Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in the air are to be .20 atm and .80 atm respectively, calculate the mole fractions of oxygen and nitrogen in the air dissolved in water at 298 k. The solubilities of oxygen and nitrogen in water at 298 k are 1.3 x 10^-3 mol/lxatm and 6.8 x 10^-4 mol/lxatm respectively.
To calculate the mole fractions of oxygen and nitrogen in the air dissolved in water, we can use Henry's law. According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
The mole fraction (X) of a gas in a mixture is defined as the ratio of the number of moles of the gas to the total number of moles of all gases present.
First, let's calculate the number of moles of oxygen and nitrogen in the air:
Number of moles of oxygen = Partial pressure of oxygen x Solubility of oxygen
= 0.20 atm x (1.3 x 10^-3 mol/lxatm)
= 2.6 x 10^-4 mol/l
Number of moles of nitrogen = Partial pressure of nitrogen x Solubility of nitrogen
= 0.80 atm x (6.8 x 10^-4 mol/lxatm)
= 5.44 x 10^-4 mol/l
Next, we need to calculate the total number of moles of all gases present. Since we are assuming only oxygen and nitrogen are present, the total number of moles is the sum of moles of oxygen and moles of nitrogen:
Total number of moles = Number of moles of oxygen + Number of moles of nitrogen
= 2.6 x 10^-4 mol/l + 5.44 x 10^-4 mol/l
= 7.04 x 10^-4 mol/l
Finally, we can calculate the mole fractions of oxygen and nitrogen:
Mole fraction of oxygen = Number of moles of oxygen / Total number of moles
= (2.6 x 10^-4 mol/l) / (7.04 x 10^-4 mol/l)
= 0.369 (rounded to three decimal places)
Mole fraction of nitrogen = Number of moles of nitrogen / Total number of moles
= (5.44 x 10^-4 mol/l) / (7.04 x 10^-4 mol/l)
= 0.774 (rounded to three decimal places)
Therefore, the mole fraction of oxygen in the air dissolved in water at 298 K is approximately 0.369, and the mole fraction of nitrogen is approximately 0.774.
To calculate the mole fractions of oxygen and nitrogen in the air dissolved in water, we can use Henry's law, which states that the concentration of a gas in a liquid is directly proportional to the partial pressure of that gas in the gas phase.
The mole fraction of a gas can be calculated by dividing the moles of that gas by the total moles of all gases present. Therefore, we need to first calculate the moles of oxygen and nitrogen dissolved in water at 298 K.
Using Henry's law, we can calculate the concentration of oxygen and nitrogen in water using the solubilities provided:
Concentration of oxygen in water = 1.3 x 10^-3 mol/lxatm * 0.20 atm = 2.6 x 10^-4 mol/L
Concentration of nitrogen in water = 6.8 x 10^-4 mol/lxatm * 0.80 atm = 5.44 x 10^-4 mol/L
Next, we need to convert these concentrations to moles. We can do this by multiplying the concentrations by the volume of water in liters. Since the volume of water is not provided, we cannot determine the actual number of moles. However, we can still calculate the mole fractions using the given concentrations.
Let's assume we have 1 L of water, then the moles of oxygen and nitrogen would be:
Moles of oxygen = 2.6 x 10^-4 mol/L * 1 L = 2.6 x 10^-4 mol
Moles of nitrogen = 5.44 x 10^-4 mol/L * 1 L = 5.44 x 10^-4 mol
Now we can calculate the mole fractions:
Mole fraction of oxygen = moles of oxygen / total moles
= 2.6 x 10^-4 mol / (2.6 x 10^-4 mol + 5.44 x 10^-4 mol)
= 2.6 x 10^-4 mol / 8.04 x 10^-4 mol
= 0.323
Mole fraction of nitrogen = moles of nitrogen / total moles
= 5.44 x 10^-4 mol / (2.6 x 10^-4 mol + 5.44 x 10^-4 mol)
= 5.44 x 10^-4 mol / 8.04 x 10^-4 mol
= 0.677
Therefore, the mole fraction of oxygen in the air dissolved in water at 298 K is approximately 0.323, and the mole fraction of nitrogen is approximately 0.677.
c = p*K
For O2, c = 1.3E-3 mol/L*atm x (0.2) and
for N2, c = 6.8E-4 mol/L*atm x (0.8).
Add to find total mols, then
XO2 = nO2/total mols
XN2 = nN2/total mols.