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Problems · Problem 2.1

Q.The solubility of N₂ gas in water at 25 ⁰C and 1 bar is 6.85 × 10⁻⁴ mol L⁻¹. Calculate

(a) Henry's law constant
(b) molarity of N₂ gas dissolved in water under atmospheric conditions when partial pressure of N₂ in atmosphere is 0.75 bar.
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✓ Free question

KH=S/P=6.85×10−4K_H = S/P = 6.85 \times 10^{-4} mol L⁻¹ bar⁻¹ at 1 bar; at 0.75 bar, S=KHP=5.138×10−4S = K_H P = 5.138 \times 10^{-4} mol L⁻¹.

Step 1. Henry's law: S=KHPS = K_H P, so KH=S/PK_H = S/P.

Step 2 (a). With S = 6.85×10−46.85 \times 10^{-4} mol L⁻¹ at P = 1 bar: KH=6.85×10−4 mol L−11 bar=6.85×10−4K_H = \dfrac{6.85 \times 10^{-4}\ \text{mol L}^{-1}}{1\ \text{bar}} = 6.85 \times 10^{-4} mol L⁻¹ bar⁻¹.

Step 3 (b). Under atmospheric conditions the partial pressure of N₂ is 0.75 bar, so S=KHP=6.85×10−4×0.75=5.1375×10−4S = K_H P = 6.85 \times 10^{-4} \times 0.75 = 5.1375 \times 10^{-4}, which rounds to 5.138×10−45.138 \times 10^{-4} mol L⁻¹ (the book's printed figure).

✓Final answer

  1. KH=6.85×10−4K_H = 6.85 \times 10^{-4} mol L⁻¹ bar⁻¹
  2. molarity of dissolved N₂ = 5.138×10−45.138 \times 10^{-4} mol L⁻¹

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