Answer :
The vapor pressure of water at 20 degrees Celcius is 0.02307427 atm. The millimolar solubility of oxygen gas in water can be determined through the ideal gas law assumption.
millimolar solubility is (n/V); assuming ideal gas mixture:
PV = nRT
n/V = P/RT
where P is (P_O2 - P_H2O)
n/V = (1.00 atm - 0.02307427 atm) / (8.314 Latm/molK) * (20+273)
n/V = 2436 mol/L
millimolar solubility is (n/V); assuming ideal gas mixture:
PV = nRT
n/V = P/RT
where P is (P_O2 - P_H2O)
n/V = (1.00 atm - 0.02307427 atm) / (8.314 Latm/molK) * (20+273)
n/V = 2436 mol/L
Answer: The mili molar solubility of oxygen gas is 41.57 mmol/L
Explanation:
Milimolar solubility is defined as the number of mili moles present in liter of solution.
To calculate the molar solubility, we use the ideal gas equation:
[tex]PV=nRT[/tex]
or,
[tex]\frac{n}{V}=\frac{P}{RT}[/tex]
where,
P = pressure = 1 atm
V = volume
n = Number of moles
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = [tex]20^oC=[273+20]=293K[/tex]
Putting values in above equation, we get:
[tex]\frac{n}{V}=\frac{1atm}{0.0821\text{ L atm }mol^{-1}K^{-1}\times 293K}\\\\\frac{n}{V}=0.04157mol/L[/tex]
To convert this into milimoles, we use the conversion factor:
1 mol = 1000 mmol
So, [tex]0.04157mol/L\times 1000=41.57mmol/L[/tex]
Hence, the mili molar solubility of oxygen gas is 41.57 mmol/L