Question

What is the concentration of O, in a fresh water stream in equilibrium with air at 25°C and 1.0 bar ? Given, K (Henry's law constant) of O₂ 1.3 × = 10-3 mol/kg bar at 25°C.​

Answer 1

Answer:

The Questions and Answers of What is the concentration of 02 in a fresh water stream in equilibrium with air at 25° C and 1.0 bar? Given, KH (Henry's law constant) of 02 = 1.3 x 10-3 mol/kg bar at 25° C.a)8.736 x 10-3g/kgb)4.16 x 10-2g/kgc)24.04g/kgd)114.5g/kgCorrect answer is option 'A'.

Answer 2

Answer:

The concentration of $O_{2}$ is $8.73\times10^{-3}g/kg$

Explanation:

The correct question is-"What is the concentration of $O_{2}$, in a fresh water stream in equilibrium with air at 25°C and 1.0 bar ? Given, K (Henry's law constant) of O₂ is 1.3 × = 10-3 mol/kg bar at 25°C.​"

Given that,

Pressure of air is $P_{air}=1bar$

We know that mole fraction of oxygen in air is 21% hence,

$X_{O}_{2}=0.21$

Therefore,according to dalton`s law of partial pressure the partial pressure of oxygen is product of its mole fraction and pressure of air.Hence,

$P_{O_{2}=P_{air}\times X_{O_{2}}=1\times0.21=0.21bar$

Now according to Henry`s law,The solubility of $O_{2}$ is

$S_{O_{2}}=K_{H}\times P_{O_{2}}=1.3\times10^{-3}\times0.21\\=2.73\times10^{-4}mol/kg$

To obtain the concentration of oxygen we shall multiply this solubility with Molar mass of Oxygen.Therefore we get

$C=2.73\times10^{-4}mol/kg\times32g/mol=8.73\times10^{-3}g/kg$

Therefore,The concentration of $O_{2}$ is $8.73\times10^{-3}g/kg$

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