What is the concentration of O2 in a freshwater stream in equilibrium with air at 30°C and 2.0 bar? Given, KH (Henry’s law constant) of O2=2.0× 10-3 mol/kg bar at 30°C

Question

What is the concentration of O2 in a freshwater stream in equilibrium with air at 30°C and 2.0 bar? Given, KH (Henry’s law constant) of O2=2.0× 10-3 mol/kg bar at 30°C (A) 8.736 × 10-3 g/kg (B) 1.344 × 10-3 g/kg (C) 24.04 g/kg (D) 114.5 g/kg

Tardigrade Admin · Accepted Answer

In this case, KH has been given in the unit of mole/kg bar Thus, solubility of O2 (or concentration) =KH Pgas chio2 (in air) = 0.21 ∴ po2=p textair chio2=0.21 bar ∴ Solubility =(2.0×10-3mol/kg bar)×0.21 bar =4.2×10-4×32 g/ kg =1.344×10-3 g/ kg

Q. What is the concentration of $O_{2}$ in a freshwater stream in equilibrium with air at $3 0^{\circ} C$ and 2.0 bar? Given, $K_{H}$ (Henry’s law constant) of $O_{2} = 2.0 \times 1 0^{- 3} m o l / k g$ bar at $3 0^{\circ} C$

$8.736 \times 1 0^{- 3} g / k g$

$1.344 \times 1 0^{- 3} g / k g$

$24.04 g / k g$

$114.5 g / k g$

Q. What is the concentration of O2​ in a freshwater stream in equilibrium with air at 30∘C and 2.0 bar? Given, KH​ (Henry’s law constant) of O2​=2.0×10−3mol/kg bar at 30∘C

8.736×10−3g/kg

1.344×10−3g/kg

24.04g/kg

114.5g/kg

In this case, KH​ has been given in the unit of mole/kg bar Thus, solubility of O2​ (or concentration) =KH​Pgas​ χo2​​ (in air) = 0.21 ∴po2​​=pair​χo2​​=0.21bar ∴ Solubility =kgbar2.0×10−3mol​×0.21bar =4.2×10−4×32g/kg =1.344×10−3g/kg

Q. What is the concentration of $O_{2}$ in a freshwater stream in equilibrium with air at $3 0^{\circ} C$ and 2.0 bar? Given, $K_{H}$ (Henry’s law constant) of $O_{2} = 2.0 \times 1 0^{- 3} m o l / k g$ bar at $3 0^{\circ} C$

$8.736 \times 1 0^{- 3} g / k g$

$1.344 \times 1 0^{- 3} g / k g$

$24.04 g / k g$

$114.5 g / k g$