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The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38×10−3 M. Part A Using Henry's law, calculate the molar concentration of O2 in the surface water of a mountain lake saturated with air at 20 ∘C and an atmospheric pressure of 665 torr . Express your answer using two significant figures. nothing

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IthaloAbreu

Answer:

[tex]1.21x10^{-3}[/tex] M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x[tex]10^{-3}[/tex]

H = [tex]\frac{0.21}{1.38x10^{-3} }[/tex]

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = [tex]1.21x10^{-3}[/tex] M

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