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, [tex]2.67\times 10^2mol/L[/tex]
Explanation :
As we know that,
[tex]C_{O_2}=k_H\times p_{O_2}[/tex]
where,
[tex]C_{O_2}[/tex] = molar solubility of [tex]O_2[/tex] = ?
[tex]p_{O_2}[/tex] = partial pressure of [tex]O_2[/tex] = 0.2 atm = 1.97×10⁻⁶ Pa
[tex]k_H[/tex] = Henry's law constant = 4.34 × 10⁹ g/L.Pa
Now put all the given values in the above formula, we get:
[tex]C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)[/tex]
[tex]C_{O_2}=8.55\times 10^3g/L[/tex]
Now we have to molar concentration of oxygen.
Molar concentration of oxygen = [tex]\frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L[/tex]
Therefore, the molar concentration of oxygen is, [tex]2.67\times 10^2mol/L[/tex]
