Respuesta :
Answer:
At a partial pressure of 2.470 atm, solubility of [tex]O_{2}[/tex] is 1.03 M
Explanation:
The given problem can be solved by using Henry's law
According to Henry's law at a particular temperature for a particular gas and solvent: [tex]\frac{S_{1}}{S_{2}}=\frac{P_{1}}{P_{2}}[/tex]
where, [tex]S_{1}[/tex] and [tex]S_{2}[/tex] are solubility of the gas at a partial pressure of [tex]P_{1}[/tex] and [tex]P_{2}[/tex] respectively.
Here, [tex]S_{1}=0.0500M[/tex], [tex]P_{1}=0.120atm[/tex] and [tex]P_{2}=2.470atm[/tex]
So, [tex]S_{2}=\frac{S_{1}P_{2}}{P_{1}}=\frac{(0.0500M)\times (2.470atm)}{0.120atm}=1.03M[/tex]
Hence, at a partial pressure of 2.470 atm, solubility of [tex]O_{2}[/tex] is 1.03 M
Taking into account the Henry's law, at a partial pressure of 2.470 atm, the solubility of O₂ is 1.03 M.
Henry's law
Henry's law states that for a gas in contact with a solvent, at a constant temperature, its concentration in the solvent is directly proportional to its pressure.
That is, Henry's law states that at constant temperature, the solubility of a gas in a liquid is proportional to the pressure of the gas, provided that no chemical reactions take place between the gas and the liquid.
This law is expressed as follows:
Sg = K× Pg
where:
- Sg= solubility or concentration of the gas
- K= Henry's law constant (specific to each gas)
- Pg= partial pressure of the gas
This case
According to Henry's law at a particular temperature for a particular gas and solvent you know:
S₁ = K₁× P₁ → [tex]K_{1} =\frac{S_{1} }{P_{1} }[/tex]
and
S₂ = K₂× P₂ → [tex]K_{2} =\frac{S_{2} }{P_{2} }[/tex]
where, S₁ and S₂ are solubility of the gas at a partial pressure of P₁ and P₂ respectively.
Since the Henry's law constant is the same since it is the same gas, then:
K₁= K₂
So:
[tex]\frac{S_{1} }{P_{1} }= \frac{S_{2} }{P_{2} }[/tex]
You know:
- S₁ = 0.0500 M
- P₁ = 0.120 atm
- S₂ = ?
- P₂ = 2.470 atm
Replacing in Henry's law:
[tex]\frac{0.0500 M}{0.120 atm}= \frac{S_{2} }{2.470 atm }[/tex]
Solving:
[tex]S_{2} =2.470 atmx\frac{0.0500 M}{0.120 atm}[/tex]
S₂= 1.03 M
Finally, at a partial pressure of 2.470 atm, the solubility of O₂ is 1.03 M.
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