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If a gas has a proportionality constant of 4.32 x 10-4 mol at room temperature for a particular solvent, what will the
solubility of the gas be in that solvent when it exerts a partial pressure of 92.4kPa?
LkPa
• Your answer should have three significant figures.

Respuesta :

[tex]0.0467 X 10^{-4} M/kPa[/tex] is the solubility of the gas when it exerts a partial pressure of 92.4kPa.

What is Henry's law?

Mathematically, we can get this from Henry's law

From Henry law;

Concentration = Henry constant × partial pressure

Thus Henry constant = [tex]\frac{Concentration}{partial \;pressure}[/tex]

Henry constant = [tex]\frac{4.32 \;X \;10^{-4} mol}{92.4kPa}[/tex]

[tex]= 0.0467 X 10^{-4} M/kPa[/tex]

Hence, [tex]0.0467 X 10^{-4} M/kPa[/tex] is the solubility of the gas when it exerts a partial pressure of 92.4kPa.

Learn more about the Henry's law here:

https://brainly.com/question/16222358

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