I would go with option #4, reduced use of resources!
:V
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, 
Explanation :
As we know that,

where,
= molar solubility of
= ?
= partial pressure of
= 0.2 atm = 1.97×10⁻⁶ Pa
= Henry's law constant = 4.34 × 10⁹ g/L.Pa
Now put all the given values in the above formula, we get:


Now we have to molar concentration of oxygen.
Molar concentration of oxygen = 
Therefore, the molar concentration of oxygen is, 
the answer is 0.000097 KM
Answer:
Option A. 107 mL
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 150 mL
Initial pressure (P₁) = 500 mmHg
Final pressure (P₂) = 700 mmHg
Temperature = constant
Final volume (V₂) =?
The final volume of the gas can be obtained by using the Boyle's law equation as shown below:
P₁V₁ = P₂V₂
500 × 150 = 700 × V₂
75000 = 700 × V₂
Divide both side by 700
V₂ = 75000 / 700
V₂ = 107 mL
Therefore, the final volume of the gas is 107 mL.