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Assuming the partial pressure of oxygen in air (0.20 atm) and nitrogen in air (0.80 atm). Calculate the mole fractions of oxygen and nitrogen in water at 298 K. FOR THIS QUESTION report the mole fraction of OXYGEN

Sagot :

batolisis

Answer:

oxygen = 4.7 * 10^-6

Nitrogen = = 9.7 * 10^6

Explanation:

partial pressure of oxygen = 0.20 atm

partial pressure of Nitrogen = 0.80 atm

calculate the mole fractions of oxygen and Nitrogen in water

Temp = 298k

applying henry's law

molar conc of oxygen in water ( Coxygen )

= Kp = 1.3 * 10^-3 Mol/L.atm * 0.20 atm = 2.6 * 10^-4 Mol

molar conc of Nitrogen in water ( Cnitrogen )

= Kp = 6.8 * 10^-4 Mol/L.atm * 0.80 atm = 5.4 * 10^-4

next Given that the number of moles in 1 liter of water = 55.5 mol  

therefore the mole fraction of oxygen

= 2.6 * 10^-4 / 55.5

= 4.7 * 10^-6

mole fraction of Nitrogen

= 5.4 * 10^-4  / 55.5

= 9.7 * 10^6

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