Question

Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2saturated water. (4 points)

Answer 1

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is,
`2.67* 10^2mol/L`

Explanation :

As we know that,

`C_(O_2)=k_H* p_(O_2)`

where,

`C_(O_2)` = molar solubility of
`O_2` = ?

`p_(O_2)` = partial pressure of
`O_2` = 0.2 atm = 1.97×10⁻⁶ Pa

`k_H` = Henry's law constant = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

`C_(O_2)=(4.34* 10^9g/L.Pa)* (1.97* 10^(-6)Pa)`

`C_(O_2)=8.55* 10^3g/L`

Now we have to molar concentration of oxygen.

Molar concentration of oxygen =
`(8.55* 10^3g/L)/(32g/mol)=2.67* 10^2mol/L`

Therefore, the molar concentration of oxygen is,
`2.67* 10^2mol/L`