Respuesta :
Answer:
Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)
Explanation:
Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P
Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm
=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.
∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = 3.45 x 10⁻² grams O₂(g) in 4L water.
0.164 grams of oxygen gas was dissolved in 4.00 Liters of water.
0.0343 grams of oxygen gas was dissolved in 4.00 Liters of water.
Explanation:
- Henry's law states that the solubility of a gas in a given volume of a liquid is directly proportional to the partial pressure of that gas above the liquid.
[tex]S=k_H\times p_o[/tex]
Where:
S = Solubility of gas in liquid
[tex]k_H[/tex]= Henry's law constant
[tex]p_o[/tex]= Partial pressure of a gas
Given:
The Henry's law constant for oxygen gas in water at 20°C is [tex]1.28 \times 10^{-3} mol/(Latm)[/tex]
To find:
a) Mass of oxygen gas in 4.00 Liter of water with pure oxygen at 1.00 atm.
b) Mass of oxygen gas in 4.00 Liter of water with oxygen gas at a partial pressure of 0.209 atm.
Solution:
a)
The Henry's law constant for oxygen gas in water at 20°C = [tex]k_H[/tex]= [tex]1.28 \times 10^{-3} mol/(Latm)[/tex]
The pressure of pure oxygen gas above water = [tex]p_o=1.00 atm[/tex]
The solubility of the oxygen gas in water:
[tex]S=1.28\times 10^{-3} mol/(Latm)\times 1.00 atm\\S=1.28\times 10^{-3} mol/L[/tex]
There are [tex]1.28\times 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:
[tex]=1.28\times 10^{-3}\times 4.00 mol=5.12\times 10^{-3}mol[/tex]
Mass of [tex]5.12\times 10^{-3}[/tex] moles of oxygen gas:
[tex]=5.12\times 10^{-3}mol\times 31.998 g/mol=0.164 g[/tex]
0.164 grams of oxygen gas was dissolved in 4.00 Liters of water.
b)
The Henry's law constant for oxygen gas in water at 20°C = [tex]k_H[/tex]= [tex]1.28 \times 10^{-3} mol/(Latm)[/tex]
The partial pressure of oxygen gas above water = [tex]p_o=0.209atm[/tex]
The solubility of the oxygen gas in water:
[tex]S=1.28\times 10^{-3} mol/(Latm)\times 0.209atm\\S=2.68\times 10^{-4} mol/L[/tex]
There are [tex]2.68\times 10^{-4}[/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:
[tex]=2.68\times 10^{-4}\times 4.00 mol=1.072\times 10^{-3}mol[/tex]
Mass of [tex]1.072\times 10^{-3}[/tex] moles of oxygen gas:
[tex]=1.072\times 10^{-3}mol\times 31.998 g/mol=0.0343g[/tex]
0.0343 grams of oxygen gas was dissolved in 4.00 Liters of water.
Learn more about Henry's law here:
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