Calculate the proportion of O2 and N2 dissolved in water at 298 K. Whe

1.	Calculate the proportion of O2 and N2 dissolved in water at 298 K. When air containing 20% O2 and 80% N2 by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are KH(O2) = 4.6 x atm and KH(N2) 8.5 x 104 atm.
Answer» C₁V₁ = C₂V₂ 6M(V₁) = 0.25M x 500 ml V₁ = \(\frac{0.25\times500}{6}\) V₁ = 20.3 mL

Calculate the proportion of O2 and N2 dissolved in water at 298 K. When air containing 20% O2 and 80% N2 by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are KH(O2) = 4.6 x atm and KH(N2) 8.5 x 104 atm.

Answer»

C₁V₁ = C₂V₂

6M(V₁) = 0.25M x 500 ml

V₁ = \(\frac{0.25\times500}{6}\)

V₁ = 20.3 mL

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Calculate the proportion of O2 and N2 dissolved in water at 298 K. When air containing 20% O2 and 80% N2 by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are KH(O2) = 4.6 x atm and KH(N2) 8.5 x 104 atm.

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