To effectively stop polymerization, certain inhibitors require the presence of a small amount of O2. At equilibrium with 1 atm of air, the concentration of O₂ dissolved in the monomer acrylic acid (CH₂=CHCOOH) is 1.64x10⁻³ M.(c) What is the mole fraction?

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shaikhmohd0119 shaikhmohd0119 · Answer 1 · 2022-09-01T22:50:42+02:00

Mole fraction of O₂ is 2.67

Given that,

Solubility of O₂ in acrylic acid = 1.64x10⁻³ M

Partial pressure of O₂ = 0.2095 atm

Now,

According to Henry's Law

[tex]S_{gas}[/tex] = [tex]k_{H}[/tex]× [tex]P_{gas}[/tex]

Where,

[tex]S_{gas\\[/tex] = Solubility of Gas in the solution

[tex]k_{H}[/tex] = Henry's Constant

[tex]P_{gas}[/tex] = Partial pressure of the gas

Hence,

Solubility of O₂ = Henry's constant × Partial pressure of O₂

1.64x10⁻³ M = [tex]k_{H}[/tex] x 0.2095

Hence,

Henry's constant = 0.007828 mol/ L.atm

Now,

For mole fraction,

From Henry's Law we know that,

Partial pressure of gas = henry's constant × mole fraction of gas

hence,

0.2095 = 0.007828 × mole fraction of O₂

Hence,

mole fraction of O₂ = 0.02095÷0.007828

mole fraction of O₂ = 2.67

Thus from the above conclusion we can say that , Mole fraction of O₂ is 2.67.

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