A gaseous mixture of 50% 0, and 50% N2 (by volume) is bubbled through water at 2

Question

A gaseous mixture of 50% 0, and 50% N2 (by volume) is bubbled through water at 25°C and a total pressure of 1 atm. The vapor pressure of pure water at 25°C is 0.0313 atm. Assuming that the gaseous mixture behaves ideally, compute the partial pressure of O2 under these conditions. Recall that Dalton's Law says that the total pressure of a mixture of gases is the sum of the partial pressures of the components in the mixture. Compute the concentration of O2 in mole/L dissolved in the water. The Henry's Law constant for 02 at 25°C is KH 1.3 x 103 M/atm. a. b.

Explanation / Answer

a) According to Dalton's law of partial pressure, total pressure,P=x(N2)*p(N2) +x(O2)*p(O2)+x(H2O)*p(H2O)

where x represnts mole fraction of the gas in vapor phase

As both O2 and N2 are equal in amount (=50% of total volume),so both has equal mole fraction.

as volume is proportional to mole/mole fraction

x(N2)=x(O2)

and both behave ideally , p=nRT/V

so ,pO2=pN2 [for equal amount]

P=x(N2)*p(N2 +x(O2)*p(O2)+x(H2O)*p(H2O)

P=1 atm

x(H2O)*p(H2O)=0.0313 atmn      [water vapor pressure]

So 1 atm=2 *x(O2)*p(O2)+0.0313 atm

or, x(O2)*p(O2)=partial pressure of O2 or N2=0.484 atm

partial pressure of O2=0.484 atm

b) given, henry's law constant =KH=1.3*10^-3 M/atm

pO2=0.484 atm

According to henry's law,

Concentration of O2=CO2=KH*pO2=(1.3*10^-3 M/atm)*0.484 atm=6.296*10^-4 M or mol/L

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