times 10^{-4}$ mol L⁻¹
$K_H = S/P = 6.85\times10^{-4}/1 = 6.85\times10^{-4}$ mol L⁻¹ bar⁻¹. At 0.75 bar: $S = K_H P = 6.85\times10^{-4} \times 0.75 = 5.14\times10^{-4}$ mol L⁻¹.
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