102.3936× [tex]10^{-3}[/tex]g of oxygen gas was dissolved in 2.50 Liters of water.
The Henry's law constant for oxygen gas in water at 20°C =k = 1.28 ×[tex]10^{-3}[/tex]mol/(Latm)
The pressure of pure oxygen gas above water = p=1 atm
The solubility of the oxygen gas in water:
S= 1.28 ×[tex]10^{-3}[/tex]mol/(Latm) ×1.00atm
S= 1.28 ×[tex]10^{-3}[/tex]mol/L
There are 1.28 ×[tex]10^{-3}[/tex] moles of oxygen gas in 1 liter of water
The volume of the water = 4.00 L
Moles of oxygen gas in 4.00 L of water:
1.28 ×[tex]10^{-3}[/tex]×2.50mol = 3.2× [tex]10^{-3[/tex]mol
Mass of 3.2× [tex]10^{-3[/tex] moles of oxygen gas:
=3.2× [tex]10^{-3[/tex] mol× 31.998 g/mol =102.3936× [tex]10^{-3}[/tex]g
To know more about Henry's Law refer to https://brainly.com/question/12823901
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